# 7.4: Solve Rational Equations - Mathematics

Learning Objectives

• Solve rational equations
• Solve rational equations by clearing denominators
• Identify extraneous solutions in a rational equation
• Solve for a variable in a rational formula
• Applications of rational equations
• Identify the components of a work equation
• Solve a work equation
• Define and write a proportion
• Solve proportional problems involving scale drawings
• Define direct variation, and solve problems involving direct variation
• Define inverse variation and solve problems involving inverse variation
• Define joint variation and solve problems involving joint variation

Equations that contain rational expressions are called rational equations. For example, (frac{2x+1}{4}=frac{x}{3}) is a rational equation. Rational equations can be useful for representing real-life situations and for finding answers to real problems. In particular, they are quite good for describing a variety of proportional relationships.

One of the most straightforward ways to solve a rational equation is to eliminate denominators with the common denominator, then use properties of equality to isolate the variable. This method is often used to solve linear equations that involve fractions as in the following example:

Solve (frac{1}{2}x-3=2-frac{3}{4}x) by clearing the fractions in the equation first.

Multiply both sides of the equation by 4, the common denominator of the fractional coefficients.

(egin{array}{c}frac{1}{2}x-3=2-frac{3}{4}x 4left(frac{1}{2}x-3 ight)=4left(2-frac{3}{4}x ight) ext{},,,,4left(frac{1}{2}x ight)-4left(3 ight)=4left(2 ight)+4left(-frac{3}{4}x ight)2x-12=8-3xunderline{+3x},,,,,,,,,,,,,,,,,,,,,,,underline{+3x}5x-12=8,,,,,,,,,,,,,,,underline{+12},,,,underline{+12} 5x=20x=4end{array})

We could have found a common denominator and worked with fractions, but that often leads to more mistakes. We can apply the same idea to solving rational equations. The difference between a linear equation and a rational equation is that rational equations can have polynomials in the numerator and denominator of the fractions. This means that clearing the denominator may sometimes mean multiplying the whole rational equation by a polynomial. In the next example, we will clear the denominators of a rational equation with a terms that has a polynomial in the numerator.

### Example

Solve the equation (2cdot2cdot2), or 8, will be the LCM.

(egin{array}{l}4=2cdot28=2cdot2cdot2cdot2 ext{LCM}=2cdot2cdot2 ext{LCM}=8end{array})

The LCM of 4 and 8 is also the lowest common denominator for the two fractions.

Multiply both sides of the equation by the common denominator, 8, to keep the equation balanced and to eliminate the denominators.

(egin{array}{r}8cdot frac{x+5}{8}=frac{7}{4}cdot 8,,,,,,,\frac{8(x+5)}{8}=frac{7(8)}{4},,,,,,\frac{8}{8}cdot (x+5)=frac{7(4cdot 2)}{4}\frac{8}{8}cdot (x+5)=7cdot 2cdot frac{4}{4}1cdot (x+5)=14cdot 1,,,end{array})

Simplify and solve for x.

(egin{array}{r}x+5=14x=9,,,end{array})

Check the solution by substituting 9 for x in the original equation.

(egin{array}{r}frac{x+5}{8}=frac{7}{4}\frac{9+5}{8}=frac{7}{4}\frac{14}{8}=frac{7}{4}\frac{7}{4}=frac{7}{4}end{array})

(x=9)

In the next example, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they don’t share any common factors.

### Example

Solve the equation (frac{8}{x+1}=frac{4}{3}).

[hidden-answer a=”331190″]Clear the denominators by multiplying each side by the common denominator. The common denominator is (3 ext{ and }x+1) don’t have any common factors.

(egin{array}{c}3left(x+1 ight)left(frac{8}{x+1} ight)=3left(x+1 ight)left(frac{4}{3} ight)end{array})

Simplify common factors.

(egin{array}{c}3cancel{left(x+1 ight)}left(frac{8}{cancel{x+1}} ight)=cancel{3}left(x+1 ight)left(frac{4}{cancel{3}} ight)24=4left(x+1 ight)24=4x+4end{array})

Now this looks like a linear equation, and we can use the addition and multiplication properties of equality to solve it.

(egin{array}{c}24=4x+4underline{-4},,,,,,,,,,,,,,underline{-4}20=4x,,,,,,,,x=5,,,,,,,,,end{array})

Check the solution in the original equation.

(egin{array}{r},,,,,frac{8}{left(x+1 ight)}=frac{4}{3}\frac{8}{left(5+1 ight)}=frac{4}{3}\frac{8}{6}=frac{4}{3}end{array})

Reduce the fraction (frac{8}{6}) by simplifying the common factor of 2:

(largefrac{cancel{2}cdot4}{cancel{2}cdot3} ormalsize=largefrac{4}{3})

(x=1)

You could also solve this problem by multiplying each term in the equation by 3 to eliminate the fractions altogether. Here is how it would look.

### Example

Solve the equation (frac{x}{3}+1=frac{4}{3}).

[hidden-answer a=”950823″]Both fractions in the equation have a denominator of 3. Multiply both sides of the equation (not just the fractions!) by 3 to eliminate the denominators.

(3left( frac{x}{3}+1 ight)=3left( frac{4}{3} ight))

Apply the distributive property and multiply 3 by each term within the parentheses. Then simplify and solve for x.

(egin{array}{r}3left( frac{x}{3} ight)+3left( 1 ight)=3left( frac{4}{3} ight)\cancel{3}left( frac{x}{cancel{3}} ight)+3left( 1 ight)=cancel{3}left( frac{4}{cancel{3}} ight) x+3=4,,,,,,,,,,,,,,,underline{-3},,,,,underline{-3}x=1end{array})

(x=1)

In the video that follows we present two ways to solve rational equations with both integer and variable denominators. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=132

## Excluded Values and Extraneous Solutions

Some rational expressions have a variable in the denominator. When this is the case, there is an extra step in solving them. Since division by 0 is undefined, you must exclude values of the variable that would result in a denominator of 0. These values are called excluded values. Let’s look at an example.

### Example

Solve the equation (frac{2x-5}{x-5}=frac{15}{x-5}).

[hidden-answer a=”266674″]Determine any values for x that would make the denominator 0.

(frac{2x-5}{x-5}=frac{15}{x-5})

5 is an excluded value because it makes the denominator (x-5) equal to 0.

Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for x.

(egin{array}{r}2x-5=152x=20x=10end{array})

Check the solution in the original equation.

(egin{array}{r}frac{2x-5}{x-5}=frac{15}{x-5},,\frac{2(10)-5}{10-5}=frac{15}{10-5}\frac{20-5}{10-5}=frac{15}{10-5}\frac{15}{5}=frac{15}{5},,,,,,,,,end{array})

(x=10)

In the following video we present an example of solving a rational equation with variables in the denominator. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=132

You’ve seen that there is more than one way to solve rational equations. Because both of these techniques manipulate and rewrite terms, sometimes they can produce solutions that don’t work in the original form of the equation. These types of answers are called extraneous solutions. That’s why it is always important to check all solutions in the original equations—you may find that they yield untrue statements or produce undefined expressions.

### Example

Solve the equation (frac{16}{m+4}=frac

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{m+4}).

[hidden-answer a=”450589″]Determine any values for m that would make the denominator 0. (m+4) equal to 0.

Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for m.

(egin{array}{l}16=m^{2},,,0=

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-16,,,0=left( m+4 ight)left( m-4 ight)end{array})

(egin{array}{c}0=m+4,,,,,, ext{or},,,,,,0=m-4m=-4,,,,,, ext{or},,,,,,m=4m=4,-4end{array})

Check the solutions in the original equation.

Since (m=−4) leads to division by 0, it is an extraneous solution.

(egin{array}{c}frac{16}{m+4}=frac

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{m+4}\frac{16}{-4+4}=frac

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{-4+4}\frac{16}{0}=frac{16}{0}end{array})

(-4) is excluded because it leads to division by 0.

(egin{array}{c}frac{16}{4+4}=frac

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{4+4}\frac{16}{8}=frac{16}{8}end{array})

(m=4)

## Rational formulas

Rational formulas can be useful tools for representing real-life situations and for finding answers to real problems. Equations representing direct, inverse, and joint variation are examples of rational formulas that can model many real-life situations. As you will see, if you can find a formula, you can usually make sense of a situation.

When solving problems using rational formulas, it is often helpful to first solve the formula for the specified variable. For example, work problems ask you to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, (W=rt). The amount of work done (W) is the product of the rate of work (r) and the time spent working (t). Using algebra, you can write the work formula 3 ways:

(W=rt)

Find the time (t): (t=frac{W}{r})(divide both sides by r)

Find the rate (r): (r=frac{W}{t})(divide both sides by t)

### Example

The formula for finding the density of an object is (D=frac{m}{v}), where D is the density, m is the mass of the object and v is the volume of the object. Rearrange the formula to solve for the mass (m) and then for the volume (v).

(D=frac{m}{v})

Multiply both side of the equation by v to isolate m.

(vcdot D=frac{m}{v}cdot v)

Simplify and rewrite the equation, solving for m.

(egin{array}{l}vcdot D=mcdot frac{v}{v}vcdot D=mcdot 1vcdot D=mend{array})

To solve the equation (D=frac{m}{v}) in terms of v, you will need do the same steps to this point, and then divide both sides by D.

(egin{array}{r}frac{vcdot D}{D}=frac{m}{D}\frac{D}{D}cdot v=frac{m}{D}1cdot v=frac{m}{D}v=frac{m}{D}end{array})

(v=frac{m}{D})

Now let’s look at an example using the formula for the volume of a cylinder.

### Example

The formula for finding the volume of a cylinder is (V=pi{r^{2}}h), where V is the volume, r is the radius and h is the height of the cylinder. Rearrange the formula to solve for the height (h).

(V=pi

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h)

Divide both sides by (pi

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) to isolate h.

(frac{V}{pi

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}=frac{pi

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h}{pi

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})

Simplify. You find the height, h, is equal to (frac{V}{pi

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}).

(frac{V}{pi

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}=h)

(h=frac{V}{pi

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})

In the following video we give another example of solving for a variable in a formula, or as they are also called, a literal equation. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=132

## Applications of Rational Equations

Rational equations can be used to solve a variety of problems that involve rates, times and work. Using rational expressions and equations can help you answer questions about how to combine workers or machines to complete a job on schedule. A Good Day’s Work

### Work

A “work problem” is an example of a real life situation that can be modeled and solved using a rational equation. Work problems often ask you to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, (d=rt).) The amount of work done (W) is the product of the rate of work (r) and the time spent working (t). The work formula has 3 versions.

(egin{array}{l}W=rt\,,,,,t=frac{W}{r}\,,,,,r=frac{W}{t}end{array})

Some work problems include multiple machines or people working on a project together for the same amount of time but at different rates. In that case, you can add their individual work rates together to get a total work rate. Let’s look at an example.

### Example

Myra takes 2 hours to plant 50 flower bulbs. Francis takes 3 hours to plant 45 flower bulbs. Working together, how long should it take them to plant 150 bulbs?

[hidden-answer a=”550322″]Think about how many bulbs each person can plant in one hour. This is their planting rate.

Myra: (frac{25,, ext{bulbs}}{1,, ext{hour}})

Francis: (frac{15,, ext{bulbs}}{1,, ext{hour}})

Combine their hourly rates to determine the rate they work together.

Myra and Francis together:

(frac{25,, ext{bulbs}}{1,, ext{hour}}+frac{15,, ext{bulbs}}{1,, ext{hour}}=frac{40,, ext{bulbs}}{1,, ext{hour}})

Use one of the work formulas to write a rational equation, for example (r=frac{W}{t}). You know r, the combined work rate, and you know W, the amount of work that must be done. What you don’t know is how much time it will take to do the required work at the designated rate.

(frac{40}{1}=frac{150}{t})

Solve the equation by multiplying both sides by the common denominator, then isolating t.

(egin{array}{c}frac{40}{1}cdot 1t=frac{150}{t}cdot 1t40t=150 =frac{150}{40}=frac{15}{4} =3frac{3}{4} ext{hours}end{array})

It should take 3 hours 45 minutes for Myra and Francis to plant 150 bulbs together. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=132

Other work problems go the other way. You can calculate how long it will take one person to do a job alone when you know how long it takes people working together to complete the job.

### Example

Joe and John are planning to paint a house together. John thinks that if he worked alone, it would take him 3 times as long as it would take Joe to paint the entire house. Working together, they can complete the job in 24 hours. How long would it take each of them, working alone, to complete the job?

[hidden-answer a=”593775″]Choose variables to represent the unknowns. Since it takes John 3 times as long as Joe to paint the house, his time is represented as 3x.

Let x = time it takes Joe to complete the job

3x = time it takes John to complete the job

The work is painting 1 house or 1. Write an expression to represent each person’s rate using the formula (r=frac{W}{t}).

Joe’s rate: (frac{1}{x})

John’s rate: (frac{1}{3x})

Their combined rate is the sum of their individual rates. Use this rate to write a new equation using the formula (W=rt).

combined rate: (frac{1}{x}+frac{1}{3x})

The problem states that it takes them 24 hours together to paint a house, so if you multiply their combined hourly rate (left( frac{1}{x}+frac{1}{3x} ight)) by 24, you will get 1, which is the number of houses they can paint in 24 hours.

(egin{array}{l}1=left( frac{1}{x}+frac{1}{3x} ight)241=frac{24}{x}+frac{24}{3x}end{array})

Now solve the equation for x. (Remember that x represents the number of hours it will take Joe to finish the job.)

(egin{array}{l},,,1=frac{3}{3}cdot frac{24}{x}+frac{24}{3x}\,,,1=frac{3cdot 24}{3x}+frac{24}{3x}\,,,1=frac{72}{3x}+frac{24}{3x}\,,,1=frac{72+24}{3x}\,,,1=frac{96}{3x}3x=96\,,,x=32end{array})

Check the solutions in the original equation.

(egin{array}{l}1=left( frac{1}{x}+frac{1}{3x} ight)241=left[ frac{ ext{1}}{ ext{32}}+frac{1}{3 ext{(32})} ight]241=frac{24}{ ext{32}}+frac{24}{3 ext{(32})}1=frac{24}{ ext{32}}+frac{24}{96}1=frac{3}{3}cdot frac{24}{ ext{32}}+frac{24}{96}1=frac{72}{96}+frac{24}{96}[end{array})

The solution checks. Since (x=32), it takes Joe 32 hours to paint the house by himself. John’s time is 3x, so it would take him 96 hours to do the same amount of work.

It takes 32 hours for Joe to paint the house by himself and 96 hours for John the paint the house himself.

In the video that follows, we show another example of finding one person’s work rate given a combined work rate. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=132

As shown above, many work problems can be represented by the equation (frac{t}{a}+frac{t}{b}=1), where t is the time to do the job together, a is the time it takes person A to do the job, and b is the time it takes person B to do the job. The 1 refers to the total work done—in this case, the work was to paint 1 house.

The key idea here is to figure out each worker’s individual rate of work. Then, once those rates are identified, add them together, multiply by the time t, set it equal to the amount of work done, and solve the rational equation.

We present another example of two people painting at different rates in the following video. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=132

## Proportions Matryoshka, or nesting dolls.

A proportion is a statement that two ratios are equal to each other. There are many things that can be represented with ratios, including the actual distance on the earth that is represented on a map. In fact, you probably use proportional reasoning on a regular basis and don’t realize it. For example, say you have volunteered to provide drinks for a community event. You are asked to bring enough drinks for 35-40 people. At the store you see that drinks come in packages of 12. You multiply 12 by 3 and get 36 – this may not be enough if 40 people show up, so you decide to buy 4 packages of drinks just to be sure.

This process can also be expressed as a proportional equation and solved using mathematical principles. First, we can express the number of drinks in a package as a ratio:

(frac{12 ext{ drinks }}{1 ext{ package }})

Then we express the number of people who we are buying drinks for as a ratio with the unknown number of packages we need. We will use the maximum so we have enough.

(frac{40 ext{ people }}{x ext{ packages }})

We can find out how many packages to purchase by setting the expressions equal to each other:

(frac{12 ext{ drinks }}{1 ext{ package }}=frac{40 ext{ people }}{x ext{ packages }})

To solve for x, we can use techniques for solving linear equations, or we can cross multiply as a shortcut.

(egin{array}{l},,,,,,,frac{12 ext{ drinks }}{1 ext{ package }}=frac{40 ext{ people }}{x ext{ packages }} ext{}xcdotfrac{12 ext{ drinks }}{1 ext{ package }}=frac{40 ext{ people }}{x ext{ packages }}cdot{x} ext{},,,,,,,,,,,,,,,,,,,,,,,,,12x=40 ext{},,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,x=frac{40}{12}=frac{10}{3}=3.33end{array})

We can round up to 4 since it doesn’t make sense to by 0.33 of a package of drinks. Of course, you don’t write out your thinking this way when you are in the grocery store, but doing so helps you to be able to apply the concepts to less obvious problems. In the following example we will show how to use a proportion to find the number of people on teh planet who don’t have access to a toilet.

Example

As of March, 2016 the world’s population was estimated at 7.4 billion. . According to water.org, 1 out of every 3 people on the planet lives without access to a toilet. Find the number of people on the planet that do not have access to a toilet.

Read and Understand: We can use a proportion to find the unknown number of people who live without a toilet since we are given that 1 in 3 don’t have access, and we are given the population of the planet.

Define and Translate: We know that 1 out of every 3 people don’t have access, so we can write that as a ratio (fraction)

(frac{1 ext{ doesn't }}{3 ext{ do }}).

Let the number of people without access to a toilet be x. The ratio of people with and without toilets is then

(frac{x ext{ don't }}{7.4 ext{ billion do}})

Notice how it helps to use descriptions or units to know where to place the given numbers in the proportion.

Write and Solve: Equate the two ratios since they are representing the same fractional amount of the population.

(frac{1}{3}=frac{x}{7.4 ext{ billion }})

Solve:

(egin{array}{l}frac{1}{3}=frac{x}{7.4} ext{}7.4cdotfrac{1}{3}=frac{x}{7.4}cdot{7.4} ext{}2.46=xend{array})

Interpret: The original units were billions of people, so our answer is (2.46) billion people don’t have access to a toilet. Wow, that’s a lot of people.

In the next example, we will use the length of a person’t femur to estimate their height. This process is used in forensic science and anthropology, and has been found in many scientific studies to be a very good estimate.

Example

It has been shown that a person’s height is proportional to the length of their femur . Given that a person who is 71 inches tall has a femur length of 17.75 inches, how tall is someone with a femur length of 16 inches?

Read and Understand: Height and femur length are proportional for everyone, so we can define a ratio with the given height and femur length. We can then use this to write a proportion to find the unknown height.

Define and Translate: Let x be the unknown height. Define the ratio of femur length and height for both people using the given measurements.

Person 1: (frac{ ext{femur length}}{ ext{height}}=frac{17.75 ext{inches}}{71 ext{inches}})

Person 2: (frac{ ext{femur length}}{ ext{height}}=frac{16 ext{inches}}{x ext{inches}})

Write and Solve: Equate the ratios, since we are assuming height and femur length are proportional for everyone.

(frac{17.75 ext{inches}}{71 ext{inches}}=frac{16 ext{inches}}{x ext{inches}})

Solve by using the common denominator to clear fractions. The common denominator is (71x)

(egin{array}{c}frac{17.75}{71}=frac{16}{x}71xcdotfrac{17.75}{71}=frac{16}{x}cdot{71x}17.75cdot{x}=16cdot{71}x=frac{16cdot{71}}{17.75}=64end{array})

Interpret: The unknown height of person 2 is 64 inches. In general, we can reduce the fraction (frac{17.75}{71}=0.25=frac{1}{4}) to find a general rule for everyone. This would translate to saying for every one femur length, a person’s height is 4 times that length.

Another way to describe the ratio of femur length to height that we found in the last example is to say there’s a 1:4 ratio between femur length and height, or 1 to 4.

Ratios are also used in scale drawings. Scale drawings are enlarged or reduced drawings of objects, buildings, roads, and maps. Maps are smaller than what they represent and a drawing of dendritic cells in your brain is most likely larger than what it represents. The scale of the drawing is a ratio that represents a comparison of the length of the actual object and it’s representation in the drawing. The image below shows a map of the us with a scale of 1 inch representing 557 miles. We could write the scale factor as a fraction (frac{1}{557}) or as we did with the femur-height relationship, 1:557. Map with scale factor

In the next example we will use the scale factor given in the image above to find the distance between Seattle Washington and San Jose California.

Example

Given a scale factor of 1:557 on a map of the US, if the distance from Seattle, WA to San Jose, CA is 1.5 inches on the map, define a proportion to find the actual distance between them.

Read and Understand: We need to define a proportion to solve for the unknown distance between Seattle and San Jose.

Define and Translate: The scale factor is 1:557, and we will call the unknown distance x. The ratio of inches to miles is (frac{1}{557}).

We know inches between the two cities, but we don’t know miles, so the ratio that describes the distance between them is (frac{1.5}{x}).

Write and Solve: The proportion that will help us solve this problem is (frac{1}{557}=frac{1.5}{x}).

Solve using the common denominator (557x) to clear fractions.

(egin{array}{c}frac{1}{557}=frac{1.5}{x}557xcdotfrac{1}{557}=frac{1.5}{x}cdot{557x}x=1.5cdot{557}=835.5)

Interpret: We used the scale factor 1:557 to find an unknown distance between Seattle and San Jose. We also check our answer of 835.5 miles with Google maps, and found that the distance is 839.9 miles, so we did pretty well!

In the next example, we will find a scale factor given the length between two cities on a map, and their actual distance from each other.

Example

Two cities are 2.5 inches apart on a map. Their actual distance from each other is 325 miles. Write a proportion to represent and solve for the scale factor for one inch of the map.

Read and Understand: We know that for each 2.5 inches on the map, it represents 325 actual miles. We are looking for the scale factor for one inch of the map.

Define and Translate: The ratio we want is (frac{2.5}{325})

Write and Solve: We can use a proportion to equate the two ratios and solve for the unknown distance.

(egin{array}{c}frac{1}{x}=frac{2.5}{325}325xcdotfrac{1}{x}=frac{2.5}{325}cdot{325x}325=2.5xx=130)

Interpret: The scale factor for one inch on the map is 1:130, or for every inch of map there are 130 actual miles.

The video that follows show another example of finding an actual distance using the scale factor from a map. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=132

In the video that follows, we present an example of using proportions to obtain the correct amount of medication for a patient, as well as finding a desired mixture of coffees. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=132

## Variation So many cars, so many tires.

## Direct Variation

Variation equations are examples of rational formulas and are used to describe the relationship between variables. For example, imagine a parking lot filled with cars. The total number of tires in the parking lot is dependent on the total number of cars. Algebraically, you can represent this relationship with an equation.

( ext{number of tires}=4cdot ext{number of cars})

The number 4 tells you the rate at which cars and tires are related. You call the rate the constant of variation. It’s a constant because this number does not change. Because the number of cars and the number of tires are linked by a constant, changes in the number of cars cause the number of tires to change in a proportional, steady way. This is an example of direct variation, where the number of tires varies directly with the number of cars.

You can use the car and tire equation as the basis for writing a general algebraic equation that will work for all examples of direct variation. In the example, the number of tires is the output, 4 is the constant, and the number of cars is the input. Let’s enter those generic terms into the equation. You get (y=kx). That’s the formula for all direct variation equations.

( ext{number of tires}=4cdot ext{number of cars} ext{output}= ext{constant}cdot ext{input})

### Example

Solve for k, the constant of variation, in a direct variation problem where (x=10).

[hidden-answer a=”714779″]Write the formula for a direct variation relationship.

(y=kx)

Substitute known values into the equation.

(300=kleft(10 ight))

Solve for k by dividing both sides of the equation by 10.

(egin{array}{l}frac{300}{10}=frac{10k}{10}\,,,,30=kend{array})

The constant of variation, k, is 30.

In the video that follows, we present an example of solving a direct variation equation. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=132

## Inverse Variation

Another kind of variation is called inverse variation. In these equations, the output equals a constant divided by the input variable that is changing. In symbolic form, this is the equation (y=frac{k}{x}).

One example of an inverse variation is the speed required to travel between two cities in a given amount of time.

Let’s say you need to drive from Boston to Chicago, which is about 1,000 miles. The more time you have, the slower you can go. If you want to get there in 20 hours, you need to go 50 miles per hour (assuming you don’t stop driving!), because (frac{1,000}{40}=25).

The equation for figuring out how fast to travel from the amount of time you have is (d=rt). If you solve (r=frac{d}{t}), or (speed=frac{miles}{time}).

In the case of the Boston to Chicago trip, you can write (y=frac{k}{x}).

### Example

Solve for k, the constant of variation, in an inverse variation problem where (y=25).

[hidden-answer a=”752007″]Write the formula for an inverse variation relationship.

(y=frac{k}{x})

Substitute known values into the equation.

(25=frac{k}{5})

Solve for k by multiplying both sides of the equation by 5.

(egin{array}{c}5cdot 25=frac{k}{5}cdot 5125=frac{5k}{5}125=k,,,end{array})

The constant of variation, k, is 125.

In the next example, we will find the water temperature in the ocean at a depth of 500 meters. Water temperature is inversely proportional to depth in the ocean. Water temperature in the ocean varies inversely with depth.

Example

The water temperature in the ocean varies inversely with the depth of the water. The deeper a person dives, the colder the water becomes. At a depth of 1,000 meters, the water temperature is 5º Celsius. What is the water temperature at a depth of 500 meters?

[hidden-answer a=”700119″]You are told that this is an inverse relationship, and that the water temperature (y) varies inversely with the depth of the water (x).

(egin{array}{l},,,,,,,,,,y=frac{k}{x} emp=frac{k}{depth}end{array})

Substitute known values into the equation.

(5=frac{k}{1,000})

Solve for k.

(egin{array}{l}1,000cdot5=frac{k}{1,000}cdot 1,000\,,,,,,,,5,000=frac{1,000k}{1,000}\,,,,,,,,5,000=kend{array})

Now that k, the constant of variation is known, use that information to solve the problem: find the water temperature at 500 meters.

(egin{array}{l}temp=frac{k}{depth} emp=frac{5,000}{500} emp=10end{array})

At 500 meters, the water temperature is 10º C.

In the video that follows, we present an example of inverse variation. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=132

## Joint Variation

A third type of variation is called joint variation. Joint variation is the same as direct variation except there are two or more quantities. For example, the area of a rectangle can be found using the formula (A=lw), where l is the length of the rectangle and w is the width of the rectangle. If you change the width of the rectangle, then the area changes and similarly if you change the length of the rectangle then the area will also change. You can say that the area of the rectangle “varies jointly with the length and the width of the rectangle.”

The formula for the volume of a cylinder, (pi).

### Example

The area of a triangle varies jointly with the lengths of its base and height. If the area of a triangle is 30 inches(^{2}) when the base is 10 inches and the height is 6 inches, find the variation constant and the area of a triangle whose base is 15 inches and height is 20 inches.

[hidden-answer a=”264626″]You are told that this is a joint variation relationship, and that the area of a triangle (A) varies jointly with the lengths of the base (b) and height (h).

(egin{array}{l},,,,,,,,,,y=kxzArea=k(base)(height)end{array})

Substitute known values into the equation, and solve for k.

(30=kleft(10 ight)left(6 ight)30=60k\frac{30}{60}=frac{60k}{60}\frac{1}{2}=k)

Now that k is known, solve for the area of a triangle whose base is 15 inches and height is 20 inches.

(egin{array}{l}Area=k(base)(height)Area=(15)(20)(frac{1}{2})Area=frac{300}{2}Area=150,, ext{square inches}end{array})

The constant of variation, k, is (frac{1}{2}), and the area of the triangle is 150 square inches.

Finding k to be (A=frac{1}{2}bh). The (frac{1}{2}) that you calculated in this example!

In the following video, we show an example of finding the constant of variation for a jointly varying relation. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=132

### Direct, Joint, and Inverse Variation

k is the constant of variation. In all cases, (k eq0).

• Direct variation: (y=kx)
• Inverse variation: (y=frac{k}{x})
• Joint variation: (y=kxz)

## Summary

Rational formulas can be used to solve a variety of problems that involve rates, times, and work. Direct, inverse, and joint variation equations are examples of rational formulas. In direct variation, the variables have a direct relationship—as one quantity increases, the other quantity will also increase. As one quantity decreases, the other quantity decreases. In inverse variation, the variables have an inverse relationship—as one variable increases, the other variable decreases, and vice versa. Joint variation is the same as direct variation except there are two or more variables.

## Summary

You can solve rational equations by finding a common denominator. By rewriting the equation so that all terms have the common denominator, you can solve for the variable using just the numerators. Or, you can multiply both sides of the equation by the least common multiple of the denominators so that all terms become polynomials instead of rational expressions.

An important step in solving rational equations is to reject any extraneous solutions from the final answer. Extraneous solutions are solutions that don’t satisfy the original form of the equation because they produce untrue statements or are excluded values that make a denominator equal to 0.

## Open Resources for Community College Algebra

In your algebra studies, you have learned how to solve linear equations, quadratic equations, and radical equations. In this section, we examine some similarities among the processes for solving these equations. Understanding these similarities can improve your general equation solving ability, even into the future with new equations that are not of these three types.

Figure 7.4.1. Alternative Video Lesson

### Subsection 7.4.1 Equations Where the Variable Appears Once

Here are some examples of equations that all have something in common: the variable only appears once.

For equations like this, there is a strategy for solving them that will keep you from overcomplicating things. In each case, according to the order of operations, the variable is having some things “done” to it in a specific order.

then that result is added to (1)

and this result is a number, (7)

then that result is squared

and this result is a number, (36)

then that result has (3) subtracted from it

then that result has a square root applied

and this result is a number, (3)

Because there is just one instance of the variable, and then things happen to that value in a specific order according to the order of operations, then there is a good strategy to solve these equations. We can just undo each step in the opposite order.

###### Example 7.4.2 .

Solve the equation (2x+1=7 ext<.>)

The actions that happen to (x) are multiply by (2 ext<,>) and then add (1 ext<.>) So we will do the opposite actions in the opposite order to each side of the equation. We will subtract (1) and then divide by (2 ext<.>)

You should check this solution by substituting it into the original equation.

###### Example 7.4.3 .

Solve the equation ((x+4)^2=36 ext<.>)

The actions that happen to (x) are add (4 ext<,>) and then square. So we will do the opposite actions in the opposite order to each side of the equation. We will apply the Square Root Property and then subtract (4 ext<.>)

You should check these solutions by substituting them into the original equation.

###### Example 7.4.4 .

Solve the equation (sqrt<2x-3>=3 ext<.>)

The actions that happen to (x) are multiply by (2 ext<,>) and then subtract (3 ext<,>) and then apply the square root. So we will do the opposite actions in the opposite order to each side of the equation. We will square both sides, add (3) and then divide by (2 ext<.>)

You should check this solution by substituting it into the original equation.

### Subsection 7.4.2 Equations With More Than One Instance of the Variable

Now consider equations like

In these examples, the variable appears more than once. We can't exactly dive in to the strategy of undoing each step in the opposite order. For each of these equations, remind yourself that you can apply any operation you want, as long as you apply it to both sides of the equation. In many cases, you will find that there is some basic algebra move you can take that will turn the equation into something more “standard” that you know how to work with.

With (5x+1=3x+2 ext<,>) we have a linear equation. If we can simply reorganize the terms to combine like terms, a solution will be apparent.

With (x^2+6x=-8 ext<,>) adding (8) to both sides would give us a quadratic equation in standard form. And then the quadratic formula can be used.

With (sqrt=sqrt-1 ext<,>) the complication is those two radicals. We can take any action we like as long as we apply it to both sides, and squaring both sides would remove at least one radical. Maybe after that we will have a simpler equation.

###### Example 7.4.5 .

Solve the equation (5x+1=3x+2 ext<.>)

We'll use basic algebra to rearrange the terms.

You should check this solution by substituting it into the original equation.

###### Example 7.4.6 .

Solve the equation (x^2+6x=-8 ext<.>)

Adding (8) to each side will give us a quadratic equation in standard form, and then we may apply The Quadratic Formula.

You should check these solutions by substituting them into the original equation.

###### Example 7.4.7 .

Solve the equation (sqrt=sqrt-1 ext<.>)

Hoping to obtain a simpler equation, we will square each side. This will eliminate at least one radical, which may help.

You should check this solution by substituting it into the original equation. It is especially important to do this when the equation was a radical equation. At one point, we squared both sides, and this can introduce extraneous solutions (see Remark 6.4.4).

### Subsection 7.4.3 Solving For a Variable in Terms of Other Variables

In the examples so far in this section, there has been one variable (but possibly more than one instance of that variable). This leaves out important situations in science applications where you have a formula with multiple variables, and you need to isolate one of them. Fortunately these situations are not more difficult than what we have explored so far, as long as you can keep track of which variable you are trying to solve for.

###### Example 7.4.8 .

In physics, there is a formula for converting a Celsius temperature to Fahrenheit:

Solve this equation for (C) in terms of (F ext<.>)

The variable we are after is (C ext<,>) and that variable only appears once. So we will apply the strategy of undoing the things that are happening to (C ext<.>) First (C) is multiplied by (frac<9><5> ext<,>) and then it is added to (32 ext<.>) So we will undo these actions in the opposite order: subtract (32) and then multiply by (frac<5><9>) (or divide by (frac<9><5>) if you prefer).

We are satisfied, because we have isolated (C) in terms of (F ext<.>)

###### Example 7.4.9 .

In physics, when an object of mass (m) is moving with a speed (v ext<,>) its “kinetic energy” (E) is given by:

Solve this equation for (v) in terms of the other variables.

The variable we are after is (v ext<,>) and that variable only appears once. So we will apply the strategy of undoing the things that are happening to (v ext<.>) First (v) is squared, then it is multiplied by (m) and by (frac<1><2> ext<.>) So we will undo these actions in the opposite order: multiply by (2 ext<,>) divide by (m ext<,>) and apply the square root.

At the very end, we chose the positive square root, since a speed (v) cannot be negative. We are satisfied, because we have isolated (v) in terms of (E) and (m ext<.>)

When there is only one instance of a variable in an equation, describe a strategy for solving the equation.

You can do whatever algebra you like to the sides of an equation, as long as you do what?

## Course Curriculum

Training and Certifications Make An Instructor, But Experience, Dedication And Character Builds Master Teachers.

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My idea for TabletClass began in my classroom while I was earning my Master of Science in Educational Technology. I wanted to leverage my technical knowledge of the internet and computer programming to create something special for my students. I started my effort to help students learn outside of class by making math videos and posting them on a website. I found that my web resource was working extremely well for my students in and out of school. I kept improving my web resource by using my students' suggestions and feedback I learned and refined a way to teach with videos that was very effective. I used my website in creative ways, like introducing new lessons online to those students who were ready to move forward, while at the same time focusing my attention on those students who were not. My students found the videos engaging and easy to understand, but more importantly I saw a dramatic increase in their abilities and skills—this was truly exciting! My classroom experiences gave way to the desire to build the ultimate online math program, so I decided to create TabletClass Math.

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## Solving Rational Equations with Like Denominators - Concept Solving rational equations is substantially easier with like denominators. When solving rational equations, first multiply every term in the equation by the common denominator so the equation is "cleared" of fractions. Next, use an appropriate technique for solving for the variable.

Remember that an equation means you have an equal sign. And you're solving you're tying to find the value of the variable. In our case I'm probably going to be using x's like in this problem. Okay so when you start looking at equations that have denominators it can be really tricky unless all of the denominators are the same. If I have all the denominators the same like this, there's an idea I want you guys to think about, if I were to go through and multiply all 3 of these terms by 2, it'll be pretty much the same thing as just crossing out those denominators. Right if I multiply everything by 2 in this original equation then I would get this for my 3 terms. And now there's no fractions so I'm a happy camper.
I would subtract 3x from both sides and I can really easily get that x is equals to negative 1. That's a really great tool I want you guys to remember, when you're looking at your homework. I'm going to say it one more time, if all of your denominators are exactly the same, you can cross them out which is essentially like dividing all of those terms by that denominator. You can cross them out and just work with the numerators. Let me show you what that would look like if I had my variable in the denominator, you can do the same idea if I had like 3x over x plus 5 over x is equal to 8x squared over x. A lot of students who didn't know this trick would go through and they would reduce this fraction which is clever, they're on the right track. Other students might try to combine these fractions yeah because those are combinable fractions they have the same denominator.
But you guys this is the trick, these 3 terms all have exactly the same denominator, so I'm going to multiply everything by x and then just work with the tops. That can make your problems much easier when you're approaching them. Keep that in mind when you're going through solving rational equations. Especially if they all have like denominators.

## Choose Online Math Guide

Let us study about list of rational numbers. The basic of mathematics is said to be as clear knowledge of numbers and its operations.
The math includes various forms of numbers as integers, whole numbers, natural numbers, real rational and irrational numbers.
The rational numbers are defined as the form of fractional numbers that are with the two simple integers. Some of the examples for list of rational numbers are below.

List of rational numbers – example 1:

Find out the rational numbers that are in the following number series (1, sqrt(2) , 1/2 , 0.3333…, -5/2 , 7/4 , sqrt(9) , 55/10 , 4.44232…) and list them in the order that smallest to the largest rational numbers?

The number series given is as follows: (1, sqrt(2) , 1/2 , 0.3333…, -5/2 , 7/4 , sqrt(9) , 55/10 , 4.44232…)
All the possible rational numbers that present in the given number series is found to be as follows:
(1, 1/2 , -5/2 , 7/4 , sqrt(9) and 55/10 )
Therefore the rational numbers in the order from the smallest to the largest numbers are as follows:
(-5/2 , 1/2 , 1, 7/4 , sqrt(9) , 55/10 )
(-2.5, 0.5, 1, 1.75, 3, 5.5)

List of rational numbers – example 2:

Find out the rational numbers that are in the following number series (1/5 , sqrt(16) , 0.0222…, 3/2 , 8/4 , 5/100 , 4.2, 343.234…) and list them in the order that smallest to the largest rational numbers?

The number series given is as follows: (1/5 , sqrt(16) , 0.0222…, 3/2 , 8/4 , 5/100 , 4.2, 343.234…)
All the possible rational numbers that present in the given number series is found to be as follows:
(1/5 , sqrt(16) , 3/2 , 8/4 , 5/100 , 4.2)
Therefore the rational numbers in the order from the smallest to the largest numbers are as follows:
(5/100 , 1/5 , 3/2 , 8/4, sqrt(16) , 4.2)
(0.05, 1.2, 1.5, 2, 4, 4.2)

I am planning to write more post on algebra 2 help free, word problems algebra 2. Keep checking my blog.

List of rational numbers – exercises:

Find out the rational numbers that are in the following number series (-11/2 , sqrt(6) , sqrt(4), 0.02, 7/3 , 28/4 , 1/1000 , 43.234…) and list them in the order that smallest to the largest rational numbers?(Answer:-5.2, 0.001, 0.02, sqrt(4) , 28/4 )
Is the number 23.3434…. is a rational number? (Answer: no)
Is the number sqrt(25) is a rational number? (Answer: yes)

## Input Arguments

### Eqns — Rational equations vector of symbolic equations | vector of symbolic expressions

Rational equations, specified as a vector of symbolic equations or symbolic expressions. A rational equation is an equation that contains at least one fraction in which the numerator and the denominator are polynomials.

The relation operator == defines symbolic equations. If a symbolic expression eqn in eqns has no right side, then a symbolic equation with a right side equal to 0 is assumed.

## Cardano’s formula for solving cubic equations Let $a_<3>x^<3>+a_<2>x^<2>+a_<1>x+a_<0>=0$, $a_<3> eq 0$ be the cubic equation. By dividing the equation with $a_3$ we obtain:

The equation above is called a normalized cubic equation.

The square member we remove by the substitution $x= y – frac<3>$. Now we have:

In addition to tags $p = b-frac><3>$ and $q = frac<2a^<3>> <27>– frac <3>+ c$ we obtain the canonical form of the cubic equation:

It is enough to solve the cubic equation of this type.

Cardano’s formula

The solution of the cubic equation $y^ <3>+ py +q = 0$ we search in form:

These solution must satisfy the initial equation, that is:

After transformation of the previous equation, we obtain:

We choose $3vw + p =0$ as an additional requirement, because each number is possible in the infinite way to display in the form of the sum of two numbers.

Therefore, we need to solve the following system of equations:

Systems of equations (1) and (2) are not equivalent. The solution of the system (1) is the solution of the system (2), however, the reversal does not have to be valid. Therefore, we choose solutions which satisfy the equation $v w= – frac <3>$.

By the Vieta’s formulas, solutions of the system (2) are roots of the quadratic equation

By using the formula for solutions of the quadratic equation, we obtain:

The solution of the equation $x^<3>+px+q=0$ we write in the following form:

The formula above is called the Cardano’s formula.

The expression $left (frac <2> ight) ^ <2>+left (frac <3> ight) ^<3>$ which appears in the Cardano’s formula is called the discriminant of the cubic equation $x^<3>+px+q=0$. The discriminant of the cubic equation we will denote as $Delta$.

If $Delta > 0$, then the cubic equation has one real and two complex conjugate roots if $Delta = 0$, then the equation has three real roots, whereby at least two roots are equal if $Delta < 0$ then the equation has three distinct real roots.

Modified Cardano’s formula

be any value of third root and

Then the solutions of the cubic equation $x^<3>+px+q=0$ we can write in the form:

Example 1. Solve the following equation

The discriminant of the given equation is equal to:

Therefore, $Delta>0$ and equation has one real and two complex conjugate solutions.

By the Cardano’s formula we have:

The solutions of the given equation are:

$x_2 = v_1 epsilon + w_1 epsilon^2$

$= 2 cdot left (- frac<1> <2>– frac><2>i ight) + 1 cdot left ( -frac<1> <2>+ frac><2>i ight)$

$x_3 =v_1 epsilon^2 +w_1 epsilon = 2 cdot left ( -frac<1> <2>+ frac><2>i ight) + 1 cdot left (- frac<1> <2>– frac><2>i ight) = – frac<3> <2>+ frac> <2>i.$

We can use formulas above when $Delta > 0$ and $Delta =0$. When $Delta < 0$, we have a different situation, because in the Cardano’s formula appears the square root of a negative number, that is, we have complex numbers.

For example, how to solve the equation $x^<3>-15x-4=0$?

By using the Cardano’s formula, we obtain:

$z=sqrt<125>(cosvarphi + i sin varphi), quad quad w=sqrt<125>left[cos(2pi – varphi) + i sin (2pi – varphi) ight],$

because $z$ and $w$ are complex conjugate numbers and they have the same modulus.

It is valid: $z_1 w_3=z_2 w_2= z_3 w_1=5$ and solutions of the equation $x^<3>-15x-4=0$ are real numbers:

The angle $varphi$ we can eliminate in the following way. We know that

and $an varphi$ we can obtain from the coefficients of the equation:

where, if $varphi<0$ then the solutions are changing the sign.

$an varphi= frac<2sqrt<121>> <4>=frac<22><4>=5.5 Longrightarrow varphi = arctan (5.5) = 1.3909428270 ldots rad.$

Therefore, the one solution of the given equation is

$= 2 cdot 2.2360679774 cdot 0.8944271909$

In general, the solutions of the cubic equation $x^3 + px +q=0$, where $Delta < 0$ and $z=-frac<2>+sqrti$, are:

where $varphi= arctan left[ an left(-frac<2sqrt<-Delta>> ight) ight]$ and if $an left(-frac<2sqrt<-Delta>> ight) < 0$, then solutions are changing the sign.

A rational number is a number that can be expressed as a fraction. It can also be represented by decimal numbers that are either terminating or repeating.

Characteristics of Rational Numbers:

• Rational numbers are infinite.
• Between two rational numbers there are infinite rational numbers.
• Rational numbers contain whole and natural numbers.
• It can be expressed in fraction or in decimal form.

Rational vs. Irrational

• Rational number is a number that can be expressed as a ratio of two integers while irrational number is a number that cannot be expressed that way.
• A number that has repeating and terminating decimal is a rational number while a number that has non-repeating and non-terminating decimal is an irrational number.

SOLVING WORD PROBLEMS

In solving word problems involving rational numbers, we should always apply PEMDAS rule:

Parentheses Any parts of the equation that are written inside a set of parentheses should be done first from the inside out.
Exponent Next step in solving is to evaluate the exponents.
Multiplication, Division The third step is to multiply and divide the terms in order as you read them from left to right.
Addition, Subtraction The last step is the addition and subtraction of terms in order as you read them from left to right.

## 7.4: Solve Rational Equations - Mathematics

The total job can be represented as being a single job, which translates mathematically to a "1." Consequently, each hour of the job must constitute a third of the job, assuming Bridget works at the same speed throughout the entire job.

Now, if the question was changed so that it took five hours to do the job, we would see.

To fully understand this relationship, we should realize a three-hour job requires 1 &frasl3 of the work to be completed per hour. A five-hour job demands that 1 &frasl5 of the job gets handled per hour. Using mathematical terms, this relationship is called the reciprocal. So, a seven-hour job would translate to 1 &frasl7 of the work completed per hour.

To solve this equation, see what happens when we multiply all the terms of this equation by the denominators. If we multiply all the terms by seven and four, we get this equation.

We can cancel factors when a denominator and numerator contain the same factor, as shown in this next step.

After canceling on the left side and multiplying (7)(4)(1) on the right side, we get this remaining equation.

Now, we need to combine terms on the left side to get this next step.

Our last step involves dividing both sides by 11. So, the final answer is x = 2.5454 in a repeating pattern, which rounds to.

As we can see, when the denominators were canceled, the remaining equation is easily solved.

Notice how the table includes a "1" within the bottom row. The "1" stands for the single job that will be completed and it is a standard for all work problems.

Second, we need to start filling the table. Discussed within the work basics section, we will formulate the amount of work they do within one hour. We do this by finding the reciprocals of their work times. For Nancy, the reciprocal of 3 is 1 &frasl3. For Ben, the reciprocal of 4 is 1 &frasl4. These new figures will be placed within the "Work Per Time" column.

We would like to know how long they need to work together to finish the job. Since they will both begin and end at the same time, they will work for the same amount of time, which is known. So, we will call this unknown "x" and place it within the table.

The next step to solving this problem demands that we find values for the last column, "Work." Imagine if Ben worked for 2 hours. We would multiply 1 &frasl4 times 2 and get 1 &frasl2, which would mean only 1 &frasl2 of the job was done. So, if we multiply "Work Per Time" times "Time," we will gain "Work" within our table.

We will multiply across and represent work as algebraic expressions for each row.

Last, we need to develop an equation to solve this problem. We can get an equation if we think of Nancy and Ben's combined work effort. They are working together to get one problem done. Therefore, the last column holds the equation.

The sum of Nancy and Ben's work must be set equal to "1" for one job.

As was done within the solving simple rational equations section, we will follow the same steps. We will multiply all the terms by "3" and "4" to get.

Continuing on by adding like terms, we get this next equation.

After we divide both sides of the equation by "7," we get this solution once it is rounded to the nearest thousandth.

This means if Nancy and Ben work together, they will take 1.714 hours to do the job. This equates to 1 hour and 0.714(60) minutes or 1 hour and roughly 43 minutes to do the job.

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