Solving Rational Equations

Rational equations can be used in everyday life, from calculating speeds to calculating average costs. A rational equation is a type of equation involving one or more rational expressions. While solving these equations, multiplications, divisions, additions, or subtractions can be used. In this article, we will solve rational equations using cross multiplying and using least common denominators (LCD) methods. Also, we will go through these methods with examples and daily life problems.

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A rational equation is a type of fraction where either the numerator or denominator is a polynomial.

Solving Rational Equations Using Cross multiplying

Cross multiplying is a practical way to solve a rational equation when there is a single rational expression on each side of the equation.

Cross multiplication, StudySmarter Originals

With cross multiplying, we see that $a.d=b.c$. This method is really useful for solving rational equations. Let's take a look at the example.

Solve $\frac{4}{x+2}=\frac{8}{4x-2}$

Solution:

Steps for solving the rational equation:

Here we have a $\frac{a}{b}=\frac{c}{d}$ form, so we can solve this equation with cross-multiplication.

• Cross multiply: We can multiply 4 with$4x-2$, and 8 with $\left(x+2\right)$

$\left(x+2\right)×8=4×\left(4x-2\right)$

• Distributive property:

$8x+16=16x-8$

• Subtract $8x$ from each side:

$16=8x-8$

• Add 8 to each side:

$24=8x$

• Divide each side by 8:

$x=3$

To see if the solution works out, it should be plugged into the equation.

$\frac{4}{3+2}=\frac{8}{4×3-2}$

$\frac{4}{5}=\frac{8}{12-2}$

$\frac{4}{5}=\frac{8}{10}$

If we divide the right side by 2:

$\frac{4}{5}=\frac{4}{5}$

This equality shows that the solution works out fine!

Solve $\frac{8}{3x-2}=\frac{2}{x-1}$

Solution:

To solve the equation, cross multiplying should be performed, because it is again in the form of $\frac{a}{b}=\frac{c}{d}$

Steps for solving the rational equation:

• Cross multiplying: We should multiply 8 with $\left(x-1\right)$, and 2 with $\left(3x-2\right)$

$8×\left(x-1\right)=2×\left(3x-2\right)$

• Distributive property:

$8x-8=6x-4$

• Subtract 6x from each side:

$2x-8=-4$

• Add 8 to each side:

$2x=4$

• Divide each side by 2:

$x=2$

Now, let's plug x = 2 into the rational equation to see if it works.

$\frac{8}{3×2-2}=\frac{2}{2-1}$

$\frac{8}{6-2}=\frac{2}{1}$

$\frac{8}{4}=2$

As you can see, if we can get the same value on each side when the solution is plugged into the solution, it means it works out as in this case.

Solving Rational Equations Using Least Common Denominators (LCD)

When a rational equation is not given as a proportion as in the case of cross-multiplication, we can solve the equation by using the least common denominators method. The least common denominator, or LCD, is the simplest common denominator to use. We factor the expressions and multiply all of the unique factors to determine the LCD of two rational expressions.

To find the LCD you start by listing the multiples of the fractions. You then find the lowest multiple that they have in common.

Each side of the equation should be multiplied by the least common denominator to solve the rational equation.

To understand this method, let's solve some examples.

What is the solution of $\frac{3}{x}+\frac{5}{4}=\frac{8}{x}$?

Solution:

Steps for solving the rational equation:

In this example, we have denominators as x, 4, and again x. So, the least common denominator can be determined as the multiplication of x and 4, which is 4x. We should use the LCD to multiply with the equation on each side to find the solution.

• Multiply each side by the LCD which is $4x$: $4x\left(\frac{3}{x}+\frac{5}{4}\right)=4x\left(\frac{8}{x}\right)$
• We should distribute $4x$ to each side and multiply it with the values in the parentheses.

$\left(4x×\frac{3}{x}\right)+\left(4x×\frac{5}{4}\right)=\left(4x×\frac{8}{x}\right)$

$\left(4\overline{)x}×\frac{3}{\overline{)x}}\right)+\left(\overline{)4}x×\frac{5}{\overline{)4}}\right)=\left(4\overline{)x}×\frac{8}{\overline{)x}}\right)$

$\left(4×3\right)+\left(5x\right)=\left(4×8\right)$

$12+5x=32$

• Simplify : $12+5x=32$

To simplify this, we should subtract 12 from each side:

$\overline{)12}+5x-\overline{)12}=32-12$

$5x=20$

• Divide each side by 5 :

$\frac{5x}{5}=\frac{20}{5}$

$\frac{\overline{)5}x}{\overline{)5}}=\frac{20}{5}$$x=4$

So the solution for the rational equation is $x=4$, yet we should be sure if it works out. So, again, we should plug this solution into the equation:

$\frac{3}{4}+\frac{5}{4}=\frac{8}{4}$

$\frac{8}{4}=\frac{8}{4}$

Since we have equality again, this solution seems to be working out.

Solve the equation by using LCD: $\frac{1}{2x}+\frac{3}{x+7}=\frac{-1}{x}$

Solution:

Steps for solving the rational equation:

We should first determine what the LCD is for this case. We see that the denominators are 2x, which is basically x times 2, $x+7$, and x. Since there is x involved in 2x anyway, we can take x and $x+7$ to determine the LCD by multiplying them.

So, it gives the LCD as $x×\left(x+7\right)$.

To solve the rational equation, we should multiply each side by the LCD.

• Multiply each side by the LCD which is $x\left(x+7\right)$: $x\left(x+7\right)\left(\frac{1}{2x}+\frac{3}{x+7}\right)=x\left(x+7\right)\left(\frac{-1}{x}\right)$
• Distributive property: We should distribute the LCD to each side and multiply it with the values in the parentheses.

$\left(x×\left(x+7\right)×\frac{1}{2x}\right)+\left(x×\left(x+7\right)×\frac{3}{x+7}\right)=\left(x×\left(x+7\right)×\frac{-1}{x}\right)$

• We can simplify this:

$\left(\overline{)x}×\left(x+7\right)×\frac{1}{2\overline{)x}}\right)+\left(x×\left(\overline{)x+7}\right)×\frac{3}{\overline{)x+7}}\right)=\left(\overline{)x}×\left(x+7\right)×\frac{-1}{\overline{)x}}\right)$

$\frac{x+7}{2}+3x=-1×\left(x+7\right)$

• We should distribute the minus sign to the right side:

$\frac{x+7}{2}+3x=-x-7$

• To make it easier to solve the equation, we can multiply each side by 2:

$2×\left(\frac{x+7}{2}\right)+2×3x=2×\left(-x-7\right)$

• We should distribute 2 in the parentheses with multiplication:

$\overline{)2}×\left(\frac{x+7}{\overline{)2}}\right)+2×3x=2×\left(-x-7\right)$

$x+7+6x=-2x-14$

$7x+7=-2x-14$

• We can add $2x$ to each side to simplify the x value:

$7x+7+2x=-2x-14+2x$

$7x+7+2x=\overline{)-2x}-14+\overline{)2x}$

$9x+7=-14$

• We can subtract 7 from each side to simplify again:

$9x+7-7=-14-7$

$9x=-21$

• Divide each side to 9 to leave x value alone:

$\frac{9x}{9}=\frac{-21}{9}$

$\frac{\overline{)9}x}{\overline{)9}}=\frac{-21}{9}$

$x=\frac{-21}{9}=\frac{-7}{3}$

The solution for the rational equation is then $x=\frac{-7}{3}$. Let's see if it works in the equation:

$\frac{-3}{14}+\frac{9}{14}=\frac{3}{7}$

$\frac{6}{14}=\frac{3}{7}$

We have the equality, so it means that our solution works.

Solving rational equations with two solutions

Sometimes we can have two solutions after solving the rational equations. These two solutions may both work, or just one of them may be the solution. Also, in some cases, there may be no solution at all. In this section, we will work on each case with different examples.

Solve $2-\frac{16}{x-5}=\frac{6}{x}$

Solution:

Here, we have denominators $\left(x-5\right)$and x. The multiplication of those will give us the LCD which is $x×\left(x-5\right)$. To find the solutions.

• Multiply each side by the LCD:

$x×\left(x-5\right)×\left(2-\frac{16}{x-5}\right)=x×\left(x-5\right)×\frac{6}{x}$

• Distribute the LCD to the parentheses:

$\left(2×x×\left(x-5\right)\right)-\left(x×\left(x-5\right)×\frac{16}{x-5}\right)=x×\left(x-5\right)×\frac{6}{x}$

• We can do simplification:

$\left(2×x×\left(x-5\right)\right)-\left(x×\overline{)\left(x-5\right)}×\frac{16}{\overline{)x-5}}\right)=\overline{)x}×\left(x-5\right)×\frac{6}{\overline{)x}}$

• Do the multiplications:

$2x×x-2x×5-16x=6x-30$

$2{x}^{2}-10x-16x=6x-30$

• Subtract $6x$ from each side to simplify:

$2{x}^{2}-10x-16x-6x=\overline{)6x}-30-\overline{)6x}$

$2{x}^{2}-32x=-30$

• Add 30 to each side:

$2{x}^{2}-32x+30=\overline{)-30}+\overline{)30}$

$2{x}^{2}-32x+30=0$

• Write in standard form: $2{x}^{2}-32x+30=0$

• Divide it by 2:

$\frac{\overline{)2}{x}^{2}}{\overline{)2}}-\frac{32x}{2}+\frac{30}{2}=0$

${x}^{2}-16x+15=0$

• At this point, we should factor the equation to find the solutions. The important thing is that we have to find two values making up 15, factor also ${x}^{2}$ . The relative addition should be giving the middle value of the equation which is $-16x$.

${x}^{2}-16x+15=0\phantom{\rule{0ex}{0ex}}x⇔-15\phantom{\rule{0ex}{0ex}}x⇔-1$

We can divide x2 as x times x, and 15 as - 15 and - 1 . When we cross multiply x with - 15 , and the other x with - 1, and add them all together, we should be getting $-16x$ which is satisfied in this way.

As a result, we can write the equation as: $\left(x-15\right)×\left(x-1\right)=0$

The values making the parentheses zero are our solutions!

$x-15=0\to x=15\phantom{\rule{0ex}{0ex}}x-1=0\to x=1$

Let's check if these two solutions both work:

• For $x=15$

$2-\frac{16}{15-5}=\frac{6}{15}$

$2-\frac{16}{10}=\frac{6}{15}$

$\frac{2×10-16}{10}=\frac{6}{15}$

$\frac{20-16}{10}=\frac{6}{15}$

$\frac{4}{10}=\frac{6}{15}$

Divide the left side by 2, and the right side by 3:

$\frac{\overline{)4}}{\overline{)10}}=\frac{\overline{)6}}{\overline{)15}}$

$\frac{2}{5}=\frac{2}{5}$

So, $x=15$ is one of the solutions.

• For $x=1$ :
$2-\frac{16}{1-5}=\frac{6}{1}$

$2-\frac{16}{-4}=6$

$2+4=6$

which again works as the solution.

Solve the equation $\frac{x+3}{x-3}+\frac{x}{x-5}=\frac{x+5}{x-5}$

Solution:

Steps for solving the rational equation:

We can solve this example with the LCD method since the equation is not in the form of

$\frac{a}{b}=\frac{c}{d}$

But how can we find the LCD? We should look at the denominators for that: $\left(x-3\right)$,$\left(x-5\right)$ and again $\left(x-5\right)$. So the least common denominator would be the multiplication of $\left(x-3\right)$ and $\left(x-5\right)$. To find the solutions, we should multiply each side by the LCD.

• Multiply each side by the LCD which is $\left(x-3\right)\left(x-5\right)$:
$\left(x-3\right)×\left(x-5\right)×\left(\frac{x+3}{x-3}+\frac{x}{x-5}\right)=\left(x-3\right)×\left(x-5\right)×\left(\frac{x+5}{x-5}\right)$
• Distribute the LCD to the parentheses:
$\left(\left(x-3\right)×\left(x-5\right)×\left(\frac{x+3}{x-3}\right)\right)+\left(\left(x-3\right)×\left(x-5\right)×\left(\frac{x}{x-5}\right)\right)=\left(\left(x-3\right)×\left(x-5\right)×\left(\frac{x+5}{x-5}\right)\right)$
• Simplify:

$\left(\overline{)\left(x-3\right)}×\left(x-5\right)×\left(\frac{x+3}{\overline{)x-3}}\right)\right)+\left(\left(x-3\right)×\overline{)\left(x-5\right)}×\left(\frac{x}{\overline{)x-5}}\right)\right)=\left(\left(x-3\right)×\left(\overline{)x-5}\right)×\left(\frac{x+5}{\overline{)x-5}}\right)\right)$

$\left(\left(x-5\right)×\left(x+3\right)\right)+\left(x×\left(x-3\right)\right)=\left(\left(x-3\right)×\left(x+5\right)\right)$

• Do the multiplications:
$\left(x×x+3x-5x-5×3\right)+\left(x×x-3x\right)=\left(x×x+5x-3x-5×3\right)$

$\left({x}^{2}-2x-15\right)+\left({x}^{2}-3x\right)=\left({x}^{2}+2x-15\right)$

$2{x}^{2}-5x-15={x}^{2}+2x-15$

• Cross the values on the right side to the left side:

$2{x}^{2}-5x-15-{x}^{2}-2x+15=0$

• Simplify:

${x}^{2}-7x=0$

• Since x is common in the equation, we can factor the equation as: $x×\left(x-7\right)=0$
• To find the solutions, we should find the values making the multipliers zero in the equation, which are $x=0$ and $x=7$

When $x=0$ and $x=7$ are both plugged into the equation, it can be seen that they both work out and are both the solutions for the equation:

For $x=0$ :

$\frac{0+3}{0-3}+\frac{0}{0-5}=\frac{0+5}{0-5}$ $\frac{3}{-3}+0=\frac{5}{-5}$

$-1=-1$

For $x=7$:

$\frac{7+3}{7-3}+\frac{7}{7-5}=\frac{7+5}{7-5}$

$\frac{10}{4}+\frac{7}{2}=\frac{12}{2}$

$\frac{10}{4}+\frac{14}{4}=6$

$\frac{24}{4}=6$

Solving rational equations with extraneous solutions

In the examples above, we have seen that we may obtain two solutions from the rational equations. We were checking if they work by plugging them into the equation. If the solutions do not work in the equations, these solutions are called extraneous solutions.

Solve $\frac{4}{x-1}=\frac{8{x}^{2}}{{x}^{2}-1}-\frac{x}{x+1}$

Solution:

We can solve this type of equation with the LCD method. When we look at the denominators, they are $x-1$, ${x}^{2}-1$ which is the multiplication of $x-1$ and $x+1$. So, the least common denominator should be the multiplication of both $x-1$ and $x+1$ , which is $\left(x-1\right)×\left(x+1\right)$. Each side of the equation should be multiplied by the LCD to find the solutions.

Steps for solving the rational equation:

• Multiply each side by LCD:
$\left(x-1\right)×\left(x+1\right)×\frac{4}{x-1}=\left(x-1\right)×\left(x+1\right)×\left(\frac{8{x}^{2}}{{x}^{2}-1}\right)-\left(x-1\right)×\left(x+1\right)×\left(\frac{x}{x+1}\right)$
• Simplify:

$\left(\overline{)x-1}\right)×\left(x+1\right)×\frac{4}{\overline{)x-1}}=\left(\overline{)x-1}\right)×\left(\overline{)x+1}\right)×\left(\frac{8{x}^{2}}{\overline{){x}^{2}-1}}\right)-\left(x-1\right)×\left(\overline{)x+1}\right)×\left(\frac{x}{\overline{)x+1}}\right)$

$4×\left(x+1\right)=8{x}^{2}-x×\left(x-1\right)$

• Distributive property:

$4x+4=8{x}^{2}-{x}^{2}+x$

• Subtract $4x+4$ from each side:

$4x+4-4x-4=7{x}^{2}+x-4x-4$

$7{x}^{2}-3x-4=0$

We should factor the equation to find solutions. We can divide 7x2 as the multiplication of 7x and x, and - 4 as the multiplication of + 4 and - 1. The goal here is to find the middle value which is - 3x.

$7{x}^{2}-3x-4=0\phantom{\rule{0ex}{0ex}}7x⇔+4\phantom{\rule{0ex}{0ex}}x⇔-1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Through cross multiplying and addition like this, we can obtain the middle value.

• Factor: $\left(7x+4\right)×\left(x-1\right)=0$

Solutions will be $x=-\frac{4}{7}$ or $x=1$

However, when$x=1$is substituted in the equation, it results in division by zero which gives an undefined result. So, $x=-\frac{4}{7}$remains the only solution that works out. $x=1$ is the extraneous solution!

Solve the equation and check for extraneous solutions: $\frac{18}{{x}^{2}-3x}-\frac{6}{x-3}=\frac{5}{x}$

Solution:

We can solve this example with the LCD method. When we look at the denominators, they are: ${x}^{2}-3x$ which is the multiplication of $x-3$ and x. So, the least common denominator is $x×\left(x-3\right)$. The solutions can be found by multiplying the equation with the LCD.

Steps for solving the rational equation:

• Multiply each side by LCD which is $x×\left(x-3\right)$:

$x×\left(x-3\right)×\left(\frac{18}{{x}^{2}-3x}-\frac{6}{x-3}\right)=x×\left(x-3\right)×\left(\frac{5}{x}\right)$

• Distribute the LCD to the parentheses:

$\left(x×\left(x-3\right)×\left(\frac{18}{{x}^{2}-3x}\right)\right)-\left(x×\left(x-3\right)×\left(\frac{6}{x-3}\right)\right)=x×\left(x-3\right)×\left(\frac{5}{x}\right)$

• Simplify:

$\left(\overline{)x}×\left(\overline{)x-3}\right)×\left(\frac{18}{\overline{){x}^{2}-3x}}\right)\right)-\left(x×\left(\overline{)x-3}\right)×\left(\frac{6}{\overline{)x-3}}\right)\right)=\overline{)x}×\left(x-3\right)×\left(\frac{5}{\overline{)x}}\right)$

$18-6x=5×\left(x-3\right)$

$18-6x=5x-15$

• Subtract $18-6x$ from each side:

$18-6x-18+6x=5x-15-18+6x$

$0=11x-33$

• Add 33 to each side:

$11x=33$

• Divide each side by 11:

$x=3$

To check for extraneous solutions, the solution found should be plugged into the equation. However, when $x=3$ is plugged in, the equation results as undefined because there is $x-3$ in the denominators. So, there is no solution for the equation!

Solving Rational Equations given as functions

In these kinds of questions, we will see rational equations given as functions with the domains. We will again solve the rational equation in the same ways, we will obtain the solutions. However, we will check if the solutions work out by looking at if the solutions stay in the domain. If the solution is not in the domain, it is ignored.

For instance, we may have a function like $F\left(x\right)=\frac{4{x}^{2}+10}{2{x}^{2}-5}$. The $F\left(x\right)$ value may be given in the questions and it should be inserted in the function. We may have a domain for the solutions ranging from a to b. After solving the equation again with methods like cross multiplying and the LCD, we should check the domain for the solutions and see if they are extraneous.

Let's look at an example:

The total sales S (in millions of dollars) of a laptop can be modeled by

$S\left(t\right)=\frac{8{t}^{2}+20}{4{t}^{2}-10}$ ; and the domain is $0\le t\le 7$

where $t$ is the number of consumers (in thousands). For how many consumers were the total sales of computers about $6 million? Solution: To solve this equation, 6 should be inserted in $S\left(t\right)$ since it is representing the sales because$S\left(t\right)$ represents the total sales and it is given as$6 million in the example. We should solve the equation to find the $t$ values, and then check if they are in the domain given.

Steps for solving the rational equation:

• Write the equation:

$6=\frac{8{t}^{2}+20}{4{t}^{2}-10}$

• To solve the equation, we can cross multiply :
$6×\left(4{t}^{2}-10\right)=8{t}^{2}+20$
• Distributive property:
$\left(6×4{t}^{2}\right)-6×10=8{t}^{2}+20$

$24{t}^{2}-60=8{t}^{2}+20$

• Subtract $8{t}^{2}+20$ from each side:
$24{t}^{2}-60-8{t}^{2}-20=8{t}^{2}+20-8{t}^{2}-20$

$16{t}^{2}-80=0$

• Add 80 to each side:
$16{t}^{2}-80+80=80$

$16{t}^{2}=80$

• Divide each side by 16:

$\frac{\overline{)16}{t}^{2}}{\overline{)16}}=\frac{80}{16}$

${t}^{2}=5$

• Take square roots of each side: $±2.24\approx t$

Since -2.24 is not in the domain ($0\le t\le 7$), the only solution remains as +2.24.

Thus, the total sales of the computers about \$6 million for 2240 consumers.

Solving Rational Equations - Key takeaways

• If a solution does not work out in the equation or makes the equation undefined, it is called an extraneous solution and it is ignored.

• If a domain for the solution is given, and the solution is found to be not in the domain, then it is ignored.

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What is rational equation example?

5/5x = x/4

How to solve rational equations?

To solve rational equations you can either use:

• Cross multiplication
• Least common denominators

What is the first step in solving a rational equation?

Depending on what method you use to solve the equation will depend on your first step:

• Cross multiplication - start by cross multiplying the equation.
• Least common denominators - start by finding the LCD

How to solve rational equations with extraneous solutions?

Rational equations with extraneous solutions can be solved using the LCD method.

What are the steps to solve rational equations?

Depending on the method you use to solve the rational equation will depend on what steps you need to take. When using the Least Common Denominator (LCD) method you will first find the LCD for each side of the expression before multiplying them by the LCD. The equation can then be simplified to find the solution. When using the cross multiplication method to solve the equation you can start by cross multiplying each side of the equation. Then you can work through the equation simplifying it step by step.

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