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12.8: Rational Equations Using Proportions

Difficulty Level: Basic Created by: CK-12
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Suppose you were traveling on a paddle boat at a constant speed. In 6 minutes, you traveled \begin{align*}x\end{align*}x meters, and in 10 minutes, you traveled \begin{align*}x+4\end{align*}x+4 meters. Could you find the value of \begin{align*}x\end{align*}x in this scenario? If so, how would you do it? After completing this Concept, you'll be able to solve rational equations using proportions so that you can handle this type of problem.

Solution of Rational Equations

You are now ready to solve rational equations! There are two main methods you will learn to solve rational equations:

  • Cross products
  • Lowest common denominators

In this Concept you will learn how to solve using cross products.

Solving a Rational Proportion

When two rational expressions are equal, a proportion is created and can be solved using its cross products.

Example A

For example, to solve \begin{align*}\frac{x}{5}=\frac{(x+1)}{2}\end{align*}x5=(x+1)2, cross multiply and the products are equal.

\begin{align*}\frac{x}{5} = \frac{(x+1)}{2} \rightarrow 2(x)=5(x+1)\end{align*}x5=(x+1)22(x)=5(x+1)

Solve for \begin{align*}x\end{align*}x:

\begin{align*}2(x) &= 5(x+1) \rightarrow 2x=5x+5\\ 2x-5x &= 5x-5x+5\\ -3x &= 5\\ x &= -\frac{5}{3}\end{align*}2(x)2x5x3xx=5(x+1)2x=5x+5=5x5x+5=5=53

Example B

Solve \begin{align*}\frac{2x}{x+4}=\frac{5}{x}\end{align*}2xx+4=5x.


\begin{align*}\frac{2x}{x+4} &= \frac{5}{x} \rightarrow 2x^2=5(x+4)\\ 2x^2 &= 5(x+4) \rightarrow 2x^2=5x+20\\ &2x^2-5x-20 = 0\end{align*}2xx+42x2=5x2x2=5(x+4)=5(x+4)2x2=5x+202x25x20=0

Notice that this equation has a degree of two; that is, it is a quadratic equation. We can solve it using the quadratic formula.

\begin{align*}x=\frac{5 \pm \sqrt{185}}{4} \Rightarrow x \approx -2.15 \ \text{or} \ x \approx 4.65\end{align*}x=5±1854x2.15 or x4.65

Example C

Solve \begin{align*}\frac{3x}{5x+2}=\frac{1}{x}\end{align*}3x5x+2=1x.


Start by cross multiplying:

\begin{align*}\frac{3x}{5x+2}=\frac{1}{x} \Rightarrow 3x^2=5x+2 \Rightarrow 3x^2-5x-2=0\end{align*}3x5x+2=1x3x2=5x+23x25x2=0

Since this equation has a squared term as its highest power, it is a quadratic equation. We can solve this by using the quadratic formula, or by factoring.

1. Since there are no common factors, start by finding the product of the coefficient in front of the squared term and the constant:

\begin{align*}3\cdot -2=-6\end{align*}32=6

2. What factors of -6 add up to 5? That would be -6 and 1, since -6+1=-5.

3. Factor, beginning by breaking up the middle term, \begin{align*}-5x\end{align*}5x, as above:

\begin{align*} 0&=3x^2-5x-2=3x^2-6x+1x-2\\ &=3x(x-2)+1(x-2)= (3x+1)(x-2) \end{align*}0=3x25x2=3x26x+1x2=3x(x2)+1(x2)=(3x+1)(x2)

4. Use the Zero Product Principle:

\begin{align*}(3x+1)(x-2)=0 \Rightarrow 3x+1=0 \text{ or } x-2=0 \Rightarrow x=-\frac{1}{3} \text{ or } x=2\end{align*}(3x+1)(x2)=03x+1=0 or x2=0x=13 or x=2

Guided Practice

Solve \begin{align*} -\frac{x}{2}=\frac{3x-8}{x}\end{align*}x2=3x8x.


\begin{align*} \text{Cross multiply:} && -\frac{x}{2}=\frac{3x-8}{x} \Rightarrow x^2&=-2(3x-8)\\ \text{Set one side equal to zero to get a quadratic equation:} && 0&=x^2+2(3x-8)\\ \text{Simplify by distributing:} && 0&=x^2+6x-16\\ \text{Factor by determining } -16=8\cdot -2 \text{ and } 6=8+(-2): && 0&=x^2-2x+8x-16=x(x-2)+8(x-2)=(x+8)(x-2)\\ \text{Use the zero product principle:} && 0&=(x+8)(x-2) \Rightarrow x+8=0 \text{ or } x-2=0 \Rightarrow x=-8 \text{ or } x=2 \end{align*}Cross multiply:Set one side equal to zero to get a quadratic equation:Simplify by distributing:Factor by determining 16=82 and 6=8+(2):Use the zero product principle:x2=3x8xx20000=2(3x8)=x2+2(3x8)=x2+6x16=x22x+8x16=x(x2)+8(x2)=(x+8)(x2)=(x+8)(x2)x+8=0 or x2=0x=8 or x=2


Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Solving Rational Equations (12:57)

Solve the following equations.

  1. \begin{align*}\frac{2x+1}{4}=\frac{x-3}{10}\end{align*}2x+14=x310
  2. \begin{align*}\frac{4x}{x+2}=\frac{5}{9}\end{align*}4xx+2=59
  3. \begin{align*}\frac{5}{3x-4}=\frac{2}{x+1}\end{align*}53x4=2x+1
  4. \begin{align*}\frac{7x}{x-5}=\frac{x+3}{x}\end{align*}7xx5=x+3x
  5. \begin{align*}\frac{2}{x+3}-\frac{1}{x+4}=0\end{align*}2x+31x+4=0
  6. \begin{align*}\frac{3x^2+2x-1}{x^2-1}=-2\end{align*}3x2+2x1x21=2

Mixed Review

  1. Divide: \begin{align*}-2 \frac{9}{10} \div - \frac{15}{8}\end{align*}2910÷158.
  2. Solve for \begin{align*}g: -1.5 \left(-3 \frac{4}{5}+g \right)=\frac{201}{20}\end{align*}.
  3. Find the discriminant of \begin{align*}6x^2+3x+4=0\end{align*} and determine the nature of the roots.
  4. Simplify \begin{align*}\frac{6b}{2b+2}+3\end{align*}.
  5. Simplify \begin{align*}\frac{8}{2x-4}- \frac{5x}{x-5}\end{align*}.
  6. Divide: \begin{align*}(7x^2+16x-10) \div (x+3)\end{align*}.
  7. Simplify \begin{align*}(n-1)*(3n+2)(n-4)\end{align*}.

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proportion A statement in which two fractions are equal: \frac{a}{b} = \frac{c}{d}.
Zero Product Property The only way a product is zero is if one or more of the terms are equal to zero: a\cdot b=0 \Rightarrow a=0 \text{ or } b=0.
Rational Expression A rational expression is a fraction with polynomials in the numerator and the denominator.

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Feb 24, 2012
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