Introduction

If we have rational expressions on both sides of an equation, say, \dfrac{{\color{red}x-1}}{{\color{blue}2}} = \dfrac{{\color{blue}2x-1}}{{\color{red}5}}, then we can eliminate the rational expressions by applying cross-multiplication:

  • we cross-multiply the left-hand side numerator ({\color{red}x-1}) with the right-hand side denominator ({\color{red}5}), and

  • we cross-multiply the left-hand side denominator ({\color{blue}2}) with the right-hand side numerator ({\color{blue}2x-1}).

Doing this, we get the following equation:

({\color{red}x-1}) \cdot ({\color{red}5}) = ({\color{blue}2}) \cdot ({\color{blue}2x - 1})

After we've cross-multiplied, we proceed as usual to solve the equation: \begin{align*} (x-1) \cdot 5 &= 2 \cdot (2x - 1) \\[3pt] 5x-5 &= 4x-2 \\[3pt] 5x-5+5 &= 4x-2+5 \\[3pt] 5x &= 4x+3 \\[3pt] 5x-4x &= 4x+3-4x \\[3pt] x &= 3 \end{align*}

Note: Cross-multiplication might seem magical, but it is really just a shortcut for multiplying both sides of the equation by each denominator:

\begin{align*} \require{cancel} \dfrac{{\color{red}x-1}}{{\color{blue}2}} &= \dfrac{{\color{blue}2x-1}}{{\color{red}5}} \\[5pt] {\color{blue}2} \cdot \dfrac{{\color{red}x-1}}{{\color{blue}2}} &= {\color{blue}2} \cdot \dfrac{{\color{blue}2x-1}}{{\color{red}5}} \\[5pt] \cancel{\color{blue}2} \cdot \dfrac{{\color{red}x-1}}{\cancel{\color{blue}2}} &= {\color{blue}2} \cdot \dfrac{{\color{blue}2x-1}}{{\color{red}5}} \\[5pt] {\color{red}x-1} &= {\color{blue}2} \cdot \dfrac{{\color{blue}2x-1}}{{\color{red}5}} \\[5pt] ({\color{red}x-1}) \cdot {\color{red}5} &= {\color{blue}2} \cdot \dfrac{{\color{blue}2x-1}}{{\color{red}5}} \cdot {\color{red}5} \\[5pt] ({\color{red}x-1}) \cdot {\color{red}5} &= {\color{blue}2} \cdot \dfrac{{\color{blue}2x-1}}{\cancel{\color{red}5}} \cdot \cancel{\color{red}5} \\[5pt] ({\color{red}x-1}) \cdot ({\color{red}5}) &= ({\color{blue}2}) \cdot ({\color{blue}2x - 1}) \end{align*}

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Example: Cross-Multiplying with One Binomial

Solve the equation \dfrac{2x+5}{30} = \dfrac{1}{2}.

EXPLANATION

First, we cross-multiply:

\begin{align*} \dfrac{{\color{red}2x+5}}{{\color{blue}30}} &= \dfrac{{\color{blue}1}}{{\color{red}2}}\\[5pt] ({\color{red}2x+5}) \cdot {\color{red}2} &= {\color{blue}30}\cdot {\color{blue}1} \end{align*}

Then, we proceed as usual to solve the equation:

\begin{align*} (2x+5) \cdot 2 &= 30\cdot 1 \\[5pt] 4x+10 &=30 \\[5pt] 4x+10-10 &=30-10 \\[5pt] 4x &=20\\[5pt] \dfrac{4x}{4} &=\dfrac{20}{4}\\[5pt] x &=5 \end{align*}

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Practice: Cross-Multiplying with One Binomial

If $\dfrac {4t +3}{14} = \dfrac 5 2$, then $t =$

a
 $8$
b
 $\dfrac{1}{4}$
c
 $4$
d
 $\dfrac{1}{8}$
e
 $\dfrac{2}{3}$
Practice: Cross-Multiplying with One Binomial

If $\dfrac{3x + 7}{8} = \dfrac{x}{4},$ then $x =$

a
b
c
d
e
Practice: Cross-Multiplying with One Binomial

Solve for $s$ where $\dfrac{3s-4}{36}=\dfrac{s}{4}.$

a
 $s=-\dfrac{2}{3}$
b
 $s=\dfrac{4}{3}$
c
 $s=-\dfrac{3}{2}$
d
 $s=-\dfrac{1}{3}$
e
 $s=3$
Example: Cross-Multiplying with Two Binomials

Solve for x , where \dfrac{6-2x}{2} = \dfrac{6-x}{3}.

EXPLANATION

First, we cross-multiply:

\begin{align} \dfrac{6-2x}{2} &= \dfrac{6-x}{3}\\[5pt] (6-2x) \cdot 3 &= 2 \cdot (6-x) \end{align}

Then, we proceed as usual to solve the equation:

\begin{align} (6-2x) \cdot 3 &= 2 \cdot (6-x)\\[5pt] 18-6x &=12-2x\\[5pt] 18-6x-18 &= 12-2x-18 \\[5pt] -6x &= -2x-6 \\[5pt] -6x+2x &= -2x - 6 + 2x \\[5pt] -4x &= -6 \\[5pt] \dfrac{-4x}{-4} &= \dfrac{-6}{-4} \\[5pt] x &= \dfrac{6}{4} \\[5pt] x &= \dfrac{3}{2} \end{align}

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Practice: Cross-Multiplying with Two Binomials

If $\dfrac{2x + 3}{4} = \dfrac{3x + 1}{2},$ then $x =$

a
b
c
d
e
Practice: Cross-Multiplying with Two Binomials

Solve for $j$ where $\dfrac{j-6}{3}=\dfrac{10+j}{2}.$

a
 $j=-18$
b
 $j=42$
c
 $j=-\dfrac{42}{5}$
d
 $j=-42$
e
 $j=18$
Cross-Multiplying with a Negative Sign

Consider the following equation:

\dfrac{x+1}{2} = -\dfrac{x}{3}

Even though one of the fractions is negative, we can still use cross-multiplication. All we have to do is transfer the negative into the numerator of the fraction, and proceed with cross-multiplication:

\begin{align*} \dfrac{x+1}{2} &= -\dfrac{x}{3} \\[5pt] \dfrac{x+1}{2} &= \dfrac{-x}{3} \\[5pt] (x+1) \cdot 3 &= 2 \cdot (-x) \end{align*}

Then, we proceed as usual to solve the equation:

\begin{align*} (x+1) \cdot 3 &= 2 \cdot (-x) \\[3pt] 3x+3 &= -2x \\[3pt] 3x+3-3x &= -2x-3x \\[3pt] 3 &= -5x \\[3pt] \dfrac{3}{-5} &= \dfrac{-5x}{-5} \\[3pt] -\dfrac{3}{5} &= x \end{align*}

Therefore, x = -\dfrac{3}{5}.

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Example: Cross-Multiplying with a Negative Sign

Solve for u , where \dfrac{u+2}{3} = - \dfrac{3-u}{2}.

EXPLANATION

We transfer the negative into the numerator of the fraction, and proceed with cross-multiplication:

\begin{align*} \dfrac{u+2}{3} &= - \dfrac{3-u}{2} \\[5pt] \dfrac{u+2}{3} &= \dfrac{-(3-u)}{2} \\[5pt] \dfrac{u+2}{3} &= \dfrac{-3+u}{2} \\[5pt] (u+2) \cdot 2 &= 3 \cdot (-3+u) \end{align*}

Then, we proceed as usual to solve the equation:

\begin{align*} (u+2) \cdot 2 &= 3 \cdot (-3+u) \\[3pt] 2u + 4 &= -9 + 3u \\[3pt] 2u + 4 - 2u &= -9 + 3u - 2u \\[3pt] 4 &= -9 + u \\[3pt] 4 + 9 &= -9 + u + 9 \\[3pt] 13 &= u \end{align*}

Therefore, u=13.

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Practice: Cross-Multiplying with a Negative Sign

If $\dfrac{5x+3}{3} = -\dfrac{3}{2},$ then $x =$

a
b
c
d
e
Practice: Cross-Multiplying with a Negative Sign

Solve for $k$ where $\dfrac{3-k}{5} = -\dfrac{5k}{6}.$

a
 $k=-\dfrac{18}{19}$
b
 $k=-1$
c
 $k=-\dfrac{35}{2}$
d
 $k=-6$
e
 $k=1$
Practice: Cross-Multiplying with a Negative Sign

If $\dfrac{x+3}{6} = -\dfrac{1-4x}{2},$ then $x =$

a
b
c
d
e
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