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Pair of Linear Equations: Cross-Multiplication and Reducible Equations

The formula

For $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ with a unique solution, cross-multiplication gives both variables in one line:

\dfrac{x}{b_1c_2-b_2c_1} = \dfrac{y}{c_1a_2-c_2a_1} = \dfrac{1}{a_1b_2-a_2b_1}

Worked example

Solve $2x+y=5$ and $3x+2y=8$ by cross-multiplication.

Solution: Rewrite as $2x+y-5=0$ and $3x+2y-8=0$ , so $a_1=2,b_1=1,c_1=-5$ and $a_2=3,b_2=2,c_2=-8$ :

\dfrac{x}{(1)(-8)-(2)(-5)} = \dfrac{y}{(-5)(3)-(-8)(2)} = \dfrac{1}{(2)(2)-(3)(1)}

\dfrac{x}{2} = \dfrac{y}{1} = \dfrac{1}{1} \;\Rightarrow\; x=2, \; y=1

Worked example: an equation that isn't linear yet

Solve $\dfrac{1}{2x}+\dfrac{1}{3y}=2$ and $\dfrac{1}{3x}+\dfrac{1}{2y}=\dfrac{13}{6}$ .

Solution: Neither equation is linear in $x$ and $y$ , but substituting $p=\tfrac{1}{x}$ and $q=\tfrac{1}{y}$ makes them linear in $p$ and $q$ :

\dfrac{p}{2}+\dfrac{q}{3}=2 \;\Rightarrow\; 3p+2q-12=0, \qquad \dfrac{p}{3}+\dfrac{q}{2}=\dfrac{13}{6} \;\Rightarrow\; 2p+3q-13=0

Cross-multiplying:

\dfrac{p}{(2)(-13)-(3)(-12)} = \dfrac{q}{(-12)(2)-(-13)(3)} = \dfrac{1}{(3)(3)-(2)(2)}

\dfrac{p}{10} = \dfrac{q}{15} = \dfrac{1}{5} \;\Rightarrow\; p=2, \; q=3

Converting back, $\tfrac{1}{x}=2$ gives $x=\tfrac{1}{2}$ , and $\tfrac{1}{y}=3$ gives $y=\tfrac{1}{3}$ .

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