Question

Based on cross-multiplication method, solve the following pairs of equations by cross multiplication rule.

$\frac{x}{2}-\frac{y}{3}+4=0,\frac{x}{2}-\frac{5y}{3}+12=0$

Then $x+y$ is equal to

Answer 1

Given systems of equations are:
$\frac{x}{2}-\frac{y}{3}+4=0$ ....(1)

$\frac{x}{2}-\frac{5y}{3}+12=0$ ...(2)
Now from eq1 and eq2 will get:
$3x-2y+24$ ...(3)
$3x-10y+72$ ...(4)
To solve this pair of equations for x and y using cross-multiplication, we arrange the variables x and y and their coefficients a1, a2, b1 and b2, and the constants c1 and c2 as:
$x=\frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1}$

$y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}$

$\Rightarrow x=\frac{(-2\times 72)-(-10\times 24)}{(3\times 10)-(3\times -2)}$

$\Rightarrow y=\frac{(24\times 3)-(72\times 3)}{(3\times 10)-(3\times -2)}$

$\Rightarrow x=\frac{-144-\left(240\right)}{-30-(-6)}$

$\Rightarrow x=\frac{96}{-24}=-4$

$\Rightarrow y=\frac{72-216}{-30-(-6)}$

$\Rightarrow y=\frac{-144}{-24}=6$

$\Rightarrow x=-4,y=6$

$x+y=$ $6-4=$2