Question

Find the value of $x$ and $y$ using cross multiplication method:
$x+y=15$ and $x-y=3$

• A. $(9,-6)$
• B. $(-9,6)$
• C. $(9,6)$
• D. $(-9,-6)$

Answer 1

Answer: (C)

$(9,6)$

$x+y=15$ ----- $\left(1\right)$
$x-y=3$ ----- $\left(2\right)$
First use formula for cross multiplication method:
$\frac{x}{b_1{c}_2-b_2{c}_1}=\frac{y}{c_1{a}_2-a_1{c}_2}=\frac{-1}{a_2{b}_2-a_2{b}_1}$
So, from equation (1) and (2) we can write the value of $a,b$ and $c$.
$\frac{x}{1\times 3-(-1)\times 15}=\frac{y}{15\times 1-1\times 3}=\frac{-1}{1\times (-1)-1\times 1}$
$\frac{x}{3+15}=\frac{y}{15-3}=\frac{-1}{-1-1}$
$\frac{x}{18}=\frac{y}{12}=\frac{-1}{-2}$
$\frac{x}{18}=\frac{-1}{-2}$
$x=\frac{-18}{-2}=9$
$\frac{y}{12}=\frac{-1}{-2}$
$y=\frac{-12}{-2}$
$y=6$
Thus the required solution is $(9,6)$