Question

Find the solution of $x$ and $y$ using cross multiplication method:

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

• A. $x=\frac{-10}{3}$, $y=\frac{1}{3}$
• B. $x=\frac{10}{3}$, $y=\frac{-1}{3}$
• C. $x=\frac{10}{3}$, $y=\frac{1}{3}$
• D. $x=\frac{1}{3}$, $y=\frac{1}{3}$

Answer 1

Answer: (C)

$x=\frac{10}{3}$, $y=\frac{1}{3}$

$x-4y=2$---(1)
$x+2y=4$---(2)
Using formula for cross multiplication method:
$\frac{x}{(b_1{c}_2-b_2{c}_1)}=\frac{y}{(c_1{a}_2-a_1{c}_2)}=\frac{-1}{(a_1{b}_2-a_2{b}_1)}$
So, from equation (1) and (2) we can write the value of $a,b$ and $c$.
$\frac{x}{-4\times 4-2\times 2}=\frac{y}{2\times 1-1\times 4}=\frac{-1}{1\times 2-1\times (-4)}$
$\frac{x}{-16-4}=\frac{y}{2-4}=\frac{-1}{2+4}$
$\frac{x}{-20}=\frac{y}{-2}=\frac{-1}{6}$
$\frac{x}{-20}=\frac{-1}{6}$
$6x=20$
$x=\frac{10}{3}$
$\frac{y}{-2}=\frac{-1}{6}$
$6y=2$
$y=\frac{1}{3}$
Therefore, $x=\frac{10}{3}$, $y=\frac{1}{3}$