Question

Solve for y if the pair of linear equations are

$\frac{5}{(x-2)}=\frac{4}{(1-y)}and$

$26x+3y+4=0$

• A. 3

Answer 1

The correct option is A 3

Consider,

$\frac{5}{(x-2)}=\frac{4}{(1-y)}$

$4(x–2)=5(1–y)$

$4x–8–5+5y=0$

$4x+5y–13=0...\left(1\right)$

Also,
$26x+3y+4=0...\left(2\right)$

Using cross multiplication method,

$\frac{x}{5\times 4-(3\times (-13\left)\right)}=\frac{y}{(-13\times 26)-(4\times 4)}=\frac{1}{4\times 3-(26\times 5)}$

$\Rightarrow \frac{x}{20+39}=\frac{y}{-338-16}=\frac{1}{12-130}$

$\Rightarrow \frac{y}{-338-16}=\frac{1}{12-130}$

$\Rightarrow y=\frac{-354}{-118}=3$