Question

We have two equations,
$\frac{1}{5}(x-2)=\frac{1}{4}(1-y)$

and

$26x+3y+4=0$

If we solve it using cross-multiplication, we use the following arrangement, after converting equations to the standard form



What are the values of A and B?

• A. A = 5, B = 3
  
• B. A = 3, B = 4
• C. A = 4, B = 26
• D. A = 13, B = 26

Answer 1

The correct option is A A = 5, B = 3


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

$4(x–2)=5(1–y)$

$4x–8–5+5y=0$

$4x+5y–13=0------\left(1\right)$

$26x+3y+4=0------\left(2\right)$

Using cross multiplication method


Hence A = 5 and B = 3.

$\frac{x}{20+39}=\frac{y}{(-13)\left(26\right)-16}=\frac{1}{12-130}$

$x=\frac{59}{-118}=\frac{-1}{2}$

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