Question

Solve the following pair of linear equation:
$\frac{4}{x-1}+\frac{5}{y-1}=2,\frac{8}{x-1}+\frac{15}{y-1}=3,x\ne 1,y\ne 1$

• A. $x=\frac{7}{3},y=-4$
• B. $x=-\frac{7}{3},y=4$
• C. $x=\frac{4}{3},y=-3$
• D. $x=\frac{8}{3},y=-5$

Answer 1

The correct option is A $x=\frac{7}{3},y=-4$

Given equations are:

$4(y-1)+5(x-1)=2(x-1)(y-1)$...(1)

$8(y-1)+15(x-1)=3(x-1)(y-1)$...(2)

Solving both the equations separately, we get,

$7x+6y-2xy-11=0$...(3)

$18x+11y-3xy-26=0$...(4)

Solving equations (3) and (4), we get,

$-15x-4y+19=0$...(5)

The values of x and y should be such that, when we equate these values in equation (5), we should get 0.

Let's take $y=-4$, we get x as,

$-15x-4(-4)+19=0$

$-15x+35=0$

$\Rightarrow x=\frac{7}{3}$

When we substitute $x=\frac{7}{3}$ and $y=-4$ in the equation (5)

We have,

$-15\left(\frac{7}{3}\right)-4(-4)+19=0$

$-35+16+19=0$

$-35+35=0$

$0=0$

$\Rightarrow x=\frac{7}{3},y=-4$