Question

Solve the following equations for $(x,y)$.

$\frac{2x+y}{3x-y}-\frac{x-y}{x+y}=2\frac{8}{15}$,
$7x+5y=29$

• A. $(1,2)$
• B. $(2,3)$
• C. $(-2,3)$
• D. $(4,2)$

Answer 1

Answer: (B)

$(2,3)$

Given equations are $\frac{2x+y}{3x-y}-\frac{x-y}{x+y}=2\frac{8}{15}$ ..........(1)

and $7x+5y=29$ ...........(2)

$\Rightarrow \frac{(2x+y)(x+y)-(3x-y)(x-y)}{(3x-y)(x+y)}=\frac{38}{15}$

$\Rightarrow \frac{2x^2+2xy+xy+y^2-{3x}^2+3xy+xy-y^2}{{3x}^2+3xy-xy-y^2}=\frac{38}{15}$

$\Rightarrow \frac{-x^2+7xy}{3x^2+2xy-y^2}=\frac{38}{15}$

$\Rightarrow -15x^2+105xy=114x^2+76xy-38y^2$

$\Rightarrow 129x^2-29xy-38y^2=0$

$\Rightarrow 129x^2-86xy+57xy-38y^2=0\Rightarrow 43x(3x-2y)+19y(3x-2y)=0$

$\Rightarrow (43x+19y)(3x-2y)=0$

$\Rightarrow 43x+19y=0$ ........(3)

$3x-2y=0$ ..........(4)

Solving (2) and (3), we get

From $\left(3\right)$ $y=-\frac{43x}{19}$

Substituting in (2), we get

$7x+5(-\frac{43x}{19})=29\Rightarrow x=-\frac{581}{82}$

Substituting back in (2) $\Rightarrow y=\frac{1247}{82}$

So, $(-\frac{581}{82},\frac{1247}{82})$ is one solution of the equation

Similarly by solving (2) and (4), we get

$x=2$ and $y=3$

So, $(2,3)$ is the other solution of the given equations.