Question

solve the following system of equations :1/2(2x+3y)+12/7(3x-2y)=1/2 7/2x+3y+4/3x -2y =2

Answer 1

if you want to check you can put the value of y and x

Answer 2

The value of x and y is $\bold{\frac{18}{7}\ {and}\ \frac{13}{7}}$.  

Given:

$\frac{1}{2(2 x+3 y)}+\frac{12}{7(3 x-2 y)}=\frac{1}{2} \text { and } \frac{7}{2 x+3 y}+\frac{4}{3 x-2 y}=2$

Solution:

$\frac{1}{2(2 x+3 y)}+\frac{12}{7(3 x-2 y)}=\frac{1}{2} \ldots \ldots \ldots \ldots \ldots(1)$

$\frac{7}{2 x+3 y}+\frac{4}{3 x-2 y}=2 \ldots \ldots \ldots \ldots \ldots(2)$

For calculation purpose consider $2 x+3 y=p ; 3 x-2 y=q \ldots \ldots \ldots \ldots(3)$

Put eqn (3) in eqn (1), we get

$\frac{1}{2 p}+\frac{12}{7 q}=\frac{1}{2}$

Take LCM

$\frac{7 q+24 p}{14 p q}=\frac{1}{2} \Rightarrow 7 q+24 p=\frac{14 p q}{2}=7 p q$

$7 q+24 p=7 p q \ldots \ldots \ldots \ldots(4)$

Also $\frac{7}{p}+\frac{4}{q}=2$

Take LCM

$7 q+4 p=2 p q \ldots \ldots \ldots \ldots \ldots(5)$

Subtracting (4) and (5), we get

$20 \mathrm{p}=5 \mathrm{pq}$

$\Rightarrow q=4$

Substitute q value in (4)

$28+24 p=28 p$

$\Rightarrow 4 p=28$

$\Rightarrow p=7$

From (3)  2x+3y = 7; 3x-2y = 4

Multiply the above equations with 3 and 2 respectively,

$6 x+3 y=21 \ldots \ldots \ldots \ldots(6)$

$6 x-4 y=8 \dots \ldots \ldots \ldots(7)$

Subtracting (6) and (7), we get

$7 y=13$

$\bold{\Rightarrow y=\frac{13}{7}}$

As 3x-2y = 4  

$\Rightarrow 3 x-2\left(\frac{13}{7}\right)=4$

$3 x=4+\frac{26}{7}=\frac{54}{7}$

$\bold{\Rightarrow x=\frac{18}{7}}$