Question

12 Solve for x and y ,1/5x+1/6y=12,1/3x-3/7y=8,x not=0,y not=0​

Answer 1

Answer:

$x=\frac{89}{4080}, \ y=\frac{89}{1512}$

Step-by-step explanation:

$\frac{1}{5x}+\frac{1}{6y}=12, \ \frac{1}{3x}-\frac{3}{7y}=8, \ where \ x,y\neq 0\\\ let \ \frac{1}{x}=a \ and \ \frac{1}{y}=b, \ substituting \ these \ values \ in \ above \ eqns. \ we \ get\\\frac{a}{5}+\frac{b}{6}=12 \ ...(1) \ and \ \frac{a}{3}-\frac{3b}{7}=8 \ ...(2)\\6a+5b=360 \ ...(A) \ and\\7a-9b=168 \ ...(B)\\Multiplying \ (A) \ with \ 7 \ and \ (B) \ with \ 6 \ and \ by \ solving \ we \ get,\\(42a+35b)-(42a-54b)=2520-1008\\\therefore \ (35+54)b=1512$

$\therefore \ 89b=1512 \implies b=\frac{1512}{89}\\\\putting \ value \ of \ b \ in \ (B)\\\therefore \ 7a-9(\frac{1512}{89})=168\\\therefore \ a=\frac{4080}{89}\\ Now,\\ x=\frac{1}{a}=\frac{89}{4080} \\y=\frac{1}{b}=\frac{89}{1512}$