Question

(v) 4/x + 2y = 8 ; 2x+3y= 12​

Answer 1

Answer:

Solution:

(i)$x=\sqrt3,y=4-\frac{2}{\sqrt3}$

(ii)$x=-\sqrt3,y=4+\frac{2}{\sqrt3}$

Step-by-step explanation:

I have applied elimination to solve the given equations

Given:

$\frac{4}{x}+2y=8$......(1)

$2x+3y=12$.......(2)​

$(1)*3\implies\frac{12}{x}+6y=24$

$(2)*2\implies4x+6y=24$

subtracting we get

$\frac{12}{x}-4x=0$

​

$\frac{12}{x}=4x$

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

$x^2=3$

$x=\sqrt3,-\sqrt3$

(2) can be written as

$y=\frac{12-2x}{3}$

$y=4-\frac{2x}{3}$

(i) when $x=\sqrt3$

$y=4-\frac{2\sqrt3}{3}$

$y=4-\frac{2}{\sqrt3}$

(i) when $x=-\sqrt3$

$y=4+\frac{2\sqrt3}{3}$

$y=4+\frac{2}{\sqrt3}$