Question

if 2x+3y=1 and x+(y/x)=3/x,then x=?​

Answer 1

GIVEN :

If 2x+3y=1 and $x+\frac{y}{x}=\frac{3}{x}$, then x=?​

TO FIND :

The value of x in the given equations

SOLUTION :

Given equations are 2x+3y=1 and $x+\frac{y}{x}=\frac{3}{x}$

$2x+3y=1\hfill (1)$

and $x+\frac{y}{x}=\frac{3}{x}\hfill (2)$

Equation (2) ⇒

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

$\frac{x^2+y}{x}==\frac{3}{x}$

$x^2+y=3\hfill (3)$

Multiply the equation (3) into 3 we get

$3x^2+3y=9\hfill (4)$

Subtracting the equations (1) and (4)

  2x+3y=1

$3x^2+3y=9$

(-)__(-)___(-)________

$2x-3x^2=-8$

$-3x^2+2x+8=0$

$-(3x^2-2x-8)=0$

$3x^2-2x-8=0$

Solving the quadratic equation,

$3x^2-6x+4x-8=0$

3x(x-2)+4(x-2)=0

(3x+4)(x-2)=0

3x+4=0 or x-2=0

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

∴ the values of x in the given equations are  $x=-\frac{4}{3}$ and x=2

Answer 2

Answer:

GIVEN :

If 2x+3y=1 and , then x=?​

TO FIND :

The value of x in the given equations

SOLUTION :

Given equations are 2x+3y=1 and  

and  

Equation (2) ⇒

Multiply the equation (3) into 3 we get

Subtracting the equations (1) and (4)

 2x+3y=1

(-)__(-)___(-)________

Solving the quadratic equation,

3x(x-2)+4(x-2)=0

(3x+4)(x-2)=0

3x+4=0 or x-2=0

and x=2

∴ the values of x in the given equations are   and