Question

please help me solve it by substitution method. please tomorrow is my exams please friends​

Answer 1

Answer:

$\frac{1}{9}$ is the answer :)

Step-by-step explanation:

$\frac{2}{x} -\frac{3}{y} = 15$  

=> $\frac{2y-3x}{xy}=15$ ------------------- (LCM)

=> $2y -3x =15xy$ --------------- (eq 1)

$\frac{8}{x} + \frac{5}{y} = 77$  

=> $\frac{8y +5x}{xy} =77$ -------------------------(LCM)

=> $8y + 5x =77xy$ ------------------- (eq 2)

now we need to do substitution method,

so, in eq 1,                                      

2y -3x =15xy

=> 2y=15xy + 3x

=>  2y - 15xy = 3x --------------- (bringing like terms together)

=> y(2 - 15x) = 3x

=> y= $\frac{3x}{2-15x}$

(NOTE:- YOU CAN FIND Y OR X VALUE FIRST IT IS YOUR CHOICE)

now, substitute y value in eq 2,

8y + 5x =77xy

=> 8($\frac{3x}{2-15x}$) + 5x = 77x(

=> $\frac{24x}{2-15x} + 5x =\frac{ 231x^{2} }{2-15x}$

=> $\frac{24x + 5x(2 -15x)}{2 -15x} =\frac{ 231x^{2} }{2-15x}$ --------------------------- ( LCM)

=> $\frac{24x + 10x -75x^{2}}{2 -15x} =\frac{ 231x^{2} }{2-15x}$

=> $\frac{34x -75x^{2}}{2 -15x} =\frac{ 231x^{2} }{2-15x}$

=> $34x -75x^{2} =231x^{2}$ ------( why i cut both denominator is because when you do cross multiplication the numerator and the denominator are same equations so instead of doing cross multiplication cut it here only )

=> $34x =231x^{2} +75x^{2}$

=> $x = \frac{231x^{2} +75x^{2}}{34}$

=> $x=$ $\frac{306x^{2}}{34}$

=> $x=9x^{2}$

=>$1=9x$ --------( LHS and RHS has x so cut one x to get the value of x)

=> $\frac{1}{9} = x$  

∴$\frac{1}{9}$ is the answer :)

So, PLS MRK AS BRAINLIEST AS I HAVE TYPED THESE ALL DOWN PLSSSS................ :)