Question

$\sf\pink{SOLVE THE FOLLOWING EQUATIONS}$

1.) $\frac{x + 1}{x - 1} = \frac{2x + 3}{2x - 5}$


2.) $(2x - 3) {}^{2} + (2x + 3) {}^{2} = (8x + 6)(x - 1) + 22$​

Answer 1

Step-by-step explanation:

This is required solution.

Answer 2

Step-by-step explanation:

1) Given that :-

(x+1)/(x-1)=(2x+3)/(2x-5)

On applying cross multiplication then

=>(x+1)(2x-5)=(2x+3)(x-1)

=>2x^2+2x-5x-5=2x^2-2x+3x-3

=>2x^2-3x-5=2x^2+x-3

=>-3x-5=x-3

=>-3x-x=-3+5

=>-4x=2

=>x=2/-4

=>x= -1/2

The value of x= -1/2

Check:-

LHS:-

(x+1)/(x-1)

=>(-1/2+1)/(-1/2-1)

=>(1/2)/(-3/2)

=> -1/3

RHS:-

(2x+3)/(2x-5)

=>[2(-1/2)+3]/[2(-1/2)-5]

=>(-1+3)/(-1-5)

=>2/-6

=>-1/3

LHS =RHS is true for x=-1/2

2) Given that:-

(2x-3)^2+(2x+3)^2=(8x+6)(x-1)+22

=>[(2x)^2-2(2x)(3)+(3)^2]+[(2x)^2+2(2x)(3)+(3)^2] = 8x^2+6x-8x-6+22

=>4x^2-12x+9+4x^2+12x+9=8x^2-2x+16

=>8x^2+18=8x^2-2x+16

=>18=-2x+16

=>18-16=-2x

=>2=-2x

=>-2x=2

=>x= 2/-2

=>x= -1

The value of x= -1

Check:-

LHS:-

(2x-3)^2+(2x+3)^2

=>[2(-1)-3)]^2+[2(-1)+3]^2

=>(-2-3)^2+(-2+3)^2

=>(-5)^2+1^2

=>25+1

=>26

RHS:-

(8x+6)(x-1)+22

=>[8(-1)+6][-1-1]+22

=>(-8+6)(-2)+22

=>(-2)(-2)+22

=>4+22

=>26

LHS=RHS is true for x= -1