Question

How to do 40

In Problems 39-42, multiply both sides of the equation by the LCD and solve the resulting quadratic equation. 39. x 15 w xwgisla 3d x - 2 x + 1 129

Answer 1

LCD is a special case of LCM .

to find LCD find LCM of denominators in fraction.

here given expression is

$\large \frac{x}{x-2}-1=\frac{3}{x+1}$

here denominator are (x-2) and (x+1)

To find LCD find Lcm of (x-2) and (x+1)

We have no common multiple for x-2 and x+1 .so to get Lcm multiply both.

so LCM LCD is (x-2)(x+1)

now multiply LCD on both side

$(x-2)(x+1)\left ( \frac{x}{x-2}-1 \right )=\frac{3}{x+1} (x-2)(x+1)$

$(x-2)(x+1) *\frac{x}{x-2}-1* (x-2)(x+1) =\frac{3}{x+1}* (x-2)(x+1)$

now cancel common terms in numerator and denominator of different parts of equation

$x(x+1)- (x-2)(x+1) ={3}(x-2)$

take (x+1) as common in left side

$(x+1)(x-x+2) ={3}(x-2)$

$2(x+1)={3}(x-2)$

now open brackets

2x+ 2 = 3x-6

3x-2x= 2+6

x= 8

so solution is x=8