Question

Hey guys! I need help with question 3.

Answer 1

So firstly, you want to multiply both sides by (x-2)(x+2) since it is the LCM, or lowest common multiplier. Your equation will look like this: $\frac{3}{x+2} (x-2)(x+2)-\frac{4}{x-2} (x-2)(x+2)=5(x-2)(x+2)$

Next, you can cancel out the denominators on 3/x+2 and 4/x-2 on the left side: $3(x-2)-4(x+2)=5(x-2)(x+2)$

Next, foil: $3x-6-4x-8=5x^2-20$

Next, combine like terms: $-x-14=5x^2-20$

Next, add x + 14 on both sides: $0=5x^2+x-6$

Next, replace x with -5x + 6x: $0=5x^2-5x+6x-6$

Next, factor 5x^2-5x and 6x-6 separately. Make sure that both factors have the same quantity within the parentheses: $0=5x(x-1)+6(x-1)$

With this, we can rewrite it as $0=(5x+6)(x-1)$ From here, we solve for x in (5x+6) and (x-1) separately and setting them to zero.

$5x+6=0\\ 5x=-6\\x=-\frac{6}{5} =-1.2$

$x-1=0\\ x=1$

In short, x = -1.2, 1.