Question

Find values of a and b that make the following equality into identity:

Answer 1

Answer:

1. $a=b=\dfrac{3}{4}$

2. $a=-3,\ b=3$

Step-by-step explanation:

1. Given:

$\dfrac{3x}{(x-2)(3x+2)}=\dfrac{a}{x-2}+\dfrac{b}{3x+2}$

To find a and b, add two fractions in right side:

$\dfrac{a}{x-2}+\dfrac{b}{3x+2}\\ \\=\dfrac{a(3x+2)+b(x-2)}{(x-2)(3x+2)}\\ \\=\dfrac{3ax+2a+bx-2b}{(x-2)(3x+2)}\\ \\=\dfrac{(3a+b)x+(2a-2b)}{(x-2)(3x+2)}$

So,

$\dfrac{3x}{(x-2)(3x+2)}=\dfrac{(3a+b)x+(2a-2b)}{(x-2)(3x+2)}$

Two fractions with same denominators are equal when they have equal numerators, so

$3x=(3a+b)x+(2a-2b)$

Equate coefficients:

$At\ x:\ \ 3=3a+b\\ \\At \ 1: 0=2a-2b$

From the second equation:

$a=b$

Substitute it into the first equation:

$3b+b=3\\ \\4b=3\\ \\b=\dfrac{3}{4}$

Hence,

$\dfrac{3x}{(x-2)(3x+2)}=\dfrac{\frac{3}{4}}{x-2}+\dfrac{\frac{3}{4}}{3x+2}$

2. Given:

$\dfrac{3}{x^2-5x+6}=\dfrac{a}{x-2}+\dfrac{b}{x-3}$

Note that $(x-2)(x-3)=x^2-3x-2x+6=x^2-5x+6$

To find a and b, add two fractions in right side:

$\dfrac{a}{x-2}+\dfrac{b}{x-3}\\ \\=\dfrac{a(x-3)+b(x-2)}{(x-2)(x-3)}\\ \\=\dfrac{ax-3a+bx-2b}{(x-2)(x-3)}\\ \\=\dfrac{(a+b)x+(-3a-2b)}{(x-2)(x-3)}$

So,

$\dfrac{3}{(x-2)(x-3)}=\dfrac{(a+b)x+(-3a-2b)}{(x-2)(x-3)}$

Two fractions with same denominators are equal when they have equal numerators, so

$3=(a+b)x+(-3a-2b)$

Equate coefficients:

$At\ x:\ \ 0=a+b\\ \\At \ 1: 3=-3a-2b$

From the first equation:

$a=-b$

Substitute it into the second equation:

$-3(-b)-2b=3\\ \\3b-2b=3\\ \\b=3\\ \\a=-3$

Hence,

$\dfrac{3}{x^2-5x+6}=\dfrac{-3}{x-2}+\dfrac{3}{x-3}$