Problem 4

Reduce to one fraction with the lowest possible denominator:

$\frac{3a+2b}{a+b}-\frac{25a^2-b^2}{a^2-b^2}=\frac{a}{2b}$

Solution

a) Write a common denominator. Since $a^2-b^2=(a+b)(a-b)$ rewrite

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

b) Combine the fractions:

$\frac{(3a+2b)(a-b)-(25a^2-b^2)}{(a+b)(a-b)}$

c) Expand the numerator. Compute

$(3a+2b)(a-b)=3a^2-3ab+-2b^2=3a^2-ab-2b^2$

and then subtract

$3a^2-ab-2b^2-(25a^2-b^2)=-22a^2-ab-b^{2.}$

d) Thus, the expression reduces to

$-\frac{22a^2+ab+b^2}{(a+b)(a-b)}=-\frac{22a^2+ab+b^2}{a^2-b^2}$

Answer: $-\frac{22a^2+ab+b^2}{a^2-b^2}$