Chapter A.3, Problem 65E

Solution

Simplify the compound fraction.

  $\frac{(15x-11)}{(2x-1)(x-3)}$ .

Given:

The given compound fraction is $\frac{2+\frac{13x-3}{x-4}}{2x+\frac{x+3}{x-4}}$ .

Concept Used:

• Compound fraction is a fraction with fractions in the numerator or denominator.
• First multiply the denominator of the fraction by the whole number. 
• Then, add that number to the numerator of the fraction, and keep the original denominator the same.

Calculation:

The given compound fraction is $\frac{2+\frac{13x-3}{x-4}}{2x+\frac{x+3}{x-4}}$ .

  $\begin{array}{c}\frac{2+\frac{13x-3}{x-4}}{2x+\frac{x+3}{x-4}}=\frac{\frac{2(x-4)+13x-3}{x-4}}{\frac{2x(x-4)+x+3}{x-4}}\\ =\frac{\frac{2x-8+13x-3}{x-4}}{\frac{2x^2-8x+x+3}{x-4}}\\ =\frac{\frac{15x-11}{x-4}}{\frac{2x^2-7x+3}{x-4}}\\ =\frac{(15x-11)(x-4)}{(2x^2-7x+3)(x-4)}\\ =\frac{(15x-11)}{(2x^2-7x+3)}\\ solve2x^2-7x+3=2x^2-(6+1)x+3\\ =2x^2-6x-x+3\\ =2x(x-3)-1(x-3)\\ =(2x-1)(x-3)\\ putthevalueabove\\ \frac{2+\frac{13x-3}{x-4}}{2x+\frac{x+3}{x-4}}=\frac{(15x-11)}{(2x^2-7x+3)}\\ =\frac{(15x-11)}{(2x-1)(x-3)}\end{array}$

Thus by simplifying the compound fraction $\frac{2+\frac{13x-3}{x-4}}{2x+\frac{x+3}{x-4}}$ the solution is $\frac{(15x-11)}{(2x-1)(x-3)}$ .