Chapter 6.1, Problem 41WE

To determine

To calculate: The simplified form of the expression $\frac{25x^2-36y^2}{10x^2+3xy-18y^2}$ and state restrictions on variable.

Answer to Problem 41WE

The simplified value of the expression $\frac{25x^2-36y^2}{10x^2+3xy-18y^2}$ is $\frac{5x+6y}{2x+3y}$ and restriction is $x\ne -\frac{3y}{2},x\ne \frac{6y}{5}$.

Explanation of Solution

Given information:

The expression $\frac{25x^2-36y^2}{10x^2+3xy-18y^2}$.

Formula used:

The simplest form of an algebraic fraction is when numerator and denominator have no common factor other than 1 and $-1$ . Simplification is done by factorizing numerator and denominator.

Calculation:

Consider the expression $\frac{25x^2-36y^2}{10x^2+3xy-18y^2}$.

Recall that the simplest form of an algebraic fraction is when numerator and denominator have no common factor other than 1 and $-1$ . Simplification is done by factorizing numerator and denominator.

So, factorize numerator using the formula $X^2-Y^2=\left(X+Y\right)\left(X-Y\right)$ and denominator using the middle term splitting method and then strike out common terms from the expression and simplify it as,

  $\begin{array}{c}\frac{25x^2-36y^2}{10x^2+3xy-18y^2}=\frac{{\left(5x\right)}^2-{\left(6y\right)}^2}{10x^2+15xy-12xy-18y^2}\\ =\frac{\left(5x+6y\right)\left(5x-6y\right)}{5x\left(2x+3y\right)-6y\left(2x+3y\right)}\\ =\frac{\left(5x+6y\right)\left(5x-6y\right)}{\left(2x+3y\right)\left(5x-6y\right)}\\ =\frac{5x+6y}{2x+3y}\end{array}$

To restrict the variable, exclude the value that makes the denominator zero.

So, denominator will be zero when,

  $\begin{array}{c}\left(2x+3y\right)\left(5x-6y\right)=0\\ 2x+3y=0,5x-6y=0\\ 2x=-3y,5x=6y\\ x=\frac{-3y}{2},x=\frac{6y}{5}\end{array}$

For denominator not to be zero, $x\ne -\frac{3y}{2},x\ne \frac{6y}{5}$ .

Thus, the simplified value of the expression $\frac{25x^2-36y^2}{10x^2+3xy-18y^2}$ is $\frac{5x+3y}{2x+3y}$ and restriction is $x\ne -\frac{3y}{2},x\ne \frac{6y}{5}$.