Chapter 5.4, Problem 39WE

To determine

To simplify the given expression.

Answer to Problem 39WE

  $\frac{\left(x+y+z\right)}{\left(x+y-z\right)}$

Explanation of Solution

Given:

Expression: $\frac{x^2-y^2-z^2-2yz}{x^2-y^2+z^2-2xz}$

Formula used:

  ${\left(a\pm b\right)}^2=a^2\pm 2ab+b^2$

  $a^2-b^2=\left(a-b\right)\left(a+b\right)$

Calculation:

Consider the given expression,

  $\Rightarrow \frac{x^2-y^2-z^2-2yz}{x^2-y^2+z^2-2xz}$

Regrouping the terms,

  $\Rightarrow \frac{x^2-y^2-z^2-2yz}{-y^2+x^2+z^2-2xz}$

Factor $(-1)$ out of $\left(-y^2-z^2-2yz\right)$ in the numerator.

  $\Rightarrow \frac{x^2-\left(y^2+z^2+2yz\right)}{-y^2+x^2+z^2-2xz}$

  $\Rightarrow \frac{x^2-{\left(y+z\right)}^2}{-y^2+{\left(x-z\right)}^2}[\because {\left(a\pm b\right)}^2=a^2\pm 2ab+b^2]$

Regrouping the terms,

  $\Rightarrow \frac{x^2-{\left(y+z\right)}^2}{{\left(x-z\right)}^2-y^2}$

  $\Rightarrow \frac{\left(x+y+z\right)\left(x-\left(y+z\right)\right)}{\left(x-z-y\right)\left(x-z+y\right)}[\because a^2-b^2=\left(a-b\right)\left(a+b\right)]$

  $\Rightarrow \frac{\left(x+y+z\right)\left(x-y-z\right)}{\left(x-y-z\right)\left(x+y-z\right)}$

  $\Rightarrow \frac{\left(x+y+z\right)}{\left(x+y-z\right)}$   [Reducing the fractions]

Conclusion:

Therefore, after simplification, the answer is $\frac{\left(x+y+z\right)}{\left(x+y-z\right)}$ .