Chapter 11, Problem 5CA

To determine

To calculate the system of equation.

Answer to Problem 5CA

The inverse pairs are,

  $y=-3x^2$ and $y=\sqrt{\frac{-1}{3}x}$ ,

  $y=-x+7$ and $y=-x+7$ ,

  $y=\frac{1}{2}x+2$ and $y=2x-4$ ,and

  $y=x^2-5$ and $y=\sqrt{x+5}$

Explanation of Solution

Given information: the functions are

  $\begin{array}{l}y=-3x^2,x\ge 0\\ y=-x+7\\ y=2x-4\\ y=\sqrt{\frac{-1}{3}x}\\ y=\frac{1}{2}x+2\\ y=x^2-5\\ y=\sqrt{x+5}\\ y=-x+7\end{array}$

Calculation:

For equation 1,

  $y=-3x^2$

Shifting the values,

  $x^2=\frac{y}{-3}$

So,

  $x=\sqrt{\frac{-1}{3}y}$

Hence, $y=-3x^2$ and $y=\sqrt{\frac{-1}{3}x}$ are inverse.

For equation 2,

  $y=-x+7$

So,

  $x=7-y$

Hence, $y=-x+7$ and $y=-x+7$ are inverse.

For equation 3,

  $y=2x-4$

So,

  $2x=y+4$

Cross-multiplying,

  $x=\frac{1}{2}y+2$

Hence, $y=\frac{1}{2}x+2$ and $y=2x-4$ are inverse

For equation 6,

  $y=x^2-5$

So,

  $x^2=y+5$

So,

  $x=\sqrt{y+5}$

Hence, $y=x^2-5$ and $y=\sqrt{x+5}$ are inverse.

So, the inverse pairs are,

  $y=-3x^2$ and $y=\sqrt{\frac{-1}{3}x}$ ,

  $y=-x+7$ and $y=-x+7$ ,

  $y=\frac{1}{2}x+2$ and $y=2x-4$ ,and

  $y=x^2-5$ and $y=\sqrt{x+5}$