Chapter SH, Problem 9.21EP

To determine

To find: the solution of the system of equation.

Answer to Problem 9.21EP

The solution of the equation is at $\text{x=0 and x=4}$ .

Explanation of Solution

Given:

  ${\text{y=x}}^{\text{2}}\text{-2x-3}$ .

  $\text{y=2x-3}$ .

Concept used:

Addition and elimination method are used to solve the two unknown variables having two equation.

In additions method the two equation will sum up and eliminating method two equation will subtract.

Substituting one another is also done to solve two unknown variables.

Calculation:

  ${\text{y=x}}^{\text{2}}\text{-2x-3}$ .......... $\left(\text{i}\right)$

  $\text{y=2x-3}$ ......... $\left(\text{ii}\right)$

Putting valuer of $\text{y}$ from equation $\left(\text{ii}\right)$ in $\left(\text{i}\right)$ :

  ${\text{2x-3=x}}^{\text{2}}\text{-2x-3}$ .

  $\Rightarrow {\text{x}}^{\text{2}}\text{-4x=0}$ .

  $\Rightarrow \text{x}\left(\text{x-4}\right)\text{=0}$ .

  $\text{x=0 and x=4}$ .

Hence, at the value of $\text{x=0 and x=4}$ the equation satisfied.