Chapter 9.6, Problem 10E

To determine

To solve the system of equation by graphing.

  $y=\frac{1}{2}{x}^2-3x+4$

  $y=x-2$

Answer to Problem 10E

The solutions are $(2,0)$ and $(6,4)$

Explanation of Solution

Given information:

The system of equation

  $y=\frac{1}{2}{x}^2-3x+4$

  $y=x-2$

Calculation:

The system of equation

  $y=\frac{1}{2}{x}^2-3x+4$

  $y=x-2$

The graph of system of equation

  

The graph of system of equation intersecting at $(2,0)$ and $(6,4)$

Put each solution in the both original equations to check that the values make both equations true

  $y=\frac{1}{2}{x}^2-3x+4$                    Original equation             $y=x-2$

  $0=\frac{1}{2}{(2)}^2-3(2)+4$                   $x=2,y=0$             $0=2-2$

  $0=\frac{1}{2}{(2)}^2-3(2)+4$                    Simplify                    $0=2-2$

  $0=0$ true                    Simplify                    $0=0$ true

  $y=\frac{1}{2}{x}^2-3x+4$                    Original equation             $y=x-2$

  $4=\frac{1}{2}{(6)}^2-3(6)+4$  $x=6,y=4$  $4=6-2$

  $4=\frac{1}{2}{(6)}^2-3(6)+4$                    Simplify                    $4=6-2$

  $4=4$ true                    Simplify                    $4=4$ true

The solutions are $(2,0)$ and $(6,4)$