Chapter 9.6, Problem 54E

a.

To determine

To calculate the system of equation using graph

Answer to Problem 54E

  $x=3,-4$ and $y=4,11$

Explanation of Solution

Given information:

  $\begin{array}{l}y=x^2-5\\ y=-x+7\end{array}$

Graph:

  

Interpretation:

Since the two lines intersect at,

  $(3,4),(-4,11)$

Hence, the solution of the system of equation is,

  $x=3,-4$ and $y=4,11$

b.

To determine

To calculate the system of equation using substitution

Answer to Problem 54E

  $x=3,-4$ and $y=4,11$

Explanation of Solution

Given information:

  $\begin{array}{l}y=x^2-5\\ y=-x+7\end{array}$

Formula used: substitution

Calculation:

Equation 1 is $y=x^2-5$

Equation 2 is $y=-x+7$

Substituting equation 1 to equation 2,

  $x^2-5=-x+7$

Shifting the values,

  $x^2-5-(-x+7)=0$

On solving,

  $x^2+x-12=0$

Factorizing,

  $x^2+4x-3x-12=0$

So,

  $x(x+4)-3(x+4)=0$

So,

  $(x+4)(x-3)=0$

So,

  $x+4=0$ and $x-3=0$

Hence,

  $x=-4$ and $x=3$

For -4, value of y, substituting the value in equation 2,

  $-(-4)+7$

On solving,

  $11$

For 3, value of y, substituting the value in equation 2,

  $-(3)+7$

On solving,

  $4$

Hence, the solution of the system of equation is,

  $x=3,-4$ and $y=4,11$

c.

To determine

To calculate the system of equation using elimination

Answer to Problem 54E

  $x=3,-4$ and $y=4,11$

Explanation of Solution

Given information:

  $\begin{array}{l}y=x^2-5\\ y=-x+7\end{array}$

Formula used: elimination

Calculation:

Equation 1 is $y=x^2-5$

Equation 2 is $y=-x+7$

For elimination, subtracting equation 1 and equation 2

  $y-y=x^2-5-(-x+7)$

On solving,

  $x^2+x-12=0$

Factorizing,

  $x^2+4x-3x-12=0$

So,

  $x(x+4)-3(x+4)=0$

So,

  $(x+4)(x-3)=0$

So,

  $x+4=0$ and $x-3=0$

Hence,

  $x=-4$ and $x=3$

For -4, value of y, substituting the value in equation 2,

  $-(-4)+7$

On solving,

  $11$

For 3, value of y, substituting the value in equation 2,

  $-(3)+7$

On solving,

  $4$

Hence, the solution of the system of equation is,

  $x=3,-4$ and $y=4,11$