Chapter 9, Problem 56SGR

To determine

To solve the system of equations.

Answer to Problem 56SGR

The value of pairs of $\left(x,y\right)$ are $\left(2,11\right)$ and $\left(-9,33\right)$ .

Explanation of Solution

Given information :

The equations provided are $y=x^2+5x-3$ and $y=15-2x$ .

Formula used :

Since there are two equations provided for y, equate the components of both with each other. This will result in a quadratic equation. Solve this equation for x in order to find the solution

The resulting equation can be solved by factoring. To do so, break up the midpoint to arrive at two factors. Then use the zero-product property to equate each factor in order to get the roots.

Calculation :

  $y=x^2+5x-3$

Let this be equation 1.

  $y=15-2x$

Let this be equation 2.

  $x^2+5x-3=15-2x$

Since both are equal to y , equate them to get a resulting equation for x

  $=>x^2+5x-3-15+2x=0$

Bringing the right-hand side to left-hand side.

  $=>x^2+7x-18=0$

This resulting equation can be simplified by factoring.

  $=>x^2+9x-2x-18=0$

Breaking the middle term.

  $=>x\left(x+9\right)-2\left(x+9\right)=0$

Arranging the common factors.

  $=>\left(x+9\right)\left(x-2\right)=0$

The two factors of the function.

Now, the first root is:

  $x+9=0$

Using zero-product property.

  $x=-9$

The second root is:

  $x-2=0$

Using zero-product property.

  $x=2$

Now, put these values of x in equation 2 to get the value of y.

  $y=15-2\left(-9\right)$

Putting $x=-9$ in equation 2.

  $=15+18$

Simplifying

  $=33$

Thus one pair of solution is $\left(-9,33\right)$ .

  $y=15-2\left(2\right)$

Putting $x=2$ in equation 2.

  $=15-4$

Simplifying

  $=11$

The other pair of solution is $\left(2,11\right)$ .