Chapter 9, Problem 60SGR

To determine

To solve the system of equations.

Answer to Problem 60SGR

The value of pairs of $\left(x,y\right)$ are $\left(-1,-10\right)$ and $\left(-9,-90\right)$ .

Explanation of Solution

Given information :

The equations provided are $y=x^2+9$ and $y=10x$ .

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 :

  $y=x^2+9$

Let this be equation 1.

  $y=10x$

Let this be equation 2.

  $x^2+9=10x$

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

  $=>x^2+10x+9=0$

Bringing the right-hand side to left-hand side. This resulting equation can be simplified by factoring.

  $=>x^2+9x+x+9=0$

Breaking the middle term.

  $=>x\left(x+9\right)+1\left(x+9\right)=0$

Arranging the common factors.

  $=>\left(x+1\right)\left(x+9\right)=0$

The two factors of the function.

Now, the first root is:

  $x+1=0$

Using zero-product property.

  $x=-1$

The second root is:

  $x+9=0$

Using zero-product property.

  $x=-9$

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

  $y=10\left(-1\right)$

Putting $x=-1$ in equation 2.

Simplifying

  $=-10$

Thus one pair of solution is $\left(-1,-10\right)$ .

  $y=10\left(-9\right)$

Putting $x=-9$ in equation 2.

Simplifying

  $=-90$

The other pair of solution is $\left(-9,-90\right)$ .