Chapter 9, Problem 22PT

To determine

To solve the system of equations.

Answer to Problem 22PT

The value of pairs of $\left(x,y\right)$ are $\left(1,-3\right)$ and $\left(4,-9\right)$ .

Explanation of Solution

Given information :

The equations provided are $y=x^2-7x+3$ and $y=-2x-1$ .

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 :

  $y=x^2-7x+3$

Let this be equation 1.

  $y=-2x-1$

Let this be equation 2.

  $-2x-1=x^2-7x+3$

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

  $x^2-7x+3=-2x-1$

  $=>x^2-7x+3+2x+1=0$

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

  $=>x^2-5x+4=0$

This resulting equation can be simplified by factoring.

  $=>x^2-x-4x+4=0$

Breaking the middle term.

  $=>x\left(x-1\right)-4\left(x-1\right)=0$

Arranging the common factors.

  $=>\left(x-4\right)\left(x-1\right)=0$

The two factors of the function.

Now, the first root is:

  $x-4=0$

Using zero-product property.

  $x=4$

The second root is:

  $x-1=0$

Using zero-product property.

  $x=1$

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

  $y=-2\times \left(1\right)-1$

Putting $x=1$ in equation 2.

  $=-2-1$

Simplifying

  $=-3$

Thus one pair of solution is $\left(1,-3\right)$ .

  $y=-2\times \left(4\right)-1$

Putting $x=4$ in equation 2.

  $=-8-1$

Simplifying

  $=-9$

The other pair of solution is $\left(4,-9\right)$ .