Chapter 9, Problem 58SGR

To determine

To solve the system of equations.

Answer to Problem 58SGR

The value of pairs of $\left(x,y\right)$ are $\left(\frac{3}{4},\frac{9}{4}\right)$ and $\left(-2,5\right)$ .

Explanation of Solution

Given information :

The equations provided are $y=4x^2+4x-3$ and $x+y=3$ . The second equation can be written as $y=3-x$

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 :

  $y=4x^2+4x-3$

Let this be equation 1.

  $y=3-x$

Let this be equation 2.

  $4x^2+4x-3=3-x$

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

  $=>4x^2+4x-3+x-3=0$

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

  $=>4x^2+5x-6=0$

This resulting equation can be simplified by factoring.

  $=>4x^2+8x-3x-6=0$

Breaking the middle term.

  $=>4x\left(x+2\right)-3\left(x+2\right)=0$

Arranging the common factors.

  $=>\left(4x-3\right)\left(x+2\right)=0$

The two factors of the function.

Now, the first root is:

  $4x-3=0$

Using zero-product property.

  $x=\frac{3}{4}$

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.

  $y=3-\left(\frac{3}{4}\right)$

Putting $x=\frac{3}{4}$ in equation 2.

  $=\frac{12-3}{4}$

Simplifying

  $=\frac{9}{4}$

Thus one pair of solution is $\left(\frac{3}{4},\frac{9}{4}\right)$ .

  $y=3-\left(-2\right)$

Putting $x=-2$ in equation 2.

  $=3+2$

Simplifying

  $=5$

The other pair of solution is $\left(-2,5\right)$ .