Chapter 9, Problem 57SGR

To determine

To solve the system of equations.

Answer to Problem 57SGR

The value of pairs of $\left(x,y\right)$ are $\left(\frac{1}{2},-\frac{1}{2}\right)$ and $\left(4,17\right)$ .

Explanation of Solution

Given information :

The equations provided are $y=2x^2-4x+1$ and $y-5x=-3$ . The second equation can be written as $y=5x-3$

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 :

  $y=2x^2-4x+1$

Let this be equation 1.

  $y=5x-3$

Let this be equation 2.

  $2x^2-4x+1=5x-3$

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

  $=>2x^2-4x+1-5x+3=0$

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

  $=>2x^2-9x+4=0$

This resulting equation can be simplified by factoring.

  $=>2x^2-8x-x+4=0$

Breaking the middle term.

  $=>2x\left(x-4\right)-1\left(x-4\right)=0$

Arranging the common factors.

  $=>\left(2x-1\right)\left(x-4\right)=0$

The two factors of the function.

Now, the first root is:

  $2x-1=0$

Using zero-product property.

  $x=\frac{1}{2}$

The second root is:

  $x-4=0$

Using zero-product property.

  $x=4$

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

  $y=5\left(\frac{1}{2}\right)-3$

Putting $x=\frac{1}{2}$ in equation 2.

  $=\frac{5}{2}-3$

Simplifying

  $=-\frac{1}{2}$

Thus one pair of solution is $\left(\frac{1}{2},-\frac{1}{2}\right)$ .

  $y=5\left(4\right)-3$

Putting $x=4$ in equation 2.

  $=20-3$

Simplifying

  $=17$

The other pair of solution is $\left(4,17\right)$ .