Chapter 7.1, Problem 12E

Solution

To calculate: The solution of the system of equation by using the substitution method.

  $\begin{array}{c}x=y+3\\ x-y^2=3y\end{array}$

The solution of system of equation is $\left(0,-3\right)$, and $\left(4,1\right)$ .

Given Information:

The given system of equation is,

  $\begin{array}{c}x=y+3\\ x-y^2=3y\end{array}$

Concept Used:

Substitute one equation value to in other equation and make the one variable equation and solve.

Calculation:

Consider the given system of equations,

  $\begin{array}{c}x=y+3\\ x-y^2=3y\end{array}$

Substitute first equation value $x=y+3$ in the second equation.

  $\begin{array}{c}y+3-y^2=3y\\ y+3-y^2-3y=0\\ -y^2-2y+3=0\\ y^2+2y-3=0\end{array}$

Factor the obtained quadratic equation and equate each factor to zero.

  $\begin{array}{c}{y}^2+3y-y-3=0\\ y\left(y+3\right)-1\left(y+3\right)=0\\ \left(y+3\right)\left(y-1\right)=0\\ y=-3,1\end{array}$

Substitute $-3$ for $y$ in the first equation.

  $\begin{array}{c}x=-3+3\\ =0\end{array}$

Substitute 1 for $y$ in the first equation.

  $\begin{array}{c}x=1+3\\ =4\end{array}$

Hence, the required solution is $\left(0,-3\right)$ and $\left(4,1\right)$ .