Chapter 7.1, Problem 30E

To determine

To find the solution for the system of equations $\{\begin{array}{c}x-2y=0\\ 3x-y^2=0\end{array}$ by the method of substitution.

Answer to Problem 30E

  $\left(0,0\right)$ and $\left(12,6\right)$ .

Explanation of Solution

Given information:

The system of equation given is $\{\begin{array}{c}x-2y=0\\ 3x-y^2=0\end{array}$ .

Calculations:

Let $x-2y=0$ … equation (1) and $3x-y^2=0$ …equation (2).

First, solve equation (1) for x and get,

  $\Rightarrow x=2y$ …equation (3)

Now substitute this expression of x in equation (2) and then solve the single variable equation in terms of y.

  $\begin{array}{l}\Rightarrow 3\left(2y\right)-\left(y^2\right)=0\\ \Rightarrow 6y-y^2=0\\ \Rightarrow y\left(6-y\right)=0\end{array}$

Solve the above equation for y and get,

  $\Rightarrow y=0,6$

Now back-substitute these values of y in equation (3) to further obtain the corresponding values of x.

For $y=0,$

  $\Rightarrow x=2\left(0\right)=0$ and,

For $y=6$ ,

  $\Rightarrow x=2\times 6=12$

Therefore, the solutions for the system of equations $\{\begin{array}{c}x-2y=0\\ 3x-y^2=0\end{array}$will be $\left(0,0\right)$ and $\left(12,6\right)$ .