Question

In the following exercises, solve the system of equations by using substitution.

$$\begin{gathered} x^2+4y^2=4 \\ 2x-y=1 \end{gathered}$$

Short answer

$$\begin{gathered} Solution: \\ \begin{array}{cc}x=\frac{16}{17},\hspace{0.17em}& y=\frac{15}{17}\end{array} \\ \begin{array}{cc}x=0,\hspace{0.17em}& y=-1\end{array} \end{gathered}$$

Step-by-step solution

Given Information 

Two equations are

$$\begin{gathered} x^2+4y^2=4 \\ 2x-y=1 \end{gathered}$$

Identify the graph and check the possible solution 

The graph of two equations

Both the graph intersects at

(0,-1) and (0.941, 0.882)

Solve the second equation for y 

$$\begin{gathered} 2x-y=1 \\ 2x-1=y \\ y=2x-1 \end{gathered}$$

Substitute it in first equation and solve for x

$$\begin{gathered} x^2+4{\left(2x-1\right)}^2=4 \\ 17x^2-16x+4=4 \\ 17x^2-16x=0 \\ x(17x-16)=0 \\ x=0 \\ 17x-16=0 \\ 17x=16 \\ x=\frac{16}{17} \end{gathered}$$

Solve for y

Substitute the x value and find out y

$$\begin{gathered} y=2x-1 \\ x=0 \\ y=2\left(0\right)-1 \\ y=-1 \\ x=\frac{16}{17} \\ y=2\left(\frac{16}{17}\right)-1 \\ y=\frac{15}{17} \end{gathered}$$

Solution is

$$\begin{gathered} (0,-1) \\ \left(\frac{16}{17},\frac{15}{17}\right) \end{gathered}$$

Check the solution

Substitute each ordered pair and verify the solution

$$\begin{gathered} (0,-1) \\ 2x-y=1 \\ 2\left(0\right)-(-1)=1,True \\ {x}^2+4y^2=4 \\ {\left(0\right)}^2+4{(-1)}^2=4,True \\ \left(\frac{16}{17},\frac{15}{17}\right) \\ 2x-y=1 \\ 2\left(\frac{16}{17}\right)-\left(\frac{15}{17}\right)=1,True \\ {x}^2+4y^2=4 \\ {\left(\frac{16}{17}\right)}^2+4{\left(\frac{15}{17}\right)}^2=4,True \end{gathered}$$