Question

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

$\left\{\begin{array}{l}{x}^2+4y^2=4\\ y=\frac{1}{2}x-1\end{array}\right.$

Short answer

The solutions of the system are$(0,-1),(2,0)$.

Step-by-step solution

Step 1. Given Information

The given system of equations is

$\left\{\begin{array}{l}{x}^2+4y^2=4\\ y=\frac{1}{2}x-1\end{array}\right.$

Step 2. Substitution

• Substitute $y=\frac{1}{2}x-1$into $x^2+4y^2=4$and simplify.

$\begin{array}{rcl}{x}^2+4{\left(\frac{1}{2}x-1\right)}^2& =& 4\\ x^2+x^2+4-4x& =& 4\\ 2x^2-4x& =& 0\\ 2x(x-2)& =& 0\end{array}$

• So, the value ofxis$x=0,2$.

Step 3. Solve for y

• Substitute $x=0$into $y=\frac{1}{2}x-1$ and solve for y.

$\begin{array}{rcl}y& =& \frac{1}{2}\times 0-1\\ & =& -1\end{array}$

• Now, substitute $x=2$into $y=\frac{1}{2}x-1$ and solve for y.

$\begin{array}{rcl}y& =& \frac{1}{2}\times 2-1\\ & =& 1-1\\ & =& 0\end{array}$

• So, the solutions of the system are$(0,-1),(2,0)$.