Question

In Problems $21$and $22$, solve each system of equations.

$\left\{\begin{array}{r}3x^2+y^2=12\\ y^2=9x\end{array}\right.$

Short answer

Solution of the system of equations is$(1,\pm 3)$

Step-by-step solution

Step 1. Given information

A system of equations is given as

$\left\{\begin{array}{r}3x^2+y^2=12\\ y^2=9x\end{array}\right.$

Step 2. Solving for x

We will use substitution method,

Divide the equation $3x^2+9x=12$by $3$,

$x^2+3x-4=0$

Now, solve as quadratic equation,

$\begin{array}{r}{x}^2-x+4x-4=0\\ x(x-1)+4(x-1)=0\\ (x+4)(x-1)=0\end{array}$

Now xhas two values but according to the system of equations xcannot be negative.

Therefore xmust be one.

Step 3. Solve for y

Now substitute the values of x,

$$\begin{gathered} y^2=9\left(1\right) \\ y=\pm 3 \end{gathered}$$