Chapter 11, Problem 21CT

To determine

The solution of the system of equations,

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

Answer to Problem 21CT

Solution:

The solution of system of equation $\{\begin{array}{l}3x^2+y^2=12\\ y^2=9x\end{array}$ are $x=1,y=\pm 3$ or $\left(1,3\right),\left(1,-3\right)$.

Explanation of Solution

Given information:

The system of equations, $\{\begin{array}{l}3x^2+y^2=12\\ y^2=9x\end{array}$.

Explanation:

Consider the system of equation,

$3x^2+y^2=12$ … (1)

$y^2=9x$… (2)

Substitute the value of $y^2=9x$ in the equation (1),

$\Rightarrow 3x^2+9x=12$

$\Rightarrow 3x^2+9x-12=0$

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

$\Rightarrow \left(x-1\right)\left(x+4\right)=0$

By using zero product property,

$\Rightarrow x-1=0$ or $x+4=0$

$\Rightarrow x=1$ or $x=-4$

Now substitute $x=1$ in equation (2),

$\Rightarrow y^2=9\left(1\right)=9$

Taking square root on both sides,

$\Rightarrow y=\pm 3$

Now substitute $x=-4$ in equation (2),

$\Rightarrow y^2=9\left(-4\right)=-36$

$\Rightarrow y=\sqrt{-36}$

This not a real number.

Thus, the solution of the system is $x=1,y=\pm 3$.

Hence, the solution of system of equation $\{\begin{array}{l}3x+y^2=12\\ y^2=9x\end{array}$ are $x=1,y=\pm 3$ or $\left(1,3\right),\left(1,-3\right)$.