Chapter 11.6, Problem 27AYU

To determine

The solution of the system $\{\begin{array}{l}y=3x+2\\ 3x^2+y^2=4\end{array}$

Answer to Problem 27AYU

Solution:

The solutions are $\left(0,2\right)$ and $\left(-1,-1\right)$.

Explanation of Solution

Given information:

The system is $\{\begin{array}{l}y=3x+2\\ 3x^2+y^2=4\end{array}$.

Explanation:

Here, the system contains two variable $x$ and $y$.

The system is $\{\begin{array}{l}y=3x+2\left(1\right)\\ 3x^2+y^2=4\left(2\right)\end{array}$.

Now from the equation (1), Substitute the value of $y=3x+2$ in the equation (2)

This gives,

$3x^2+y^2=4$

$\Rightarrow 3x^2+{\left(3x+2\right)}^2=4$

$\Rightarrow 3x^2+9x^2+12x+4=4$

$\Rightarrow 12x^2+12x=0$

$\Rightarrow x^2+x=0$

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

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

$\Rightarrow x=0$ or $x=-1$.

Now, for $x=0$,

$y=3\left(0\right)+2=2$

Now, for $x=-1$,

$y=3\left(-1\right)+2=-3+2=-1$.

Thus, the solutions are $\left(0,2\right)$ and $\left(-1,-1\right)$.