Question

Solve the following system of equationsy = x^2 + 3y = x + 5y

Answer 1

First, let's solve the second equation for y:

`\begin{gathered} y=x+5y \\ 5y-y=-x \\ 4y=-x \\ y=-(x)/(4) \end{gathered}`

Now, let's use this value of y in the first equation:

`\begin{gathered} y=x^2+3 \\ -(x)/(4)=x^2+3 \\ x^2+(x)/(4)+3=0 \\ 4x^2+x+12=0 \end{gathered}`

Let's solve this quadratic equation using the quadratic formula:

`\begin{gathered} a=4,b=1,c=12 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x_1=\frac{-1+\sqrt[]{1-192}}{8}=\frac{-1+\sqrt[]{-191}}{8} \\ x_2=\frac{-1-\sqrt[]{-191}}{8} \end{gathered}`

Now, calculating y, we have:

`\begin{gathered} y_1=-(x_1)/(4)=\frac{1-\sqrt[]{-191}}{32} \\ y_2=(-x_2)/(4)=\frac{1+\sqrt[]{-191}}{32} \end{gathered}`