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Question

# Where do the graphs of the equations $y=2 x+1$ and $2 x^2+y^2=11$ intersect?

Solution

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Given the system of equation,

\begin{aligned} y&=2x+1&&\text{Eq.1}\\ 2x^2+y^2&=11&&\text{Eq.2} \end{aligned}

Use substitution method to solve. Substitute $y=2x+1$ in second equation.

\begin{aligned} 2x^2+y^2&=11&&\text{Eq.2}\\ 2x^2+(2x+1)^2&=11\\ 2x^2+(4x^2+4x+1)&=11\\ 6x^2+4x+1+(-11)&=11+(-11)\\ 6x^2+4x-10&=0 \end{aligned}

Use the quadratic formula to get the roots.

\begin{aligned} x&=\dfrac{-(b)\pm\sqrt{b^2-4ac}}{2a}\\ &=\dfrac{-(4)\pm\sqrt{4^2-4(6)(-10)}}{2(6)}\\ &=\dfrac{-4\pm\sqrt{16+240}}{12}\\ &=\dfrac{-4\pm\sqrt{256}}{12}\\ \end{aligned}

Get the root.

\begin{aligned} x_1&=\dfrac{-4+\sqrt{1256}}{12}\\ &=\frac{-4+16}{12}\\ &=\frac{12}{12}\\ &\boxed{x_1=1}\\\\ x_2&=\dfrac{-4-\sqrt{256}}{12}\\ &=\frac{-4-16}{12}\\ &=\frac{-20}{12}\\ &\boxed{x_2=\frac{-20}{12}} \end{aligned}

Thus $x=1$ and $x=\dfrac{-20}{12}$. Use this to solve for $y$.

\begin{aligned} y&=2x+1&&\text{Eq.1}\\ y&=2(1)+1\\ &\boxed{y=3}\\\\ y&=2x+1&&\text{Eq.1}\\ y&=2\left(\frac{-20}{12}\right)+1\\ y&=\frac{-20+6}{6} &\boxed{y=-\frac{-14}{6}}\\\\ \end{aligned}

Hence the solutions are $(1,3)$ and $\left(-\dfrac{20}{12},-\dfrac{14}{6}\right)$.

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