## Related questions with answers

a. Calculate the solution set and

b. Demonstrate that your solutions are right by drawing the graphs, preferably by computer.

$x^2+2 y^2=33$\

$x^2+y^2+2 x=19$

Solution

VerifiedWe have to solve the system

$\begin{aligned} x^2+2y^2 &= 33\\ x^2+y^2+2x &= 19 \end{aligned}$

If we multiply the second equation by $-2$, we get

$-2x^2-2y^2-4x = -38$

Adding this equation to the first one, $2y^2$ cancels and we get

$\begin{aligned} x^2-2x^2-4x &= 33-38 && \text{write the equation}\\ -x^2-4x &= -5 && \text{subtract }x^2-2x^2\text{ and }33-38\\ -x^2-4x+5 &= 0 && \text{transfer }-5\text{ left}\\ x^2+4x-5 &= 0 && \text{multiply both sides by }-1\\ x^2+5x-x-5 &=0 && \text{write }4x\text{ as }5x-x\\ x(x+5)-(x+5) &= 0 && \text{extract }x\text{ and }-1\\ (x+5)(x-1) &= 0 && \text{extract }x+5 \end{aligned}$

From $x+5=0$ we get a solution $x_1=-5$ and from $x-1=0$ we get a solution $x_2=1$.

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