Question

How would I solve this? I’m having a little trouble figuring out how to use matrix and elimination

Answer 1

To solve the given system of equations we will use the substitution method.

Subtracting 2x from the first equation we get:

$\begin{gathered} 2x+y-2x=-3-2x, \\ y=-3-2x\text{.} \end{gathered}$

Substituting the above equation in the second one we get:

$x^2+(-3-2x)^2=5.$

Simplifying the above equation we get:

$\begin{gathered} x^2+(-3)^2+2\cdot(-3)(-2x)+(-2x^2)=5, \\ x^2+9+12x+4x^2=5 \end{gathered}$

Adding like terms we get:

$5x^2+12x+9=5.$

Subtracting 5 from the above equation we get:

$\begin{gathered} 5x^2+12x+9-5=5-5, \\ 5x^2+12x+4=0. \end{gathered}$

Now, notice that:

$5x^2+12x+4=(5x+2)(x+2)\text{.}$

Therefore:

$(5x+2)(x+2)=0.$

Now, we know that:

$a\cdot b=0\text{ if and only if }a=0\text{ or }b=0.$

Therefore:

$(5x+2)(x+2)=0\text{ if and only if }5x+2=0\text{ or }x+2=0$

Then:

$x=-\frac{2}{5}\text{ or }x=-2.$

Finally, substituting the above result in y=-3-2x we get:

$\begin{gathered} y=-3-2(-\frac{2}{5})=-3+\frac{4}{5}=-\frac{11}{5} \\ or \\ y=-3-2(-2)=-3+4=1. \end{gathered}$

Therefore, the solutions of the given system of equations are:

$(-\frac{2}{5},-\frac{11}{5})\text{ and }(-2,1)\text{.}$

Answer:

$(-\frac{2}{5},-\frac{11}{5})\text{ and }(-2,1)\text{.}$