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=-(2)/(5)\text{ or }x=-2.`

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

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

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

`(-(2)/(5),-(11)/(5))\text{ and }(-2,1)\text{.}`

Answer:

`(-(2)/(5),-(11)/(5))\text{ and }(-2,1)\text{.}`