Question

Use elimination I NEED IT HURRY

Answer 1

Answer:

(5,-2)

Step-by-step explanation:

The given system is:

$3x+2y=11$

$2x+6y=-2$

Both equations are in the same form. That form being standard form, $ax+by=c$.

We need one of the variable columns to contain opposite terms (we will add if this is the case) or same terms (we will subtract if this is the case).

To do this we will either need to multiply both equations or one equation.

I really like the column with the variable $y$ right now because the least common multiple of 2 and 6 is 6 and 6 is already a coefficient of one of the $y$ variables.

So This means I will need to either choose to multiply the first equation by 3 or -3. It is your choice or mine in this case.

I'm going to choose to multiply the first equation by -3.

This will give me my system as:

$-9x-6y=-33$

$2x+6y=-2$

I will now add the equations. This is method is called elimination because in this step one of the variables get eliminated.

$-7x+0y=-35$

$-7x+0=-35$

$-7x=-35$

$x=\frac{-35}{-7}$

$x=5$

So now we can find the $y-$ coordinate using of the initial equations along with $x=5$.

I choose $3x+2y=11$ along with the $x=5$:

$3(5)+2y=11$

$15+2y=11$

$2y=11-15$

$2y=-4$

$y=\frac{-4}{2}$

$y=-2$

So the solution is (5,-2).

Let's confirm the answer through a check.

I will plug our "solution" into both equations.

If both equations render true, then the "solution" is confirmed and I will refer to it as just a solution without the quotation.

$3x+2y=11$ with $(x,y)=(5,-2)$:

$3(5)+2(-2)=11$

$15+(-4)=11$

$11=11$ is true.

$2x+6y=-2$ with $(x,y)=(5,-2)$:

$2(5)+6(-2)=-2$

$10+(-12)=-2$

$-2=-2$ is true.

Since both equations ended up being true, (5,-2) is the solution of the system.