How do i go about solving the following system of equations by elimination?

r + 2s + 1 = 0 and r + 5s + 28 = 0

So,

r + 2s + 1 = 0

r + 5s + 28 = 0

There are two types of elimination: Elimination by Addition and Elimination by Substitution.

I will use Elimination by Addition.

Multiply both sides of the first equation by - 1.

-r - 2s - 1 = 0

Now add the two equations.

0r + 3s + 27 = 0

3s + 27 = 0

Now solve for s.

Subtract 27 from both sides.

3s = - 27

Divide both sides by 3.

s = - 9

Now substitute - 9 for s in the first original equation.

r + 2s + 1 = 0

r + 2 (-9) + 1 = 0

Multiply.

r - 18 + 1 = 0

Collect Like Terms.

r - 17 = 0

Add 17 to both sides.

r = 17

Check.

17 + 2 (-9) + 1 = 0

17 + - 18 + 1 = 0

17 - 18 + 1 = 0

-1 + 1 = 0

0 = 0 This checks.

Check in the second equation.

17 + 5 (-9) + 28 = 0

17 + - 45 + 28 = 0

17 - 45 + 28 = 0

-28 + 28 = 0

0 = 0 This also checks.

S = { (17,-9) }

Subtract the first equation from the second to get:

3s+27=0

Subtract 27 from both sides to get:

3s=-27

Divide by 3 to get:

s=-9

Substitute back into the first equation:

r+2 (-9) + 1=0

r-18+1=0

r-17=0

Adding 17, we see that r=17. Thus, our solution is r=17 and s=-9.