Answer :
When using substitution, you have to make sure only one variable is on one side of the equation in this case, the easiest way to do this would be to isolate s in the equation 2r+s=11.
When you do this, you should get:
s=-2r+11
Now you can plug in the value of s into the equation 6r-2s=-2. You should come up with:
6r-2(-2r+11)=-2
Now use the distributive property and solve for r:
6r+4r-22=-2
10r-22=-2
10r=10
r=1
Now that you have found the value of r, you can plug it into one of the first equations to solve for s:
2(1)+s=11
2+s=11
s=9
So the solution to the system is r=1 and s=9.
Hope I helped! :)
When you do this, you should get:
s=-2r+11
Now you can plug in the value of s into the equation 6r-2s=-2. You should come up with:
6r-2(-2r+11)=-2
Now use the distributive property and solve for r:
6r+4r-22=-2
10r-22=-2
10r=10
r=1
Now that you have found the value of r, you can plug it into one of the first equations to solve for s:
2(1)+s=11
2+s=11
s=9
So the solution to the system is r=1 and s=9.
Hope I helped! :)
