How do i solve fro r and q?
9r+q=13
3r+2q=-4

One way to solve the system of equations would be to solve for q (in terms of r) in the first equation to get:
9r + q = 13
q = 13 - 9r

Then, take the value for q you just found and insert it into the second equation:
3r + 2(13-9r) = -4

Then simplify and solve:
3r + 26 - 18r = -4
-15r = -30
r = 2

Now that you have the value for r, plug it into either of the original equations and solve for q:
9r + q = 13
9(2) + q =13
18 + q = 13
q = -5

r = 2; q = -5

Here are the steps in order:
First, you want to choose a variable that is not being used (with a number). Such as:
9r+q=13
-q      -q
q=-9r+13
Next, after you put that together (hint, if the variable was posotive to begin with, it changes the other variable to negative. If it was negative to begin with, it goes to positive) you put it into the other equation like this:
3r+2(-9r+13)=-4
As you can see, I placed the first equation into parenthases where another 'q' was. What you do now, is multiply is all.
3r-18r+26=-4
You want to get rid of the 26 now by putting in onto the other side of the equal bars.
3r-18r+26-26=-4-26
And now you put it all together, then divide.
-15r= -30
-15    -15
r=2
There you have r. Now you can move onto q.
9(2)+q=13
18+q=13
-18+18+q=13-18
q=-5
If you would like to verrify the answer, all you need to do is put them in the places of the variables.
3r+2q=-4
3(2)+2(-5)=-4
6-10=-4
-4=-4
There you have it, hope this helps! Btw, I wasn't sure if wither r=x and q=y or r=y and q=x, so I didn't put the answers in the brackets. Here are the options incase you forget though:
{(2,-5)}
or
{(-5,2)}
GOOD LUCK :D

Idon’t exactly understand but i think dbc=rst