I have this systems of equation and I'm suppose to solve it using substitution.
2a-b=17
3a+4b=-13
I have the first part of the answer: a=(81/5). Then, I plug it into the other equation: 2a-b=17, which becomes 2(81/5)-b=17.
So, I get (162/5)-b=17. This is the part where I'm completely lost. Do I subtract (162/5) from both sides, or do I multiply both sides by 5 and then subtract 162?

Question

I have this systems of equation and I'm suppose to solve it using substitution.
2a-b=17
3a+4b=-13
I have the first part of the answer: a=(81/5). Then, I plug it into the other equation: 2a-b=17, which becomes 2(81/5)-b=17. 
So, I get (162/5)-b=17. This is the part where I'm completely lost. Do I subtract (162/5) from both sides, or do I multiply both sides by 5 and then subtract 162?

anonymous · Answer

you just need to move (162/5) from left to the right
so, become -b=17-(162/5)
and then -b= -(77/5)
so, b will be  77/5. and done