SOLUTION: The instructions were to graph this equation and create a table. This is the solution I came up with, but it is wrong. Where did I make my mistakes? Problem: 4(x+2)=y+5 4(1)

Question 6077: The instructions were to graph this equation and create a table. This is the solution I came up with, but it is wrong. Where did I make my mistakes?
Problem:
4(x+2)=y+5
4(1)+8=y+5
4+8=12+5 x=1; y=12
4(0)+8=y+5
8=8+5 x=0; y=8

4(2)+8=y+5
8+8=16+5 x=2; y=16
Thank You,
TSJ

Answer by Abbey(339) (Show Source): You can put this solution on YOUR website!
It is easier to avoid mistakes by putting equations into y=mx+b format first., but you were okay for the first step:
4(x+2)=y+5
4(1)+8=y+5
Here is where you made your mistake:
4+8=12+5 x=1; y=12
This should have been x=1,y=7, because
4+8=y+5
12=y+5
subtract 5 from both sides:
7=y so the point would be (1,7)
same problem on the next equation:
you say that
4(0)+8=y+5, so:
0+8=y+5
8=y+5
subtract 5 from both sides:
3=y, so the point would be (0,3)
third problem, same issue, you are missing a step:
don't substitute in for y, until you have consolidated all like terms:
4(2)+8=y+5
8+8=y+5
16=y+5
11=y the point is (2,11)

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