Linear Relations $y=mx+b$ and graphing lines Consider the linear relation $2x - 3y = 12$.
a) Find $y$ when $x = 3$. 
b) Solve the original equation in the question for $y$. 
c) Use your equation from b) to find y when $x = 3$. 
d) Do you prefer finding the answer using step a), or do you prefer finding the answer by combining steps b) and c)? Explain 
How do I do this?!

Part a. To find $y$ when $x=3$ we begin by substituting in the value $3$ where we see $x$ in the equation: $$2x-3y=12 \Rightarrow \\2\cdot3 - 3y = 12 \Rightarrow \\6-3y = 12.$$ Next we simplify, first by moving the $6$ over by subtracting it on both sides, then by dividing out the coefficient $-3$. That is: $$6-3y=12 \Rightarrow \\ -3y = 12-6 \Rightarrow \\-3y = 6 \Rightarrow \\y = \frac{6}{-3} = -2.$$
Part b. We do the same thing, but without substituting first: $$2x - 3y = 12 \Rightarrow \\ -3y = 12 - 2x \Rightarrow \\ y = \frac{12}{-3} - \frac{2}{-3}x = \frac{2}{3}x - 4.$$
Part c. Simply plug 3 into the equation you just got, that is: $$y = \frac{2}{3}x - 4 \Rightarrow \\ y = \left(\frac{2}{3}\cdot 3\right) - 4 = 2 - 4 = -2.$$
Part d. Using b and c is typically prefered, especially if you need to check the value at multiple points, that way you aren't doing the same work over and over!
