5x + 6y = 18 solve for y

5x - 5x + 6y = 18 + 5x

6y = 18 + 5x

6y/6 = 18 + 5x/6

y = 3 + 5x

This has 2 variables, do I need to solve for x or will y=3+5x work out?

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- Jun 12th 2008, 06:16 PMmdudleyAm I doing This correctly?
5x + 6y = 18 solve for y

5x - 5x + 6y = 18 + 5x

6y = 18 + 5x

6y/6 = 18 + 5x/6

y = 3 + 5x

This has 2 variables, do I need to solve for x or will y=3+5x work out? - Jun 12th 2008, 06:43 PMMathnasium
When the problem says "solve for y," the intent is simply to get y by itself.

You're on the right track, but:

5x + 6y = 18

5x - 5x + 6y = 18 - 5x (here, you added 5x to the right hand side)

6y = 18 - 5x

6y/6 = (18-5x)/6

y = 3 - 5x/6 (When you divide by 6, you have to divide EACH TERM by 6 - this means you need to do 18/6 AND 5x/6)

So, you were on the right track, just a couple small things. - Jun 12th 2008, 07:07 PMmdudley
Thank you, a simple + or - can break me I now know this.

So when I plug in everything should it look like this?

5x + 6y = 18 solve for y

5x - 5x + 6y = 18 - 5x

6y = 18 - 5x

6y/6 = 18 + 5x/6

y = 3 - 5x/6

5x + 6(3 - 5x/6) = 18

5x + 18 - 30x/36 = 18

5x + 18 - 18 - 30x/36 = 18 -18

5x - 30x/6 = 0

30x/6 - 30x/6 = 0

x = 0

5(0) + 6y = 18

0 + 6y = 18

6y = 18

6y/6 = 18/6

y = 3

(0,3)

I will be so happy if I'm on the right trail. - Jun 12th 2008, 07:43 PMMathnasium
Not quite. When you plug it back in, you get:

5x + 6(3 - 5x/6) = 18

5x + 18 - 5x = 18 (Note that [tex]6 \times \frac{5x}{6} = \frac{30x}{6} = 5x[\math].

18 = 18.

This means you've done it right - if you ended up with something like 12 = 18, you'd know you made a mistake.

The problem is NOT asking you to find numbers for x and y. The only thing you have to do when it says "solve for y" is to get y by itself and everything else on the other side. Your final answer is:

$\displaystyle y = 3 - \frac{5x}{6}$.

That's it!

Note that this does describe a relationship between x and y, and (0,3) does satisfy this relationship. However, note that there are an infinite number of combinations of x and y that work. If you put in, say, 6 for x, you get:

$\displaystyle y = 3 - \frac{30}{6} = 3 - 5 = -2$

so you get the point (6, -2).

If you were to graph ALL the points (there are an infinite number of them) that satisfy your equation, you'd get a straight line.