I know for you guys this question will seem EASY as heck but for me like I stated above, I'm just learning this stuff.

You should never let yourself feel bad because someone else knows more than you. "Easy" is incredibly relative, and everybody has to start somewhere. As long as you give it your best shot and make a genuine effort, that's really all we can ask for.

Question is a got to make table of values for the given equation

2x + 5y = 10 and the graph it, I just need to know how to get the values so I can put them in the table, I can graph pretty well (I Think) looked on youtube, google, got confusing so I found this site

Anyway, enough of that I suppose. On to the actual problem. Making a table of values is really nothing more than finding some values of

*x* and

*y* which satisfy the given equation. Let's do that now:

- x = 1, y = 1; 2x + 5y = 2(1) + 5(1) = 7 Oops... no good
- x = 1, y = 2; 2x + 5y = 2(1) + 5(2) = 12 Uh oh, not this one either
- x = 2, y = 1; 2x + 5y = 2(2) + 5(1) = 9 Dang it! Still no good
- x = 0, y = 2; 2x + 5y = 2(0) + 5(2) = 10 Oh, this one worked
- x = 5, y = 0; 2x + 5y = 2(5) + 5(0) = 10 Another winner!

Now, we

*could* just keep going along this route, but I'm sure you can see what this is really not a good tactic. We're essentially blindly guessing and hoping it works out. For this particular problem, we can just think about it a bit and that will help us make wiser guesses, but what if the problem was more complex? What if it were, say, \(\displaystyle 3.85x + \sqrt{y} = 57.2\)? Seeing what to do in that case is a lot more tricky. So, clearly, there

*has* to be a better way!

Indeed there is. You've probably seen similar problems before where you were asked to form a table of values for a nicer equation like \(\displaystyle y = 3x\). In that case, you'd just pick a few test values for

*x* and plug them in to find the corresponding value of

*y*:

- x = -2; y = 3(-2) = -6
- x = -1; y = 3(-1) = -3
- x = 0; y = 3(0) = 0
- x = 1; y = 3(1) = 3
- x = 2; y = 3(2) = 6

So, what if we could manipulate this problem to look like that? That'd be excellent. Well, you might remember from your studies that if you have an equation, you can perform an operation to one side of the equation, and so long as you also perform that same operation to the other side, the equality still holds. What if we worked with the given equation and subtracted 2x from both sides?

- 2x + 5y = 10
- 2x + 5y - 2x = 10 - 2x
- 2x - 2x+ 5y = 10 - 2x

At this point, the 2x terms cancel out and we're left with \(\displaystyle 5y = 10 - 2x\). We can even go one step further and divide both sides by 5 to get \(\displaystyle \dfrac{5y}{5} = \dfrac{10 - 2x}{5}\) or \(\displaystyle y = \dfrac{10 - 2x}{5}\)

Great! We've reduced a new type of problem that seemed mind-spinningly complex and just downright impossible into a type of problem we've seen and solved before. I'll leave the last little bit to you to finish up. Give it a go and see what you get.