# Can someone Please help me understand this? 2x + 5y = 10 and the graph it

#### MichaelGD3

##### New member
Hey guys, new here. My name is Michael, I could tell you my whole life story but won't lol.

I recently enrolled in GED glasses, and got 3 tests done and out the way, just one to go... MATH, which I completely know not that much about.

anyways got homework, don't want anyone to help me do it obviously. but I have no idea of how to do it, teacher was in a rush and I have till Monday to get 15 pages done,

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.

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

I would appreciate any help thanks

#### Dr.Peterson

##### Elite Member
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)
There are many ways to graph such a line. If you are required to make a table of values (a good way when you are just starting out), I would solve the equation for y, and then in principle you can choose any values you want for x in the table.

Some values will be more convenient than others, though. When you look at the equation, you'll see that x has to be divided by 5; so just to keep the numbers nice, you might use only multiples of 5 for x: -10,-5,0,5,10 or something like that.

I usually start with small numbers (0 is usually very easy!) and work up until y gets too big to want to graph.

#### Denis

##### Senior Member
If you assign say 3,2,1,0,-1,-2,-3 as the x value,
what do you get as corresponding y values?

#### ksdhart2

##### Full Member
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.

#### Jomo

##### Elite Member
I can graph pretty well (I Think)

I would appreciate any help thanks
Michael, welcome to the site. If you like, you can graph it and read off some points from the graph plus any points (or point) that you used to make the graph. How many points do you need?

Alternatively, you can let x=0 and find y, then let y=0 and find x. This will give you 2 points.

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#### HallsofIvy

##### Elite Member
It helps to know (from geometry) that "two points determine a line". If you mark two points you can lay a straight edge on those two points and immediately draw the line.

So how do you find two points? As others have said, pick any convenient value for x and calculate y. Pick another value for x and calculate that value of y. That works particularly well when the line is given as "y= ax+ b".

One simple method, when the line is given as "ax+ by= c" is: take x= 0 and calculate the value of y then take y= 0 and calculate the value of x.

For example, one of the problems is to graph the line given by 2x+ 5y= 10. If x= 0, that is 5y= 10. Divide both sides of the equation by 5 to get y= 2. One point on the graph is (0, 2). If y= 0, the equation is 2x= 10. Divided both sides of the equation by 2 to get x= 5. Another point on the graph is (5, 0). Mark those two points on the graph and draw the line through them.

#### Jomo

##### Elite Member
It helps to know (from geometry) that "two points determine a line". If you mark two points you can lay a straight edge on those two points and immediately draw the line.

So how do you find two points? As others have said, pick any convenient value for x and calculate y. Pick another value for x and calculate that value of y. That works particularly well when the line is given as "y= ax+ b".

One simple method, when the line is given as "ax+ by= c" is: take x= 0 and calculate the value of y then take y= 0 and calculate the value of x.

For example, one of the problems is to graph the line given by 2x+ 5y= 10. If x= 0, that is 5y= 10. Divide both sides of the equation by 5 to get y= 2. One point on the graph is (0, 2). If y= 0, the equation is 2x= 10. Divided both sides of the equation by 2 to get x= 5. Another point on the graph is (5, 0). Mark those two points on the graph and draw the line through them.
Prof Halls, I hate to disagree with you but I prefer to let students think and actually use what they know better. Mostly all students know multiplication better than division, so when it comes to solving 5y=10 I feel it is best to have the student ask themselves 5 times what equals 10 and easily get y=2 instead of dividing.

#### mmm4444bot

##### Super Moderator
Staff member
… I'm just learning this stuff.

Question is … make table of values for the given equation

2x + 5y = 10

and [then] graph it, I just need to know how to get the values …

I can graph pretty well …
Since you already know how to graph (x,y) points and you've been instructed to make the table first, I would use the suggestions in post #2.

If you're not sure how to solve the equation for y, let us know. :cool:

#### HallsofIvy

##### Elite Member
Prof Halls, I hate to disagree with you but I prefer to let students think and actually use what they know better. Mostly all students know multiplication better than division, so when it comes to solving 5y=10 I feel it is best to have the student ask themselves 5 times what equals 10 and easily get y=2 instead of dividing.
I was responding to a question about how to graph. I wasn't saying anything about how to do arithmetic. And isn't "asking themselves 5 times what equals 10" dividing?