# Math help: 2x+6y=300,3x+5y=350, what is x and y?

I have no idea how to do problems like these, so if you can please give me the steps on how to do this also? :(

Relevance
• iceman
Lv 7
9 years ago

2x+6y=300

3x+5y=350

-6x-18y=-900

6x+10y= 700

-8y = - 200

y = 25

x = 75

• Anonymous
9 years ago

set them up right on top of eachother so that all the like terms are directly over the other:

2x + 6y = 300

3x + 5y = 350

First you want to get rid of one of the variables. You can pick either one; I'm going to pick x.

You will get rid of x by multiply the top and bottom by a number that will cancel eachother out. (for example, you'd want x to be 2 on the top and -2 on the bottom because 2-2 = 0)

So, multiply the entire top function by -3 because the coefficent of x on the bottom is 3: -3(2x + 6y = 300) = -6x - 18y = -900

Then, multiply the entire bottom function by 2. 2(3x + 5y = 350) = 6x + 10y = 700

Now set them up on top of each other again:

-6x - 18y = -900

6x + 10y = 700

Now you can add the like terms together from top to bottom.

-6x + 6x = 0 ; -18y + 10y = -8y ; -900 + 700 = -200

you are left with -8y = -200

now solve for y. -200/-8 = 25

y = 25

Now, to find x, all you do is plug in 25 for one of the given functions.

2x + 6y = 300

2x + 6(25) = 300

2x + 150 = 300

2x = 150

x = 150/2

x = 75

Now plug in the x and y values you got to check your work.

2(75) + 6(25) = 300

it does work so your answers are

x = 75 y = 25

• 9 years ago

You can use substitution or elimination. Substitution would be easiest in this problem.

You need to get one variable alone on one side of one of the equations. I will use the first one because it's easier.

2x + 6y = 300

x+3y=150

x=150-3y

Then plug in what you get for x, into the other equation where x is.

3x+5y=350

3(150-3y) + 5y = 350

450 - 9y + 5y = 350

450 - 4y= 350

-4y = -100

y=25

Then you plug in y, to find x.

2x + 6(25) = 300

2x + 150 = 300

2x = 150

x= 75

x=75 and y=25

You can plug both x and y into each equation and it should work out.

Hope this helps :) :) :)