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2007-08-31 02:31:43 · 7 answers · asked by Jevon H 1 in Science & Mathematics Mathematics

7 answers

Hi Jevon,

Let's solve this by eliminating the 'x'. This is how to do it :)

2x-y=10 ...[1]

x+3y=-2

2*(x+3y) = -2 * 2
2x + 6y = -4 ... [2]


Now, let's substract [1] from [2], so that
2x - y = 10
2x+6y = -4
----------------- -
2x - 2x - y - 6y = 10 - (-4)
0 - 7y = 14
-7y = 14
y = 14 / -7
y = -2

Now, let's insert the value of y into [1]
2x - y = 10
2x - (-2) = 10
2x + 2 = 10
2x = 8
x = 4

==================
Answer: x = 4, y = -2
==================

Hope this helps :)

Kind regard,
Ryonn

2007-08-31 02:38:02 · answer #1 · answered by Mr. Math Contributor 4 · 0 0

First,

x = -3y - 2 (substitute into other equation)

2 (-3y - 2) - y = 10

-6y - 4 - y = 10

-7y = 14

y = 14 / -7 = -2

We now just put in the value of -2 for y....

2x - y = 10

2x - - 2 = 10
2x + 2 = 10
2x = 8
x = 4

To check that x = 4 and y = -2, we put into the first equation.....

x + 3y = -2

4 + (3 * -2) = -2

4 - 6 = -2 (Which works, so we know we are correct)

2007-08-31 02:40:21 · answer #2 · answered by Doctor Q 6 · 0 0

I suggest you solve by adding. Take you two systems and multiply one of them by a number so you have the same x or y. I suggest y in this case, because one is positive and one is negative. It's just a little easier.

So you have:
x + 3y = -2
2x - y = 10

Multiply the second equation by 3 to get:

x + 3y = -2
6x - 3y = 30

Now, add them together to cancel out the y term. You get:

7x + 0y = 28

Solve for x:
x = 4

Now you can plug x into one of the equations above to solve for y. I suggest plugging it into both equations. That way you will catch it if you made an error in solving for x. So you have:

4 + 3y = -2
3y = -6
y = -2

and

2*4 - y = 10
-y = 2
y = -2

So, you now know that x = 4 and y = -2.

2007-08-31 02:37:37 · answer #3 · answered by Rev Kev 5 · 0 0

This Site Might Help You.

RE:
Solve. x+3y=-2, 2x-y=10. Show work?

2015-08-18 11:25:30 · answer #4 · answered by Josey 1 · 0 0

x+3y=-2
x=-2-3y

2x-y=10
y=2x-10

y=2(-2-3y)-10
y=-4-6y-10
y+6y=-4-10
7y=-14
y=-14/7
y=-2

x=-2-3y
x=-2-3(-2)
x=-2+6
x=4

To check:
x+3y=-2
4+3(-2)=-2
4-6=-2
-2=-2

2x-y=10
2(4)-(-2)=10
8+2=10
10=10

2007-08-31 03:57:43 · answer #5 · answered by yeyetavi 2 · 0 0

X 3y 2

2016-11-02 06:11:12 · answer #6 · answered by ? 4 · 0 0

1. x+3y=-2
2. 2x-y=10

from 1.
x+3y=-2
x= -2-3y

in 2.
2x-y=10
2(-2-3y)-y=10
-4-6y-y=10
-7y=14
y= -2


x= -2-3y
x= -2-3(-2)
x=4

2007-08-31 02:38:24 · answer #7 · answered by Gilarh 3 · 0 0

x + 3y = -2
2x - y = 10
i.e
6x - 3y = 30
adding two we get,
7x = 28
i.e.
x = 4 and
y = -2

2007-08-31 02:44:29 · answer #8 · answered by Lordsham 1 · 0 0

x + 3y = - 2
6x - 3y = 30-----ADD

7x = 28
x = 4

4 + 3y = - 2
3y = - 6
y = - 2

x = 4 , y = - 2

2007-08-31 09:01:03 · answer #9 · answered by Como 7 · 0 0

It will do no good to have other people do your work.
I will tell you how but you must do the work yourself.
You have two equations with two unknowns this you must recognize.
Place one equation above the other such that the variables line up in a column and the values on the other side of the equal sign also line up in a column.
The first thing that you should be able two recognize is that even if you subtract one from the other both variables are still there so you must get rid of one variable.
Multiply one of the equations by a number such that a variable in each equation have the same coefficient.
It does not matter which one you select.
Now rewrite the new equations so the columns line up.
Since one variable has the same coefficient as the one above it you can subtract the equations, but if you need to change the signs of one equation multiply one equation by ( -1) and add.
Now you have only one variable, solve for that variable.
When you get the answer substitute that answer in for that variable in one of the original equations and solve for the other variable. Remember that what ever you do to the left of the equal sign you must do to the right. Good Luck I hope you understand now!

2007-08-31 03:01:50 · answer #10 · answered by ? 3 · 0 0

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