HOW do i do this math??

Question

I have 2 problems that I don't understand. You have to solve these by doing the elimination method. I don't see how you can get one variable gone. please help!

4x+2y=0
9x-3y=-30


3x+2y=-11
5x+5y=-20

you don't have to solve them if you don't want to, just explain, plese. i completely forgot!

Answer 1

you have to find #'s to multiply each equation by that will make one of the variables coeffients the same, so that u can add/subtract them and eliminate.

for example, #1:

4x + 2y = 0
9x - 3y = -30

you can multiply the top one by 3 (so that the y varable will be 6)
you can multiply the bottom one by 2 (so the y variable is also 6)

3(4x+2y)= 0 (2)
2(9x-3y)=-30(2)

12x + 6y = 0
18x - 6y = -60 now Add the two to get rid of the y
__________
30 x +0y = -60

30x = -60

x = -2 now plug it in and check the answers. :)


2nd one, u have to do the same. the easiest way is again to get rid of y:

3x + 2y = -11
5x + 5y = -20

multiply the top one by 5 to make y coefficent 10
multipy the bottom one by 2 to make y coefficent also 10

5(3x+2y) = -11(5)
2(5x+5y)= = -20(2)

15x + 10y = -55
10x + 10y = -40 now u have to Subtract them to get rid of y!


15x + 10y = -55
-10x -10y = +40
_____________
5x +0y = -15

5x = -15

x = -3


tada!! :)

Answer 2

In order to get one variable gone you have to multiply one (or both) of the variables by a constant to give one of the variables the same coefficient.

3 times (4x+2y=0) is
12x+6y=0

2 times (9x-3y=30) is
18x -6y=60

Therefore making your equations:
12x+6y=0
18x -6y=60

Add:
30x =60 plus 6y minus 6y eliminates itself

Divide by 30
x=2

Input the 2 for x in one of your equations and you can find the y. Try it with the second problem to make sure you understand it.

Hope this helps

Answer 3

I'll explain the first one, hopefully it will help.

Take the 9x-3y=-30.

Move the 9x over to the -30 side by subtractin 9x from both sides. The problem should now read -3y=-30-9x.

Then, you divide both sides by -3 [-3y/-3=(-30/-3)-(9x/-3).

The problem now reads y=10-(-3x), or y=10+3x.

Since you now have a value for y, you can insert it into either equation in place of the letter y (10+3x), which will leave you with only x's to find a numerical answer for x. Then, place that # into the equation in place of x, and find the numerical equation for y.

Answer 4

At first equation you must divide the first phrase by 2 to make it : 2x + y = 0 and then second by 3 to make it 3x -y = -10 in this two phrase you can omit -y and y when you plus two phrases if you solve it you see that : x = -2 and y = 4 .
In second equation you must divide second one by 5 to make it x +y = - 4 and then multiply second phrase to -2 to make it -2x -2y = 8 and then plus two phrase and omit -2y and 2y then you see that :
x = -3 and y = -1

Answer 5

(if I don't do this right I'm sorry its been a while) in the first problem if you take
(4x+2y=0)(-3)
(9x-3y=-30)(-2)
then you get
-12x-5y=0
-18x+5y=60
________________(combine the two)
-30x=60
so x=-2
then to figure y
4(-2)+2y=0
-8
so y is -4

and to check
9(-2)-3(-4)=-30
-18 + -12=-30

Answer 6

1. 4x+2y =0 (xby3) 12x+6y=0
9x-3y=-30(xby2) 18x-6y=-60
Add 2 eqs...... 30x=-60 x=-2, y=4

2. x first equation by 5, 2nd one by 2, you do the rest

Answer 7

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Answer 8

multiply 1st equationboth l.h.s and r.h.s by 9 and
2nd equation by 4.....

then the new two equations are formed..i.e 3rd and 4th..
36x+18y=0..............3rd

36x-12y=120.........4th


subtract 3rd from 4th....so that one variable get removed and remaining one you can find out...

30y=-120

y=-40

then put the value of that variable in equtoion 1st or 2nd to get the other variable value.....

4x-80=0
x=20
answer =x=20,.....y=-40

similarly the second problem....got it