Pre-Algebra help on problems?

Question

Hey-i am in pre-algebra but they are now practicing algebra on us and its challenging. can i get help? it supposed to be using elimination? whatever that means... hehe!
heres 3 of them but theres much more i get to do... =(

x+y=4
2x-y=5

x+y=4
2x+3y=7

2x+3y=13
3x-5y=10

thank you very much in advance. =).

Answer 1

x + y = 4- - - - - -Equation 1
2x - y = 5- - - - - Equation 2
- - - - - --
3x = 9

3x/3 = 9/3

x = 9/3

x = 3

Insert the x value into equation 1

- - - - - - - - - - -- - - - - - - - - - -

x + y = 4

3 + y = 4

3 + y - 3 = 4 - 3

y = 1

Insert the y value into equation 1

- - - - - - - - - - - - - - - - - - - - --

Check for equation 1

x + y = 4

3 + 1 = 4

4 - 4

- - - - - - - -

Check for equation 2

2x - y = 5

2(3) - 1 = 5

6 - 1 = 5

5 = 5

- - - - - - - -

x + y = 4- - - - - - -Equation 1
2x + 3y = 7- - - - - Equation 2
- - - - - - - -

Multiply equation 1 by - 3

x + y = 4

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

- 3x + (- 3y) = - 12

- 3x - 3y = - 12

- - - - - - - - -

- 3x - 3y = - 12
2x + 3y = 7
- - - - - - - - - - -

- x = - 5

- 1(- x) = - 1(- 5)

x = 5

Insert the x value into equation 1

- - - - - - - - - - - - - - - - - - - - --

x + y = 4

5 + y = 4

5 + y - 5 = 4 - 5

y = - 1

Insert the y value into equation 1

- - - - - - - - - - - - - - - - - - - - - -

Check for equation 1

x + y = 4

5 + (- 1) = 4
5 - 1 = 4

4 = 4

- - - - - - -

3x + 3y = 7

2(5) + 3(- 1) = 7

10 + (- 3) = 7

10 - 3 = 7

7 = 7

- - - - - - - s-

Answer 2

x+y=4
2x-y=5
x = 4 - y

2(4-y) -y = 5
8 - 2y -y = 5
-3y = 5 -8

-3y = -3
y = 1
x = 4 - 1
x = 3
Solution: (3, 1)
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x+y=4 (-2)
2x+3y=7

-2x - 2y = -8
2x + 3y = 7
---------------
y = -1

x + y = 4
x -1 = 4
x = 4 + 1
x = 5
Solution: (5, -1)
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I) (2x+3y=13
II) (3x-5y=10

(I) 2x + 3y = 13 & (II) 3x - 5y = 10
x = (13 -3y) over 2 & x = (10 + 5y) over 3
Comparison:
(13 -3y) : 2 = (10 + 5y) : 3
3(13-3y) = 2(10 + 5y)
39 - 9y = 20 + 10y
-9y -10y = 20 - 39
-19y = -19
y = 19/19 = 1

2x+3y=13
2x + 3(1) = 13
2x = 13 - 3
2x = 10
x = 10/2
x = 5
Solution: (5, 1)
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Answer 3

Using elimination to solve problems is easy to do. What you're going to be needing to do is to add the two equations given together to eliminate one of the variables, either x or y. Sometimes you will need to multiply one or both of the equations by a number (or numbers if you need to multiply both equations) to be able to do this.

Looking at the first problem you have there, if the equations were added together right now, the y variable would be gone, so you don't need to manipulate the equations at all. Adding them together gives you will give you a third equation:

x+y=4
2x-y=5
----------
3x=9 now you will need to divide both sides by 3 to find out what x is:

3x/3 = 9/3

x=3

Substitute x=3 back into one of the original equations to get the value of y:

x+y=4

3+y=4 solve for y by subtracting 3 form both sides:

3-3+y=4-3

y=1

So the answer to the fist problem is x=3, y=1.

The second problem involves some manipulation, since neither the x nor the y would be gone if you added them together the way they are now. So you will need to multiply one of the equations by either -2 (to make the x's cancel out) or -3 (to make the y's cancel). I generally always go with the smaller number, so I'm going to multiply the top equation by -2, which will give us:

-2x - 2y = -8

now we just need to add that to the second equation (nothing needs to be done to this one this time)

-2x-2y= -8
2x+3y=7
----------------
y= -1

Now we know what y is, we can sub it into either of the equations originally given to us to find the value of x:

x + y = 4

x - 1 = 4 add 1 to both sides to solve for x:

x -1 +1 = 4 +1

x=5

So the solution for this one is x=5, y = -1.

The final problem given will have to have both equations multiplied by some number so that a variable will cancel out. You can tell this is so by looking at the numbers before the variables, although it would be possible to multiply just one of them by a fraction to get them to cancel, I've found that very few students in pre-algebra really want to do that ;)

So, I'm just going to say we'll be canceling the x variable. Since we have a 2x and a 3x, we need to think about what number these both go into. (lowest common multiple, or lcm...sound familiar?) Both 2 and 3 go into 6, so we will need to be multiplying the top equation by 3 and the second by 2 to get both of them to have a 6x...but one of them needs to be negative, it really doesn't matter which, I'm going to make the bottom equation negative, therefore I'll be multiplying it by -2, not positive 2. This will give us the following equations:

6x +9y = 39
-6x+10y= -20 now add them together:
------------------
19y=19 divide both sides by 19 to get y alone:

19y/19=19/19

y=1

Sub y=1 back into one of the equations, it doesn't matter which one:

2x+3y=13

2x + 3(1)=13

2x+3=13 solve for x, first move the 3 over to the right side:

2x=10 now divide everything thru by 2 to get x alone:

x=5

So the answer to this last one is x=5, y=1.

I hope this helped, I know it was rather long-winded :)

Answer 4

Actually it's easier to use subtitution.. But don't worry..

x+y=4 |*2| 2x+2y=8
2x-y=5 |*1| 2x - y=5
_________________ -
3y=3
y=3/3=1

x+y=4
x =4-y
x =4-1
x =3


x +y =4 |*2| 2x+2y=8
2x+3y=7 |*1| 2x+3y=7
___________________ -
y = 1

x+y=4
x=4-y
x=4-1
x=3


2x+3y=13 |*3|6x+9y=39
3x -5y=10 |*2|6x-10y=20
____________________ -
19y=19
y=19/19=1

2x+3y=13
2x+3(1)=13
2x+3=13
2x=13-3
2x=10
x=10/2=5