Algebra 2 Elimination Method.?

Question

Summer Homework and I don't have a clue about Elimination Method.

-4x+y=-12
4x+2y=6

5x+2y=12
-6x-2y=-14

5x+4y=12
7x-6y=40

5m+2n=-8
4m+3n=2

All of these don't NEED to be solved. I just need an example to go by for the other three problems. If you actually do all four, you'll definitely get the 10 points.

Answer 1

Ok
the first one is easy
when you add -4 and 4 you get 0 right
and then you add the rest
so you get
3y=-6
y=-2
and x=5/2

2. this ones the same thing but with y
so it ends up being -x=-2
so x=2 and you plug in x in one equation and solve for y
so y=1

3. this one is a little tougher
you have to multiply each equation by a different number to get the 5x and 7x opposites
for ex. 35 and -35
so everything in the first equation is multiplied by 7
and everything in the second is multiplied by -5
so they become
35x+28y=84
-35x+30y=-200
now you add the equations together
so you get 58y=-116
y=2
and x=1/2

4. this one is the same as the last one
you multiply the first one by 4 and the second by -5
so they become
20m+8n=-32
-20m-15n=-10
then you add them together
-7n=22
n= -22/7

Answer 2

OK, the thrid and fourth ones here ae the harder ones so we will work with those. As you can see in the first two, if you add the parts vertically as is, you will get (-4x +4x = 0) and in the next one (2y+-2y=0). This is your goal is to find terms that you can add together to get 0. As you can see in the third and fourth ones, if you add them together you will NOT get 0 anywhere, so ere is what you do.

For #3, pick a term to eliminate (y or x). I will pick y for no particualr reason other than it si a coller letter. You must find a greater common factor for 4 and 6. How's 12? So what do you multiply 4, and what do you multiply 6 by to get 12? 3 and 2. The whole first equation must be multipled by 3 which gives (3(5x+4y=12) = 15x+12y=36). Do the same for the next equation, but this time use 2 because remember we are trying to make the y's like terms, and we decided that the like term would be 12. YOu should them get 14x-12y=80. Readjusted equations become: 15x+12y=36,
14x-12y=80
Add vertically to get 29x+0y=116
And simply solve for x. x=4
PLug x back into one of the original equations.
5(4)+4y=12
Solve for y. y= -2

There you go. Remembe teh key is to get similar terms that are opposites (positive and negative) and add tehm together to make one of the terms 0 : )

#4, you see you will not get opposites by looking for a GCF. They will both be positive. Let's elimiate the y's. GCF of 3 and 2? I like 6. Since they are both positive, why don't you try SUBTRACTING the two equations vertically. Then one of the terms will cancel out. I hope this helps.

Answer 3

elimination by addition Method

-4x + y = - 12- - - - - Equation 1
4x + 2y = 6- - - - - -Equation 2
- - - - - - - - - -
3y = - 6

Divide both sidea of the equation by 3

3y / 3 = - 6 / 3

y = - 6 / 3

y = - 2

Insert the y value into equation 1
- - - - - - - - - - -

- 4x + y = - 12

- 4x + ( - 2) = - 12

Remove parenthesis

- 4x - 2 = - 12

Transpose 2

- 4x - 2 + 2 = - 12 + 2

- 4x = - 10

Divide both sides of the equation by - 4

- 4x / - 4 = - 10 / - 4

x = - 10 / - 4

x = 10/4

simplify

x = 5/2

Insert the x value into equation 1
- - - - - - - - -

Check for equation 1

- 4x + y = - 12

- 4(5/2) + (- 2) = - 12

- 20/2 - 2 = - 12

- 10 - 2 = - 12

- 12 = - 12

- - - - - - - - - -

Check for equation 2

4x + 2y = 6

4(5/2) + 2(- 2) = 6

20/2 + ( - 4) = 6

10 - 4 = 6

6 = 6
- - - - - - - - -

Both equations balance

The solution set { 5/2, - 2 }

- - - - - - - - --s-

Answer 4

For the first one, you add the two equations together to where the -4x and the 4x cancel out. That will they give you (y+2y)=-12+6 whic is 3y=-6
The answer will then be y=-2. Use that y to solve for x using either equation.

For the second one, you add the two equations together and the y values will cancel out.

The third and fourth ones, you need to multiply by a common factor to get the equations to cancel. For instance, in the third set, multiply the top equation by 4 and the bottom by 2. That will give you 15x+12y=36 and 14x-12y=80. Now you add them together and the ys will cancel. Then solve for x, use the x and solve for y.

Do something similar for the fourth set of equations.

Hope this helps. If you need more, let me know.

Answer 5

The first one and second one are pretty convenient, as they already have the necessary parts of the equation to eliminate. All you have to do on the top two are add the two equations together to eliminate one variable and get one equation with one variable, solve for that variable, and plug it into one of the original equations. As for the bottom two, you actually have to make modifications to one of the equations to make the parts match up for elimination. For an example, I'll do the last one.

Step 1) Modify the equations to make the parts match up for elimination.

3(5m + 2n) = 3(-8)
-2(4m + 3n) = -2(2)

which simplifies to:

15m + 6n = -24
-8m - 6n = -4

Step 2) Add the two equations together to eliminate one variable.

15m + 6n = -24
+ -8m - 6n = -4

You get:

7m + 0n = -28 or 7m = -28

Step 3) Solve for m algebraically.

7m = -28
m = -4

Step 4) Put the solved result of m back into one of the original equations and solve for n.

5(-4) + 2n = -8
-20 + 2n = -8
2n = 12
n = 6

So, your answer to the fourth problem is m = -4 and n = 6

Answer 6

a million) -4x + 4y = -8 2) x - 4y = - 7 get rid of y with the aid of including a million) and a pair of) -3x = - 15 x = +5 substitute in the two a million) or 2) to be sure the fee of y. in a million) -4*5 + 4y = -8 -20 + 4y = -8 4y = 12 y = 3

Answer 7

what are you solving for i need to know in order to get the answer