SOLUTION: Hello, can you please assist me with Solving these problems "Using Elimination" in Algebra 2? Thank you very much 1. -2x + 3y = 25 -2x + 6y = 58 2. 8x + 13y = 179

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Question 1125259: Hello, can you please assist me with Solving these problems "Using Elimination" in Algebra 2? Thank you very much
1. -2x + 3y = 25
-2x + 6y = 58


2. 8x + 13y = 179
2x - 13y = -69


3. 2x + 7y = -7
5x + 7y = 14

4. 6x + 3y = 0
-3x + 3y = 9

5. 3x - 8y = 32
-x + 8y = -16

6. 5x + 7y = -1
4x - 2y = 22
Found 5 solutions by Alan3354, PRMath, josmiceli, MathLover1, MathTherapy:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
1. -2x + 3y = 25
-2x + 6y = 58
------------------------ Subtract
==========
2. 8x + 13y = 179
2x - 13y = -69
------------------------ Add
=============
3. 2x + 7y = -7
5x + 7y = 14
----------------
4. 6x + 3y = 0
-3x + 3y = 9
---------------------
5. 3x - 8y = 32
-x + 8y = -16
------------------
6. 5x + 7y = -1
4x - 2y = 22
-----
Multiply each eqn by something to make the coefficients of either x or y the same, then subtract.
Answer by PRMath(133)   (Show Source): You can put this solution on YOUR website!
I've worked out two of these for you which may be good examples for the other problems. See what you think and if you still have problems, please contact me, or submit the other questions for help.












Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
1)
Multiply both sides of top equation by
And add the equations
2)
Simply add the equations
3)
Multiply both sides of top equation by
And add the equations
4)
Multiply both sides of the top equation by
And add the equations
5)
Simply add the equations
6)
Multiply top equation by and
Multiply bottom equation by
Then add the equations
————————————
I will do the 4th one


———————————


Plug this back into either equation




—————-
Hope this helps
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!
1.


Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations




In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get -2 and -2 to some equal number, we could try to get them to the LCM.

Since the LCM of -2 and -2 is 2, we need to multiply both sides of the top equation by -1 and multiply both sides of the bottom equation by 1 like this:

Multiply the top equation (both sides) by -1
Multiply the bottom equation (both sides) by 1


So after multiplying we get this:



Notice how 2 and -2 add to zero (ie )


Now add the equations together. In order to add 2 equations, group like terms and combine them




Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:



Divide both sides by to solve for y



Reduce


Now plug this answer into the top equation to solve for x

Plug in


Multiply



Subtract from both sides

Combine the terms on the right side

Multiply both sides by . This will cancel out on the left side.


Multiply the terms on the right side


So our answer is

,

which also looks like

(, )

Notice if we graph the equations (if you need help with graphing, check out this solver)




we get



graph of (red) (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (,). This verifies our answer.




2.


Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations




In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 8 and 2 to some equal number, we could try to get them to the LCM.

Since the LCM of 8 and 2 is 8, we need to multiply both sides of the top equation by 1 and multiply both sides of the bottom equation by -4 like this:

Multiply the top equation (both sides) by 1
Multiply the bottom equation (both sides) by -4


So after multiplying we get this:



Notice how 8 and -8 add to zero (ie )


Now add the equations together. In order to add 2 equations, group like terms and combine them




Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:



Divide both sides by to solve for y



Reduce


Now plug this answer into the top equation to solve for x

Plug in


Multiply



Reduce



Subtract from both sides

Make 175 into a fraction with a denominator of 5

Combine the terms on the right side

Multiply both sides by . This will cancel out on the left side.


Multiply the terms on the right side


So our answer is

,

which also looks like

(, )

Notice if we graph the equations (if you need help with graphing, check out this solver)




we get



graph of (red) (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (,). This verifies our answer.




3.

Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations




In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 2 and 5 to some equal number, we could try to get them to the LCM.

Since the LCM of 2 and 5 is 10, we need to multiply both sides of the top equation by 5 and multiply both sides of the bottom equation by -2 like this:

Multiply the top equation (both sides) by 5
Multiply the bottom equation (both sides) by -2


So after multiplying we get this:



Notice how 10 and -10 add to zero (ie )


Now add the equations together. In order to add 2 equations, group like terms and combine them




Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:



Divide both sides by to solve for y



Reduce


Now plug this answer into the top equation to solve for x

Plug in


Multiply



Subtract from both sides

Combine the terms on the right side

Multiply both sides by . This will cancel out on the left side.


Multiply the terms on the right side


So our answer is

,

which also looks like

(, )

Notice if we graph the equations (if you need help with graphing, check out this solver)




we get



graph of (red) (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (,). This verifies our answer.



4.


Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations




In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 6 and -3 to some equal number, we could try to get them to the LCM.

Since the LCM of 6 and -3 is -6, we need to multiply both sides of the top equation by -1 and multiply both sides of the bottom equation by -2 like this:

Multiply the top equation (both sides) by -1
Multiply the bottom equation (both sides) by -2


So after multiplying we get this:



Notice how -6 and 6 add to zero (ie )


Now add the equations together. In order to add 2 equations, group like terms and combine them




Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:



Divide both sides by to solve for y



Reduce


Now plug this answer into the top equation to solve for x

Plug in


Multiply



Subtract from both sides

Combine the terms on the right side

Multiply both sides by . This will cancel out on the left side.


Multiply the terms on the right side


So our answer is

,

which also looks like

(, )

Notice if we graph the equations (if you need help with graphing, check out this solver)




we get



graph of (red) (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (,). This verifies our answer.
-> for some reason this doesn't work
so,

-> both sides multiply by
---------------------


-----------------------add both equations





go to , substitute or





so, intersection point is at:(, )





5.


Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations




In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 3 and -1 to some equal number, we could try to get them to the LCM.

Since the LCM of 3 and -1 is -3, we need to multiply both sides of the top equation by -1 and multiply both sides of the bottom equation by -3 like this:

Multiply the top equation (both sides) by -1
Multiply the bottom equation (both sides) by -3


So after multiplying we get this:



Notice how -3 and 3 add to zero (ie )


Now add the equations together. In order to add 2 equations, group like terms and combine them




Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:



Divide both sides by to solve for y



Reduce


Now plug this answer into the top equation to solve for x

Plug in


Multiply



Subtract from both sides

Combine the terms on the right side

Multiply both sides by . This will cancel out on the left side.


Multiply the terms on the right side


So our answer is

,

which also looks like

(, )

Notice if we graph the equations (if you need help with graphing, check out this solver)




we get



graph of (red) (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (,). This verifies our answer.




6.


Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations




In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 5 and 4 to some equal number, we could try to get them to the LCM.

Since the LCM of 5 and 4 is 20, we need to multiply both sides of the top equation by 4 and multiply both sides of the bottom equation by -5 like this:

Multiply the top equation (both sides) by 4
Multiply the bottom equation (both sides) by -5


So after multiplying we get this:



Notice how 20 and -20 add to zero (ie )


Now add the equations together. In order to add 2 equations, group like terms and combine them




Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:



Divide both sides by to solve for y



Reduce


Now plug this answer into the top equation to solve for x

Plug in


Multiply



Subtract from both sides

Combine the terms on the right side

Multiply both sides by . This will cancel out on the left side.


Multiply the terms on the right side


So our answer is

,

which also looks like

(, )

Notice if we graph the equations (if you need help with graphing, check out this solver)




we get



graph of (red) (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (,). This verifies our answer.
Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!

Hello, can you please assist me with Solving these problems "Using Elimination" in Algebra 2? Thank you very much
1. -2x + 3y = 25
-2x + 6y = 58


2. 8x + 13y = 179
2x - 13y = -69


3. 2x + 7y = -7
5x + 7y = 14

4. 6x + 3y = 0
-3x + 3y = 9

5. 3x - 8y = 32
-x + 8y = -16

6. 5x + 7y = -1
4x - 2y = 22
For #s 1 - 5, ALL you have to do is either ADD the 2 equations or SUBTRACT one from the other. IGNORE the NOVEL that was written, as a NOVEL is TOTALLY UNNECESSARY. This was FAR more than OVERKILL!

I hope for your sake that you didn't look at it. No. 6 is the only one that's a little more difficult than the others. 

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