SOLUTION: I am solving by using the elimination method. -5x+y=-13 6x-5y=8 5x=8+3y 3x-4y=18 1/2x+3y=11 2x-y=5 3x+2y=2 x+3y=-4 3x-y=6 x+2y=2 4x-y=1 x+2y=7 2x+5y=-1

Algebra ->  Equations -> SOLUTION: I am solving by using the elimination method. -5x+y=-13 6x-5y=8 5x=8+3y 3x-4y=18 1/2x+3y=11 2x-y=5 3x+2y=2 x+3y=-4 3x-y=6 x+2y=2 4x-y=1 x+2y=7 2x+5y=-1       Log On


   

Question 665427: I am solving by using the elimination method.
-5x+y=-13
6x-5y=8
5x=8+3y
3x-4y=18
1/2x+3y=11
2x-y=5
3x+2y=2
x+3y=-4
3x-y=6
x+2y=2
4x-y=1
x+2y=7
2x+5y=-1
3x+4y=-5
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
solving by using the elimination method:
1.
-5x%2By=-13
6x-5y=8
Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

-5%2Ax%2B1%2Ay=-13
6%2Ax-5%2Ay=8

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 6 to some equal number, we could try to get them to the LCM.

Since the LCM of -5 and 6 is -30, we need to multiply both sides of the top equation by 6 and multiply both sides of the bottom equation by 5 like this:

6%2A%28-5%2Ax%2B1%2Ay%29=%28-13%29%2A6 Multiply the top equation (both sides) by 6
5%2A%286%2Ax-5%2Ay%29=%288%29%2A5 Multiply the bottom equation (both sides) by 5


So after multiplying we get this:
-30%2Ax%2B6%2Ay=-78
30%2Ax-25%2Ay=40

Notice how -30 and 30 add to zero (ie -30%2B30=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%28-30%2Ax%2B30%2Ax%29%2B%286%2Ay-25%2Ay%29=-78%2B40

%28-30%2B30%29%2Ax%2B%286-25%29y=-78%2B40

cross%28-30%2B30%29%2Ax%2B%286-25%29%2Ay=-78%2B40 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:

-19%2Ay=-38

y=-38%2F-19 Divide both sides by -19 to solve for y



y=2 Reduce


Now plug this answer into the top equation -5%2Ax%2B1%2Ay=-13 to solve for x

-5%2Ax%2B1%282%29=-13 Plug in y=2


-5%2Ax%2B2=-13 Multiply



-5%2Ax=-13-2 Subtract 2 from both sides

-5%2Ax=-15 Combine the terms on the right side

cross%28%281%2F-5%29%28-5%29%29%2Ax=%28-15%29%281%2F-5%29 Multiply both sides by 1%2F-5. This will cancel out -5 on the left side.


x=3 Multiply the terms on the right side


So our answer is

x=3, y=2

which also looks like

(3, 2)

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

-5%2Ax%2B1%2Ay=-13
6%2Ax-5%2Ay=8

we get



graph of -5%2Ax%2B1%2Ay=-13 (red) 6%2Ax-5%2Ay=8 (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 (3,2). This verifies our answer.


2.
5x=8%2B3y...=>..5x-3y=8
3x-4y=18
Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

5%2Ax-3%2Ay=8
3%2Ax-4%2Ay=18

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 3 to some equal number, we could try to get them to the LCM.

Since the LCM of 5 and 3 is 15, we need to multiply both sides of the top equation by 3 and multiply both sides of the bottom equation by -5 like this:

3%2A%285%2Ax-3%2Ay%29=%288%29%2A3 Multiply the top equation (both sides) by 3
-5%2A%283%2Ax-4%2Ay%29=%2818%29%2A-5 Multiply the bottom equation (both sides) by -5


So after multiplying we get this:
15%2Ax-9%2Ay=24
-15%2Ax%2B20%2Ay=-90

Notice how 15 and -15 add to zero (ie 15%2B-15=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%2815%2Ax-15%2Ax%29-9%2Ay%2B20%2Ay%29=24-90

%2815-15%29%2Ax-9%2B20%29y=24-90

cross%2815%2B-15%29%2Ax%2B%28-9%2B20%29%2Ay=24-90 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:

11%2Ay=-66

y=-66%2F11 Divide both sides by 11 to solve for y



y=-6 Reduce


Now plug this answer into the top equation 5%2Ax-3%2Ay=8 to solve for x

5%2Ax-3%28-6%29=8 Plug in y=-6


5%2Ax%2B18=8 Multiply



5%2Ax=8-18 Subtract 18 from both sides

5%2Ax=-10 Combine the terms on the right side

cross%28%281%2F5%29%285%29%29%2Ax=%28-10%29%281%2F5%29 Multiply both sides by 1%2F5. This will cancel out 5 on the left side.


x=-2 Multiply the terms on the right side


So our answer is

x=-2, y=-6

which also looks like

(-2, -6)

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

5%2Ax-3%2Ay=8
3%2Ax-4%2Ay=18

we get



graph of 5%2Ax-3%2Ay=8 (red) 3%2Ax-4%2Ay=18 (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 (-2,-6). This verifies our answer.


3.
%281%2F2%29x%2B3y=11
2x-y=5
Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition
TEST

%28%281%2F2%29%29%2Ax%2B%283%29%2Ay=11 Start with the first equation


2%28%28%281%2F2%29%29%2Ax%2B%283%29%2Ay%29=%282%29%2A%2811%29 Multiply both sides by the LCD 2



1%2Ax%2B6%2Ay=22Distribute and simplify


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



Lets start with the given system of linear equations

1%2Ax%2B6%2Ay=22
2%2Ax-1%2Ay=5

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 1 and 2 to some equal number, we could try to get them to the LCM.

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

2%2A%281%2Ax%2B6%2Ay%29=%2822%29%2A2 Multiply the top equation (both sides) by 2
-1%2A%282%2Ax-1%2Ay%29=%285%29%2A-1 Multiply the bottom equation (both sides) by -1


So after multiplying we get this:
2%2Ax%2B12%2Ay=44
-2%2Ax%2B1%2Ay=-5

Notice how 2 and -2 add to zero (ie 2%2B-2=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%282%2Ax-2%2Ax%29%2B%2812%2Ay%2B1%2Ay%29=44-5

%282-2%29%2Ax%2B%2812%2B1%29y=44-5

cross%282%2B-2%29%2Ax%2B%2812%2B1%29%2Ay=44-5 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:

13%2Ay=39

y=39%2F13 Divide both sides by 13 to solve for y



y=3 Reduce


Now plug this answer into the top equation 1%2Ax%2B6%2Ay=22 to solve for x

1%2Ax%2B6%283%29=22 Plug in y=3


1%2Ax%2B18=22 Multiply



1%2Ax=22-18 Subtract 18 from both sides

1%2Ax=4 Combine the terms on the right side

cross%28%281%2F1%29%281%29%29%2Ax=%284%29%281%2F1%29 Multiply both sides by 1%2F1. This will cancel out 1 on the left side.


x=4 Multiply the terms on the right side


So our answer is

x=4, y=3

which also looks like

(4, 3)

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

1%2Ax%2B6%2Ay=22
2%2Ax-1%2Ay=5

we get



graph of 1%2Ax%2B6%2Ay=22 (red) 2%2Ax-1%2Ay=5 (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 (4,3). This verifies our answer.


4.
3x%2B2y=2
x%2B3y=-4
Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

3%2Ax%2B2%2Ay=2
1%2Ax%2B3%2Ay=-4

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:

1%2A%283%2Ax%2B2%2Ay%29=%282%29%2A1 Multiply the top equation (both sides) by 1
-3%2A%281%2Ax%2B3%2Ay%29=%28-4%29%2A-3 Multiply the bottom equation (both sides) by -3


So after multiplying we get this:
3%2Ax%2B2%2Ay=2
-3%2Ax-9%2Ay=12

Notice how 3 and -3 add to zero (ie 3%2B-3=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%283%2Ax-3%2Ax%29%2B%282%2Ay-9%2Ay%29=2%2B12

%283-3%29%2Ax%2B%282-9%29y=2%2B12

cross%283%2B-3%29%2Ax%2B%282-9%29%2Ay=2%2B12 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:

-7%2Ay=14

y=14%2F-7 Divide both sides by -7 to solve for y



y=-2 Reduce


Now plug this answer into the top equation 3%2Ax%2B2%2Ay=2 to solve for x

3%2Ax%2B2%28-2%29=2 Plug in y=-2


3%2Ax-4=2 Multiply



3%2Ax=2%2B4 Subtract -4 from both sides

3%2Ax=6 Combine the terms on the right side

cross%28%281%2F3%29%283%29%29%2Ax=%286%29%281%2F3%29 Multiply both sides by 1%2F3. This will cancel out 3 on the left side.


x=2 Multiply the terms on the right side


So our answer is

x=2, y=-2

which also looks like

(2, -2)

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

3%2Ax%2B2%2Ay=2
1%2Ax%2B3%2Ay=-4

we get



graph of 3%2Ax%2B2%2Ay=2 (red) 1%2Ax%2B3%2Ay=-4 (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 (2,-2). This verifies our answer.


5.
3x-y=6
x%2B2y=2
Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

3%2Ax-1%2Ay=6
1%2Ax%2B2%2Ay=2

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:

1%2A%283%2Ax-1%2Ay%29=%286%29%2A1 Multiply the top equation (both sides) by 1
-3%2A%281%2Ax%2B2%2Ay%29=%282%29%2A-3 Multiply the bottom equation (both sides) by -3


So after multiplying we get this:
3%2Ax-1%2Ay=6
-3%2Ax-6%2Ay=-6

Notice how 3 and -3 add to zero (ie 3%2B-3=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%283%2Ax-3%2Ax%29-1%2Ay-6%2Ay%29=6-6

%283-3%29%2Ax-1-6%29y=6-6

cross%283%2B-3%29%2Ax%2B%28-1-6%29%2Ay=6-6 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:

-7%2Ay=0

y=0%2F-7 Divide both sides by -7 to solve for y



y=0 Reduce


Now plug this answer into the top equation 3%2Ax-1%2Ay=6 to solve for x

3%2Ax-1%280%29=6 Plug in y=0


3%2Ax%2B0=6 Multiply



3%2Ax=6-0 Subtract 0 from both sides

3%2Ax=6 Combine the terms on the right side

cross%28%281%2F3%29%283%29%29%2Ax=%286%29%281%2F3%29 Multiply both sides by 1%2F3. This will cancel out 3 on the left side.


x=2 Multiply the terms on the right side


So our answer is

x=2, y=0

which also looks like

(2, 0)

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

3%2Ax-1%2Ay=6
1%2Ax%2B2%2Ay=2

we get



graph of 3%2Ax-1%2Ay=6 (red) 1%2Ax%2B2%2Ay=2 (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 (2,0). This verifies our answer.



6.
4x-y=1
x%2B2y=7
Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

4%2Ax-1%2Ay=1
1%2Ax%2B2%2Ay=7

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 4 and 1 to some equal number, we could try to get them to the LCM.

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

1%2A%284%2Ax-1%2Ay%29=%281%29%2A1 Multiply the top equation (both sides) by 1
-4%2A%281%2Ax%2B2%2Ay%29=%287%29%2A-4 Multiply the bottom equation (both sides) by -4


So after multiplying we get this:
4%2Ax-1%2Ay=1
-4%2Ax-8%2Ay=-28

Notice how 4 and -4 add to zero (ie 4%2B-4=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%284%2Ax-4%2Ax%29-1%2Ay-8%2Ay%29=1-28

%284-4%29%2Ax-1-8%29y=1-28

cross%284%2B-4%29%2Ax%2B%28-1-8%29%2Ay=1-28 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:

-9%2Ay=-27

y=-27%2F-9 Divide both sides by -9 to solve for y



y=3 Reduce


Now plug this answer into the top equation 4%2Ax-1%2Ay=1 to solve for x

4%2Ax-1%283%29=1 Plug in y=3


4%2Ax-3=1 Multiply



4%2Ax=1%2B3 Subtract -3 from both sides

4%2Ax=4 Combine the terms on the right side

cross%28%281%2F4%29%284%29%29%2Ax=%284%29%281%2F4%29 Multiply both sides by 1%2F4. This will cancel out 4 on the left side.


x=1 Multiply the terms on the right side


So our answer is

x=1, y=3

which also looks like

(1, 3)

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

4%2Ax-1%2Ay=1
1%2Ax%2B2%2Ay=7

we get



graph of 4%2Ax-1%2Ay=1 (red) 1%2Ax%2B2%2Ay=7 (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 (1,3). This verifies our answer.


7.
2x%2B5y=-1
3x%2B4y=-5
Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

2%2Ax%2B5%2Ay=-1
3%2Ax%2B4%2Ay=-5

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 3 to some equal number, we could try to get them to the LCM.

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

3%2A%282%2Ax%2B5%2Ay%29=%28-1%29%2A3 Multiply the top equation (both sides) by 3
-2%2A%283%2Ax%2B4%2Ay%29=%28-5%29%2A-2 Multiply the bottom equation (both sides) by -2


So after multiplying we get this:
6%2Ax%2B15%2Ay=-3
-6%2Ax-8%2Ay=10

Notice how 6 and -6 add to zero (ie 6%2B-6=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%286%2Ax-6%2Ax%29%2B%2815%2Ay-8%2Ay%29=-3%2B10

%286-6%29%2Ax%2B%2815-8%29y=-3%2B10

cross%286%2B-6%29%2Ax%2B%2815-8%29%2Ay=-3%2B10 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:

7%2Ay=7

y=7%2F7 Divide both sides by 7 to solve for y



y=1 Reduce


Now plug this answer into the top equation 2%2Ax%2B5%2Ay=-1 to solve for x

2%2Ax%2B5%281%29=-1 Plug in y=1


2%2Ax%2B5=-1 Multiply



2%2Ax=-1-5 Subtract 5 from both sides

2%2Ax=-6 Combine the terms on the right side

cross%28%281%2F2%29%282%29%29%2Ax=%28-6%29%281%2F2%29 Multiply both sides by 1%2F2. This will cancel out 2 on the left side.


x=-3 Multiply the terms on the right side


So our answer is

x=-3, y=1

which also looks like

(-3, 1)

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

2%2Ax%2B5%2Ay=-1
3%2Ax%2B4%2Ay=-5

we get



graph of 2%2Ax%2B5%2Ay=-1 (red) 3%2Ax%2B4%2Ay=-5 (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 (-3,1). This verifies our answer.