SOLUTION: x + y = -4 x - y = 2 x + y = 10 y = x + 8 3x + y = 5 4x - 7y = -10 y - 2x = -5 3y - x = 5

Algebra ->  Graphs -> SOLUTION: x + y = -4 x - y = 2 x + y = 10 y = x + 8 3x + y = 5 4x - 7y = -10 y - 2x = -5 3y - x = 5       Log On


   

Question 84008: x + y = -4
x - y = 2

x + y = 10
y = x + 8

3x + y = 5
4x - 7y = -10


y - 2x = -5
3y - x = 5
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
If you want to solve the system
x+%2B+y+=+-4
x+-+y+=+2
by addition/elimination, then...

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


Lets start with the given system of linear equations

1%2Ax%2B1%2Ay=-4
1%2Ax-1%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 1 and 1 to some equal number, we could try to get them to the LCM.

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

1%2A%281%2Ax%2B1%2Ay%29=%28-4%29%2A1 Multiply the top equation (both sides) by 1
-1%2A%281%2Ax-1%2Ay%29=%282%29%2A-1 Multiply the bottom equation (both sides) by -1


So after multiplying we get this:
1%2Ax%2B1%2Ay=-4
-1%2Ax%2B1%2Ay=-2

Notice how 1 and -1 add to zero (ie 1%2B-1=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%281%2Ax-1%2Ax%29%2B%281%2Ay%2B1%2Ay%29=-4-2

%281-1%29%2Ax%2B%281%2B1%29y=-4-2

cross%281%2B-1%29%2Ax%2B%281%2B1%29%2Ay=-4-2 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:

2%2Ay=-6

y=-6%2F2 Divide both sides by 2 to solve for y



y=-3 Reduce


Now plug this answer into the top equation 1%2Ax%2B1%2Ay=-4 to solve for x

1%2Ax%2B1%28-3%29=-4 Plug in y=-3


1%2Ax-3=-4 Multiply



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

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

cross%28%281%2F1%29%281%29%29%2Ax=%28-1%29%281%2F1%29 Multiply both sides by 1%2F1. This will cancel out 1 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)

1%2Ax%2B1%2Ay=-4
1%2Ax-1%2Ay=2

we get



graph of 1%2Ax%2B1%2Ay=-4 (red) 1%2Ax-1%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 (-1,-3). This verifies our answer.


--------------------------------------------------------------------
Start with the given system
x+%2B+y+=+10
y+=+x+%2B+8+

x+%2B+%28x%2B8%29+=+10 Plug in y=x+8
2x%2B8+=+10
2x=+2
x=1
Now substitute x=1 into y=x%2B8
y+=+1+%2B+8+
y=9
So the solution is (1,9)
--------------------------------------------------------------------
If you want to solve the system
3x+%2B+y+=+5
4x+-+7y+=+-10
by addition/elimination, then...

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


Lets start with the given system of linear equations

3%2Ax%2B1%2Ay=5
4%2Ax-7%2Ay=-10

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

Since the LCM of 3 and 4 is 12, we need to multiply both sides of the top equation by 4 and multiply both sides of the bottom equation by -3 like this:

4%2A%283%2Ax%2B1%2Ay%29=%285%29%2A4 Multiply the top equation (both sides) by 4
-3%2A%284%2Ax-7%2Ay%29=%28-10%29%2A-3 Multiply the bottom equation (both sides) by -3


So after multiplying we get this:
12%2Ax%2B4%2Ay=20
-12%2Ax%2B21%2Ay=30

Notice how 12 and -12 add to zero (ie 12%2B-12=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%2812%2Ax-12%2Ax%29%2B%284%2Ay%2B21%2Ay%29=20%2B30

%2812-12%29%2Ax%2B%284%2B21%29y=20%2B30

cross%2812%2B-12%29%2Ax%2B%284%2B21%29%2Ay=20%2B30 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:

25%2Ay=50

y=50%2F25 Divide both sides by 25 to solve for y



y=2 Reduce


Now plug this answer into the top equation 3%2Ax%2B1%2Ay=5 to solve for x

3%2Ax%2B1%282%29=5 Plug in y=2


3%2Ax%2B2=5 Multiply



3%2Ax=5-2 Subtract 2 from both sides

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

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


x=1 Multiply the terms on the right side


So our answer is

x=1, y=2

which also looks like

(1, 2)

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

3%2Ax%2B1%2Ay=5
4%2Ax-7%2Ay=-10

we get



graph of 3%2Ax%2B1%2Ay=5 (red) 4%2Ax-7%2Ay=-10 (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 (1,2). This verifies our answer.


--------------------------------------------------------------------
If you want to solve the system
y+-+2x+=+-5
3y+-+x+=+5
by addition/elimination, then...

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


Lets start with the given system of linear equations

-2%2Ax%2B1%2Ay=-5
-1%2Ax%2B3%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 -1 to some equal number, we could try to get them to the LCM.

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

-1%2A%28-2%2Ax%2B1%2Ay%29=%28-5%29%2A-1 Multiply the top equation (both sides) by -1
2%2A%28-1%2Ax%2B3%2Ay%29=%285%29%2A2 Multiply the bottom equation (both sides) by 2


So after multiplying we get this:
2%2Ax-1%2Ay=5
-2%2Ax%2B6%2Ay=10

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%29-1%2Ay%2B6%2Ay%29=5%2B10

%282-2%29%2Ax-1%2B6%29y=5%2B10

cross%282%2B-2%29%2Ax%2B%28-1%2B6%29%2Ay=5%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:

5%2Ay=15

y=15%2F5 Divide both sides by 5 to solve for y



y=3 Reduce


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

-2%2Ax%2B1%283%29=-5 Plug in y=3


-2%2Ax%2B3=-5 Multiply



-2%2Ax=-5-3 Subtract 3 from both sides

-2%2Ax=-8 Combine the terms on the right side

cross%28%281%2F-2%29%28-2%29%29%2Ax=%28-8%29%281%2F-2%29 Multiply both sides by 1%2F-2. This will cancel out -2 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)

-2%2Ax%2B1%2Ay=-5
-1%2Ax%2B3%2Ay=5

we get



graph of -2%2Ax%2B1%2Ay=-5 (red) -1%2Ax%2B3%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.