SOLUTION: 3x+2y=14 x+-12y=-8 I know the answer I just don't know how to solve it by elimination, could you help?

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Question 984222: 3x+2y=14
x+-12y=-8
I know the answer I just don't know how to solve it by elimination, could you help?
Found 2 solutions by Boreal, MathLover1:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
3x+2y=14
x+-12y=-8
Elimination means you need to eliminate a variable. Let's eliminate x, and we will have just y to solve for. Once we have y, we can find x.
To eliminate x, first make the leading coefficient the same but the opposite sign. That means multiply the second equation by -3. Watch what happens
3x+2y=14
-3x+36y=24; Notice that every term in the second equation is multiplied by -3. Watch the signs.
Now you can add the two equations.
0+38y=38, because 3x-3x=0. X is gone.
38y=38
y=1
substitute into either equation.
3x+2=14
3x=12
x=4
Check (4,1) in second equation.
4-12=-8. It checks.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

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=14
1%2Ax-12%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 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=%2814%29%2A1 Multiply the top equation (both sides) by 1
-3%2A%281%2Ax-12%2Ay%29=%28-8%29%2A-3 Multiply the bottom equation (both sides) by -3


So after multiplying we get this:
3%2Ax%2B2%2Ay=14
-3%2Ax%2B36%2Ay=24

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%2B36%2Ay%29=14%2B24

%283-3%29%2Ax%2B%282%2B36%29y=14%2B24

cross%283%2B-3%29%2Ax%2B%282%2B36%29%2Ay=14%2B24 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:

38%2Ay=38

y=38%2F38 Divide both sides by 38 to solve for y



y=1 Reduce


Now plug this answer into the top equation 3%2Ax%2B2%2Ay=14 to solve for x

3%2Ax%2B2%281%29=14 Plug in y=1


3%2Ax%2B2=14 Multiply



3%2Ax=14-2 Subtract 2 from both sides

3%2Ax=12 Combine the terms on the right side

cross%28%281%2F3%29%283%29%29%2Ax=%2812%29%281%2F3%29 Multiply both sides by 1%2F3. This will cancel out 3 on the left side.


x=4 Multiply the terms on the right side


So our answer is

x=4, y=1

which also looks like

(4, 1)

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

3%2Ax%2B2%2Ay=14
1%2Ax-12%2Ay=-8

we get



graph of 3%2Ax%2B2%2Ay=14 (red) 1%2Ax-12%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 (4,1). This verifies our answer.