SOLUTION: Can you please help me solve this sytem of equations? 6x-5y=3 4x+2y=-14

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Question 153360: Can you please help me solve this sytem of equations?
6x-5y=3
4x+2y=-14

Answer by jim_thompson5910(35256) 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

6%2Ax-5%2Ay=3
4%2Ax%2B2%2Ay=-14

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

Since the LCM of 6 and 4 is 12, we need to multiply both sides of the top equation by 2 and multiply both sides of the bottom equation by -3 like this:

2%2A%286%2Ax-5%2Ay%29=%283%29%2A2 Multiply the top equation (both sides) by 2
-3%2A%284%2Ax%2B2%2Ay%29=%28-14%29%2A-3 Multiply the bottom equation (both sides) by -3


So after multiplying we get this:
12%2Ax-10%2Ay=6
-12%2Ax-6%2Ay=42

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%29-10%2Ay-6%2Ay%29=6%2B42

%2812-12%29%2Ax-10-6%29y=6%2B42

cross%2812%2B-12%29%2Ax%2B%28-10-6%29%2Ay=6%2B42 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:

-16%2Ay=48

y=48%2F-16 Divide both sides by -16 to solve for y



y=-3 Reduce


Now plug this answer into the top equation 6%2Ax-5%2Ay=3 to solve for x

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


6%2Ax%2B15=3 Multiply



6%2Ax=3-15 Subtract 15 from both sides

6%2Ax=-12 Combine the terms on the right side

cross%28%281%2F6%29%286%29%29%2Ax=%28-12%29%281%2F6%29 Multiply both sides by 1%2F6. This will cancel out 6 on the left side.


x=-2 Multiply the terms on the right side


So our answer is

x=-2, y=-3

which also looks like

(-2, -3)

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

6%2Ax-5%2Ay=3
4%2Ax%2B2%2Ay=-14

we get



graph of 6%2Ax-5%2Ay=3 (red) 4%2Ax%2B2%2Ay=-14 (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,-3). This verifies our answer.