Questions on Algebra: Systems of Linear Equations answered by real tutors!

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Question 153360: Can you please help me solve this sytem of equations?
6x-5y=3
4x+2y=-14: Can you please help me solve this sytem of equations?
6x-5y=3
4x+2y=-14
Answer by jim_thompson5910(9368) 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*x-5*y=3
4*x+2*y=-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*(6*x-5*y)=(3)*2 Multiply the top equation (both sides) by 2
-3*(4*x+2*y)=(-14)*-3 Multiply the bottom equation (both sides) by -3


So after multiplying we get this:
12*x-10*y=6
-12*x-6*y=42

Notice how 12 and -12 add to zero (ie 12+-12=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
(12*x-12*x)-10*y-6*y)=6+42

(12-12)*x-10-6)y=6+42

cross(12+-12)*x+(-10-6)*y=6+42 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*y=48

y=48/-16 Divide both sides by -16 to solve for y



y=-3 Reduce


Now plug this answer into the top equation 6*x-5*y=3 to solve for x

6*x-5(-3)=3 Plug in y=-3


6*x+15=3 Multiply



6*x=3-15 Subtract 15 from both sides

6*x=-12 Combine the terms on the right side

cross((1/6)(6))*x=(-12)(1/6) Multiply both sides by 1/6. 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*x-5*y=3
4*x+2*y=-14

we get



<BR> drawing( 500, 600, -12, 12, -13, 13,<BR> graph( 500, 600, -12, 12, -13, 13, (3-6*x)/-5, (-14-4*x)/2 ),<BR> blue( circle( -2, -3, 0.16 ) ) <BR> )<BR> graph of 6*x-5*y=3 (red) 4*x+2*y=-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.