SOLUTION: Solve the system of equations by the elimination method. (If the system is dependent, let y = c and enter a general solution in terms of c. If there is no solution, enter NO SOLUTI

Algebra ->  Inequalities -> SOLUTION: Solve the system of equations by the elimination method. (If the system is dependent, let y = c and enter a general solution in terms of c. If there is no solution, enter NO SOLUTI      Log On


   

Question 1130422: Solve the system of equations by the elimination method. (If the system is dependent, let y = c and enter a general solution in terms of c. If there is no solution, enter NO SOLUTION.)
5/6x-1/3y+-21
1/6x+2/3y+9
(x, y) =
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

%285%2F6%29x-%281%2F3%29y-21=0
%281%2F6%29x%2B%282%2F3%29y%2B9=0
%285%2F6%29x-%281%2F3%29y=21
%281%2F6%29x%2B%282%2F3%29y=-9



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

%285%2F6%29%2Ax%2B%281%2F3%29%2Ay=21 Start with the first equation


6%28%285%2F6%29%2Ax%2B%281%2F3%29%2Ay%29=%286%29%2A%2821%29 Multiply both sides by the LCD 6



5%2Ax%2B2%2Ay=126Distribute and simplify


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%281%2F6%29%2Ax%2B%282%2F3%29%2Ay=-9 Start with the second equation


6%28%281%2F6%29%2Ax%2B%282%2F3%29%2Ay%29=%286%29%2A%28-9%29 Multiply both sides by the LCD 6



1%2Ax%2B4%2Ay=-54 Distribute and simplify



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Lets start with the given system of linear equations

5%2Ax%2B2%2Ay=126
1%2Ax%2B4%2Ay=-54

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

Since the LCM of 5 and 1 is 5, we need to multiply both sides of the top equation by 1 and multiply both sides of the bottom equation by -5 like this:

1%2A%285%2Ax%2B2%2Ay%29=%28126%29%2A1 Multiply the top equation (both sides) by 1
-5%2A%281%2Ax%2B4%2Ay%29=%28-54%29%2A-5 Multiply the bottom equation (both sides) by -5


So after multiplying we get this:
5%2Ax%2B2%2Ay=126
-5%2Ax-20%2Ay=270

Notice how 5 and -5 add to zero (ie 5%2B-5=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%285%2Ax-5%2Ax%29%2B%282%2Ay-20%2Ay%29=126%2B270

%285-5%29%2Ax%2B%282-20%29y=126%2B270

cross%285%2B-5%29%2Ax%2B%282-20%29%2Ay=126%2B270 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:

-18%2Ay=396

y=396%2F-18 Divide both sides by -18 to solve for y



y=-22 Reduce


Now plug this answer into the top equation 5%2Ax%2B2%2Ay=126 to solve for x

5%2Ax%2B2%28-22%29=126 Plug in y=-22


5%2Ax-44=126 Multiply



5%2Ax=126%2B44 Subtract -44 from both sides

5%2Ax=170 Combine the terms on the right side

cross%28%281%2F5%29%285%29%29%2Ax=%28170%29%281%2F5%29 Multiply both sides by 1%2F5. This will cancel out 5 on the left side.


x=34 Multiply the terms on the right side


So our answer is

x=34, y=-22

which also looks like

(34, -22)

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

5%2Ax%2B2%2Ay=126
1%2Ax%2B4%2Ay=-54

we get



graph of 5%2Ax%2B2%2Ay=126 (red) 1%2Ax%2B4%2Ay=-54 (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 (34,-22). This verifies our answer.


answer:
(x,+y) =(34,+-22)

Answer by ikleyn(52779) About Me  (Show Source):
You can put this solution on YOUR website!
.
I notice that there are no equations at all in your post.