Question

Solve the system of equations by elimination: x - 6y = -3 and 3x + 6y = 15
A. (3, -1)
B. (3, 1)
C. (1, 3)
D. (1, -3)

Answer 1

In the method of elimination, you add some multiple of one equation to some multiple of the other equation, with the goal of one of the two variables canceling out. Note that it doesn't particularly matter which variable you try to cancel out, although in some cases it is easier to cancel one variable than the other.

In this case, the Y variable is much easier to cancel out, because there happens to be a '-6y' in the first equation and a '+6y' in the second equation. Therefore, if we let E1 to be the first equation, and E2 to be the second equation, let us add E1 to E2 directly. (We do not need to multiply either equation by anything besides 1 to cancel things out.)

E1+ E2:
(x-6y) + (3x+6y) = (-3) + (15)
Note the -6y in the first equation quickly cancels with the +6y in the second equation, giving us:
x + 3x = -3 + 15
which yields
4x = 12

Now we can solve for x:
x=3

Now that we have x=3, we can substitute it into either E1 or E2 to find y. I will substitute back into E1.

x-6y=-3
sub:
3-6y=-3

Now subtract 3 from both sides
-6y=-6
Now divide both sides by -6:
y = 1

Therefore, the solution to the system is:
x=3, y=1
As you know, such a solution can be represented as a tuple, or in parentheses as (x,y):
(3,1)

The final answer is (3,1).

(Note you could alternatively also use (-3) E1 + (1) E2 in order to eliminate the x variable, and substitute in for the y variable, but that requires multiplication by a negative constant, so it is easier to just eliminate the y variable.)

Answer 2

Please let me know if you have any questions! :)

Some more material on this subject: http://www.purplemath.com/modules/systlin5.htm

Answer 3

WOW