SOLUTION: Using the addition method, how do I solve for 1/4x+1/3Y=5 and x-y=6?

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Question 637104: Using the addition method, how do I solve for 1/4x+1/3Y=5 and x-y=6?
Found 2 solutions by ewatrrr, DrBeeee:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

1/4x+1/3y= 5    || |multiplying thru by 3 to  eliminate the y variable 

 x - y= 6             y=+x-6

1%263%2F4x+=+21

%287%2F4%29x+=+21

   x = 21%2A%284%2F7%29

    x = 12    and  y = 6

and

 12%2F4+%2B+6%2F3+=+5

Answer by DrBeeee(684) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
(1) x/4 + y/3 = 5
(2) x - y = 6
Solve by addition.
Multiply both sides of (1) by 3
3*(x/4 + y/3) = 3*(5)
3*(x/4) + 3*(y/3) = 15 yields
(3) 3/4*x + y = 15
Now add (2) + (3)
(1+3/4)*x + (1-1)*y = 6 + 15 which reduces to
7/4*x + 0*y = 21 or
(4) 7/4*x = 21
x = (4/7)*21
x = 84/7
x = 12
Then using (2)
y = x - 6
y = 6
Are these values correct? Let's see.
Put x = 12 and y = 6 into (1)
Is (12/4 + 6/3 = 5)?
Is (3 + 2 = 5)?
Is (5 = 5)? Yes
The answer is x = 12 and y = 6
PS We don't need to check our answer in equation (2) because we used it to derive the value of y.