Question 82903
how to do elimination & substitution problems.
ex.
:
Use brackets to ensure that you mean y is not part of the denominator;
-3x - (1/2)y = 10
 5x + (1/4)y = 8
:
Get rid of the fractions, mult the 1st equation by 2, and the 2nd equation by 4
Resulting in:
-6x - 1y = 20
20x + 1y = 32
:
Notice in this problem, if we just add these two equations we eliminate y:
-6x - y = 20
20x + y = 32
--------------- adding eliminates y, easy to solve for x
14x + 0 = 52
14x = 52
x = 52/14; divide both sides by 14
x = 26/7; lowest terms
:
Pick one of the equations, substitute 3 for x and solve for y:
20x + y = 32
20(26/7) + y = 32
(520/7) + y = 32
y = 32 - (520/7)
y = (224/7) - (520/7); changed 32 to 7ths
y = -296/7
:
Check solutions in one of the original equations:
-3x - (1/2)y = 10
-3(26/7) - (1/2)(-296/7) = 
(-78/7) - (-148/7) = 
(-78/7) + (148/7) =   
+70/7 = 10; proves our solution
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