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Question 29755: I'm trying to figure this equation out. Solving using the elimination method. I need to identify the solution. 3x-6y equals 0 and 2x+4y=5. I am so stuck and confused on two or more elimination methods. Either it's my instructor or I'm trying to decide which to eliminate first. The y, x or z first?
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! I'm trying to figure this equation out. Solving using the elimination method. I need to identify the solution. 3x-6y equals 0 and 2x+4y=5. I am so stuck and confused on two or more elimination methods. Either it's my instructor or I'm trying to decide which to eliminate first. The y, x or z first?
3x-6y = 0 ----(1)X2
2x+4y = 5 ----(2)X3
Therefore
6x-12y = 0 ----(3) (zero mulitplied by anything is zero)
6x+12y = 15 ----(4)
(3) +(4) implies
(6x+6x) = (0+15)
12x = 15
x = 15/12 = 3X5/3X4 = 5/4
Putting x = 5/4 in (2)
2x+4y = 5
2X(5/4)+4y =5
Multiplying by 4 through out
10+16y =20
16y =20-10
16y = 10
y = 10/16 = 5/8
Answer: x=5/4 and y= 5/8
Verification: Putting x=5/4 and y= 5/8 in (1)
3x-6y = 0 ----(1)
LHS = 3x-6y = 3X(5/4)-6X(5/8) = (15X2-6X5)/8 = (30-30)/8 = 0/8 = 0 = RHS
(finding the lcm of 4 and 8)
Note: First of all there is no z here.
x and y are the two unknowns and there are two equations in x and y,
which is fine and which is what we want to find x and y.
And there is no hard and fast rule that we have to find x first
and y next or vice versa.
Studying the problem which ever variable is easily removable,we decide upon eliminating that first and finding the other variable first. For example in some problems a particular variable will have coefficient = (1) or (-1) in which case it is easy to handle that variable for elimination and find the value of the other variable first.
When both variables have coefficients other than +1 or -1,
we select that variable whose coefficients have opposite signs,
find the lcm of the coefficients and multiply the equations by suitable numbers to equalise the numerical coefficients which after the above treatment are equal in magnitude and opposite in sign and offer ready elimination on addition (which is much more preferable to subtraction which is the procedure when coefficients are equal in magnitude and in sign)
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