System of equations...

MM

New member
Joined
Jan 3, 2006
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6
Solve each system of equations by elimination...
-4x + 3y=-1 8x + 6y=10
 
-4x + 3y=-1 [1]
8x + 6y=10 [2]

Try multiplying [1] by 2, and adding the resultant equation to [2].
 
The idea is multiply one or both equations by a number or numbers such that the coefficients of x or y in each equation will cancel when the two equations are added (or subtracted).

6x - 5y = 3 [1]

-12x + 8y = 5 [2]

"[1]" and "[2]" are just means to refer to each equation. Read "[1]" as "the first equation".

If we multiply [1] by 2, we will be able to eliminate the x's upon adding the two equations.

[1] * 2: 12x - 10y = 6 [3]

and [2] is still: -12x + 8y = 5

See that if we now add [2] and [3], the x's cancel:

Code:
     12x - 10y = 6
 +   -12x + 8y = 5
-------------------
           -2y = 11

And we have y = -11/2. Plugging into [1] or [2] gives us x.

Practise.
 
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