Problems with two unknowns
- Thread starter SILVER
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D
Deleted member 4993
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SILVER said:I just don't understand problems with two unkowns can you help me?
Post some problems - along with your work - and we'll try.
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Why do you think it is different from what you know already?
This is two numbers whose sum if 5.
3 + 2 = 5
This is two numbers whose sum if 5.
3 + X = 5
This is two numbers whose sum if 5.
Z + X = 5
Realy, how does it break down when "Z" gets in there? Addition still works, doesn't it?
This is two numbers whose sum if 5.
3 + 2 = 5
This is two numbers whose sum if 5.
3 + X = 5
This is two numbers whose sum if 5.
Z + X = 5
Realy, how does it break down when "Z" gets in there? Addition still works, doesn't it?
D
Deleted member 4993
Guest
SILVER said:
Hi Yo SILVER,
There are several ways to solve a linear system of equations in 2 variables. I don't know what you've been taught, so I'll start with a very simple way.
It's called the 'elimination method'.
The idea here is to manipulate one or both equations in such a way to make two of the same variables additive inverses of each other, thus 'eliminating' them.
[1] \(\displaystyle 80x\:+\:2y\:=\:9\)
[2] \(\displaystyle 39x\:-\:3y\:=\:11\)
Multiply [1] by 3 and multiply [2] by 2.
[3] \(\displaystyle 240x\:+\:6y\:=\:27\)
[4] \(\displaystyle 78x\:-\:6y\:=\:22\)
Now add [2] + [3]
The \(\displaystyle y\) terms drop out (add to 0) and you're left with an equation in 1 variable (x).
Solve that equation for x and substitute the answer into [1] or [2] to find y.
And you're done!