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Home Q & A Board Science-Related Homework Help Mathematics
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__Aycee __Aycee
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Posts: 108
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10 years ago
How do you solve type 4 algebra equations when the variable is on the other side?
For example, I know how to do 4(x+3)=5(x-2) but not
3(5-x)=2(5+x)
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wrote...
#1
10 years ago
Still distribute the 3 and 2!
Make it:
15-3x=10+2x
And then solve from there:
5=5x
1=x
Answer accepted by topic starter
09mohamf09mohamf
wrote...
#2
Posts: 28
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10 years ago
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#3
10 years ago
Exactly the same way!
Just as
4(x+3)=5(x-2)
==> 4x + 12 = 5x - 10
==> -x = -22
==> x = 22
you follow the same steps
3(5-x)=2(5+x)
==> 15 -3x = 10 + 2x
==> -x = -5
==> x = 5
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