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Algebra 1: Solve for a variable help?
#1
Ok... This is what they want me to do.

Solve each equation or formula for the variable specified.

v = r + at, for a.

3/5m + a = b, solve for m.

3ax - n over 5 = -4, for x.

PLEASE EXPLAIN... Its not making any sense!
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#2
(10-25-2010, 07:34 PM)L3g1tWa5te Wrote: Ok... This is what they want me to do.

Solve each equation or formula for the variable specified.

v = r + at, for a.

3/5m + a = b, solve for m.

3ax - n over 5 = -4, for x.

PLEASE EXPLAIN... Its not making any sense!

To solve for a variable means to get it alone. So if they tell you to solve for a, the outcome should look like "a = ..."

For example:

v = r + at
v - r = at
(v - r) / t = a
a = (v - r) / t


I subtracted the r from both sides to move it to the left. Then divided both sides by t. What gets left at the end is an equation solved for a
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#3
for the other two...

Code:
3/5m + a = b, solve for m.

Subtract 'a' from both sides gives you:
3/5m = b - a

then multiply both sides by the reciprocal of 3/5.. which is 5/3. it gives you
m = 3(-a + b)/5 which is  m = (-3a + 3b)/5
you dont need the parenthesis in the answer, i just put them to show it was all over 5



Code:
3ax - n over 5 = -4, for x.
or
(3ax - n)/5 = -4

First multiply both sides by 5 to get rid of the fraction.
gives you : 3ax - n = -20

add n to both sides and get: 3ax = -20 + n

divide both sides by 3a and get: x =  (-20 + n)/3a


Hope this helps, its a bit hard to explain while typing. I was going to make imgs for ya but dont have the time right now. If you dont get it, let me know i will explain in further detail.
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#4
Lol im in algebra 1 but we haven't learned this kind of stuff yet.
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