How do you solve for c=3/4y+b, for y? Or 3ax-n/5=-4, for x?
c = ?y + b; solve for y.
FIRST: subtract b from each side: c-b = ?y+b-b
c-b = ?y
NOTE: to clear fraction in front of the y just multiply bot sides by the reciprocal 4/3
4/3(c-b) = 4/3(?y) becomes 4/3(c-b) = y.
If all left side is over 5 denominator.
(3ax-n)/5 = -4......multiply both sides by 5
5[(3ax-n)/5] = 5(-4) ...... both 5s cancel left side.
3ax-n = -20
3ax-n+n = -20+n
3ax = -20+n
3ax/3a = (-20+n)/3a
x = (-20+n)/3a
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first, expand parentheses
and get rid of fractions
multiply through by 4y
now get the y terms by themselves
factor out the variable
multiply through by 5
[it will save you having to do it later]
Stephen (first reply) has the first one wrong.
c = 3/4 y + b
take b from both sides like this:
c-b =3/4 y
then multiply both sides by 4/3:
4/3 (c-b) = y
and his second one is right unless you meant the equation to read (3ax - n) / 5 = -4 in which case the answer will be different from stephen's