# 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.

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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

#1

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#2

first, expand parentheses

and get rid of fractions

c=3/4y+b

multiply through by 4y

c(4y)=3+b(4y)

now get the y terms by themselves

4cy-4by=3

factor out the variable

y(4c-4b)=3

y=3(4c-4b)

y=3/4(c-b)

3ax-n/5= -4

same pattern

multiply through by 5

[it will save you having to do it later]

15ax-n= -20

15ax=n-20

x(15a)=n-20

x=(n-20)/15a

#3

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

#4