# solving equations with fractions

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• Jan 29th 2007, 12:45 AM
Haxcake
solving equations with fractions
need help on solving these problems

$1/5x+27=22$
$1/2x-29=-22$
$3x/5+22=28$
$14-2x/3=18$
• Jan 29th 2007, 01:23 AM
ticbol
Quote:

Originally Posted by Haxcake
need help on solving these problems

$1/5x+27=22$
$1/2x-29=-22$
$3x/5+22=28$
$14-2x/3=18$

If I understand what you mean, you want to solve for x in each of those 4 equations, and 1/5x is actually (1/5)x or x/5.

So,
x/5 +27 = 22
Collect like terms, numbers with numbers, unknowns with unknowns,
x/5 = 22 -27
x/5 = -5
Clear the fraction, multiply both sides by 5,
x = -25 ----------------answer.

----------------------------------------------
x/2 -29 = -22
x/2 = -22 +29
x/2 = 7
x = 7*2
x = 14 --------------answer.

-------------------------------------------
3x/5 +22 = 28
3x/5 = 28 -22
3x/5 = 6
3x = 6*5
3x = 30
To isolate x, or so that x will be alone,
Divide both sides by 3,
x = 30/3
x = 10 -------------------answer.

------------------------------------------------
14 -2x/3 = 18
-2x/3 = 18 -14
-2x/3 = 4
Clear the fraction, multiply both sides by 3,
-2x = 12
So that x will be alone, divide both sides by -2,
x = 12 / -2
x = -6 -------------answer.

------------------------------------------
Note:
When you solve for x, or any unknown, in an equation, just do anything to isolate the x, or any unknown. Get the x, or any unknown, be alone by itself at one side of the equation.
• Jan 29th 2007, 10:23 PM
qpmathelp
Quote:

Originally Posted by Haxcake
need help on solving these problems

$1/5x+27=22$
$1/2x-29=-22$
$3x/5+22=28$
$14-2x/3=18$

use brackets

if you mean

{1/(5x) } + 27 = 22

then {1/(5x) } = 22 -27

{1/(5x) } = -5

so that x = 1/(-5*5)

or x = -1 / 25