# Thread: solving equations with fractions

1. ## solving equations with fractions

need help on solving these problems

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

2. Originally Posted by Haxcake
need help on solving these problems

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

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x/2 -29 = -22
x/2 = -22 +29
x/2 = 7
x = 7*2

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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

------------------------------------------------
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

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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.

3. Originally Posted by Haxcake
need help on solving these problems

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