Absolute
value refers to the magnitude of a number. Vertical bars, one placed
at the front and the other placed at the back of the number or
expression, are used to represent absolute value.
Examples:
 3
 = 3
 3 
= 3 the magnitude of a negative number is its
positive
 x
 = 3 what is x?
well,
we can see from the previous two examples 3 and 3 will work, so x =
{3, 3}
Let's try another
equation:
 x
+ 5  = 20
(x +
5) = 20 and – (x + 5) = 20
x +
5 – 5 = 20 – 5  x  5 = 20
x =
15 x 5 + 5 = 20 + 5
x
= 25
x = –25
so,
x = { 15, –25 }
check  15 + 5  =? 20
 –25 + 5 =? 20
 20  = 20
 –20  = 20
Always
check your answers! It is easy to make an error manipulating
negative signs! Notice that in both answers I substitute both
solutions back into the original expression in the absolute value
signs.
(That
is, I do not reintroduce a negative sign as when I solved the
equation.)
To
solve absolute value equations, you need to solve two equations: one
with the expression inside the absolute value and one with the
negative of the expression inside the absolute value.
That
is, to solve  x  = C, you need to solve x = C and –(
x ) = C
Solving an equation
that contains the variable both inside the absolute value signs and outside,
requires a bit more analysis up front.
Example, solve for x:
 2x + 3  = x + 5
x + 5 must be ≥ 0
so valid x values must be ≥ 5.
two equations to solve:
2x + 3 = x + 5
and (2x + 3) = x + 5
2x  x + 3 = x  x + 5
2x  3 = x + 5
x + 3 = 5
x 2x  3 = x + x + 5
x = 2
3x  3 = 5
3x = 5 + 3 = 8
x =  8/3 (≥ 5)
checks:

2 * 2 + 3  = 4 +3 = 7; (2) + 5 = 7
ok

2( 8/3) + 3 = 16/3 + 3 = (16+9)/3 = 7/3 = 7/3;
(8/3) + 5 = (8+15)/3 = 7/3 ok
Example, solve for
x:
 15
+ (3x – 2)  = 5x + 10
Notice that there
is an x on the right hand side of this equation and only the absolute value sign
on the left hand side. 5x + 10 ≥ 0.
This means that x must be ≥ 2.
two
equations to solve:
the
positive expression: the negative
expression:
15
+ (3x – 2) = 5x + 10 and – (15 + (3x – 2))
= 5x + 10
15 +
3x – 2 = 5x + 10 – 15 – (3x – 2) = 5x
+ 10
13 +
3x = 5x + 10 – 15 – 3x + 2 = 5x + 10
13 +
3x – 5x = 5x  5X + 10 – 5x – 13 – 3x
= 5X + 5x + 10
13 –
2x = 10
– 13 – 8x = 10
13
– 13 – 2x = 10 – 13
13 – 13 – 8x
= 13 +10
–2x
= – 3 13 – 13 – 8x = 10 + 13
–½(–2x) = –½(– 3) –8x
= 23
x = 3/2 –1/8(–8x) = –1/8(23)
x
= – 23/8 not allowed,
x
must be ≥ 2
solution is x = 3/2
check:
15 +
3(3/2) – 2 = 15 + 9/2  2 = 13 +9/2 = (26+9)/2
= 35/2 = 35/2
5(3/2) + 10 = 15/2
+ 10 = (15+20)/2 = 35/2
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