Absolute Value Linear Equations


 Math > Math Concepts >Algebra2  >  Absolute Values in Equations
 
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Solving linear equations involving

absolute value

 

        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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