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Absolute Value Equations
Consider one of the most basic absolute value equations,
| x | = 2
Clearly, there are two values of x which solve the above equation. If x = 2, we have,
| 2 | = 2 Þ 2 = 2
If x = -2, we have,
| -2 | = 2 Þ 2 = 2
This leads to the following property for solving linear absolute value equations:
| Given a is a nonnegative number, that is, a ³ 0, we have, | y | = a is equivalent to y = a or y = -a |
Example 1.
Solve the following equation: | x + 1 | = 2
We have,
x + 1 = 2 or x + 1 = -2
Solving each,
x + 1 = 2 Þ x = 1
x + 1 = -2 Þ x = -3
Example 2.
Solve the following equation: | x + 1 | - 3 = 6
We must isolate the absolute value by itself on the left before applying the property to solve. We have,
Now we can write the two equations to solve. We have,
As a check,
When x = 8: | x + 1 | - 3 = 6 becomes | 8 + 1 | - 3 = 6 Þ 9 - 3 = 6 Þ 6 = 6 ü
When x = -10: | x + 1 | - 3 = 6 becomes | -10 + 1 | - 3 = 6 Þ | -9 | - 3 = 6 Þ 9 - 3 = 6 Þ 6 = 6 ü
Example 3.
Solve the following equation: 3| x + 1 | - 1 = 5
Again, we must isolate the absolute value by itself on the left. We have,
We now have the additional step of dividing by 3 before writing the two equivalent equations. This gives,
The two equations to solve are then,
x + 1 = 2 or x + 1 = -2
The first equation has solution x = 1. The second equation has solution x = -3. The solution to the original equation is then x = 1 or x = -3.
Example 4.
Solve the following equation: | x + 1 | = -7
The absolute value of an expression can never be negative, so there is no solution.
Example 5.
Solve the following equation: | 2x + 1 | = | x + 3 |
Clearly, if both expressions within | | are equal, then a solution to the equation is obtained. We have,
2x + 1 = x + 3
2x = x + 3 - 1
2x = x + 2
2x - x = 2
x = 2
Not as obvious is a second possible solution occurring when one expression is the negative of the other. Here, we would have,
2x + 1 = - (x + 3)
2x + 1 = -x - 3
2x = - x - 3 - 1
2x = -x - 4
2x + x = -4
3x = -4
x = -4/3
Both x = 2 and x = -4/3 satisfy the original equation.
We now have another property for solving equations:
If | Ax + B | = | Cx + D | Then the equivalent equations to solve would be, Ax + B = Cx + D and Ax + B = -(Cx + D) or simplifying Ax + B = Cx + D and Ax + B = -Cx - D |
Example 6.
Solve the following equation: | x + 1 | = | x - 7 |
We have,
x + 1 = x - 7 or x + 1 = -(x - 7). The first equation results in a contradiction and has no solution. Solving the second equation yields,
x + 1 = -(x - 7)
x + 1 = -x + 7
2x = 6
x = 3
The only solution to the original equation is x = 3. Equations of this type can have two solutions, one solution, or even no solutions, depending on the original equation.