The solution of equations and inequalities involves finding the value (or values) of the variable(s) that make the mathematical statements true. The addition and multiplication properties of equations and inequalities, as well as the properties of real numbers, are used to simplify the equation/inequality as much as possible, prior to formulating the solution set for the variable in question.
Solving linear equations in one variable, i.e., equations dealing with algebraic expressions having one variable with an exponent of one, are the simplest to solve for.
EX.
2x + 3 = 7
2x + 3 + (-3) = 7 + (-3)
2x = 4
(0.5)2x = (0.5) 4
x = 2
EX.
4x + 5 = -15
4x + 5 + (-5) = -15 + (-5)
4x = -20
¼(4x) = ¼(-20)
x = -5
When the equation involves fractions, it is helpful to multiply both sides of the equation by the least common denominator of all fractions:
EX.
¾ x (40) + 4/5 (40) = 7/8 (40)
where the LCD (lowest common denominator) of 4, 5 and 8 is 40
30x + 32 = 35
30x + 32 - 32 = 35 - 32
30x = 3
x = 3 / 30 = 1/10
When the equation involves decimals, it is helpful to multiply both sides of the equation by the lowest power of 10 that converts all decimal numbers to integers:
EX.
2.5x - 3.2 = 12.5
2.5x - 3.2 + 3.2 = 12.5 + 3.2
2.5x = 15.7
By multiplying each side by 10, we get
25x = 157
x = 157/25 = 6.28
The process of solving inequalities is similar to that of solving equations, except that different additive and multiplicative properties apply.
EX.
2x - 3 > 5
2x - 3 + 3 > 5 + 3
2x > 8
½ (2x) > ½ (8)
x > 4
The solution set is { x | x > 4}
EX.
-3x - 4 < 8
-3x - 4 + 4 < 8 + 4
-3x < 12
-1/3 (-3x) > -1/3 (12)
(note the change in the inequality from < to >)
x > -4
The solution set is { x | x > -4}
EX.
1/3 x -3/4 < 1/2
1/3 x (12) -3/4 (12) < 1/2 (12)
where the LCD of 2, 3 and 4 is 12
4x - 9 < 6
4x - 9 + 9 < 6 + 9
4x < 15
x < 15/4
The solution set is { x | x < 15/4}.
If the equation or inequality involves absolute values, then its solution will involve solving two equations or inequalities.
(1) | ax + b | = k is equivalent to ax + b = k or ax + b = -k.
EX.
|2x + 1| = 5
2x + 1 = 5 or 2x + 1 = -5
2x = 4
2x = -6
x = 2 or x = -3
\ the solution set is {2 , -3}.
(2) | ax + b | < k is equal to -k < ax + b < k, where k > 0.
EX.
|3x + 2| < 7
-7 < 3x + 2 < 7
-9 < 3x < 5
-3 < x < 5/3
\ the solution set is the interval (-3, 5/3).
(3) | ax + b | > k is equal to ax + b > k or ax + b < -k, where k > 0.
EX. | 2x - 1 | > 3
2x - 1 > 3 or 2x - 1 < -3
2x > 4 or 2x < -2
x > 2 or x < -1
\ the solution set (-¥ , -1) È ( 2 , ¥ )
