Topic 2 - 2.2 -- Solving Linear Equations
Equations 1
Solve for the given variable.
5a + 3 = 23 To solve for the variable, isolate it on one side of the equation.
-3 -3 Add negative 3 to both sides of the equation.
5a = 20 Divide by 5.
5 5
a = 4
Check: 5(4) + 3 = 23
20 + 3 = 23
23 = 23
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22 - 4c = 42 Add -22 to both sides of the equation.
-22 -22
-4c = 20 Divide both sides by -4.
-4 -4
c = -5 The check is left up to the student.
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Equations 2
4(x + 3) = -5(3x - 10) Remove the parentheses by multiplying the number
4x + 12 = -15x + 50 in front of the parentheses by each number in the
- 12 -12 parentheses following it.
4x = -15x + 38 Add -12 to both sides of the equation and them
15x 15x add 15x to both sides of the equation.
19x = 38 Divide both sides of the equation by 19.
19 19
x = 2
(1/8)n + 6 = -(5/8)(n - 16) Multiply both sides of the equation by the LCD.
The LCD is 8. 1/8 is a fraction.
(8)(1/8)n + (8)(6) = (8)(-5/8)(n - 16)
n + 48 = -5(n - 16) All fractions are removed.
n + 48 = -5n + 80 Notice in this problem I added a -48 and 5n
n = -5n + 32 to both sides of the equation and divided both
6n = 32 sides of the equation by 6. I only put the result
n = 32/6 = 16/3 of each step.
Remember all fraction must be reduced to its lowest terms.