2. Manipulating Algebraic equations
• Purpose – to solve algebraic equations for an unknown variable
• Key to solving – any manipulation you do to one side of the equation must
be done to the other side of the equation.
3. Numbers added and Subtracted
• Solve the following for x:
x + 5 = 13
• Solution (isolate x):
– To get x as the only item on the left hand side of the equation, we
subtract 5 from both sides of the equation, since it was added to the
side with the x:
x + 5 – 5 = 13 – 5
– Combining like terms we get:
x + 5 – 5 = 13 – 5
x = 8
4. Numbers multiplied and divided
• Solve the following for x:
𝑥
5
= 13
• Solution (isolate x):
– To get x as the only item on the left hand side of the equation, we
multiply both sides of the equation by 5 since the x was divided by 5:
5 ∗ 𝑥
5
= 13 ∗ 5
– Combining like terms we get:
5 ∗ 𝑥
5
= 13 ∗ 5
x = 65
5. Combination of operations
• Solve the following for x:
5x + 6 = 31
• Isolate the term that contains the variable first:
5x + 6 − 6 = 31 − 6
5x = 25
• Now solve for x:
5x
5
=
25
5
x = 5
6. x in the denominator
• When x is in the denominator it is easiest to isolate the x term, then cross multiply
• Solve the following for x:
2
𝑥
−
1
5
= 2.6
• Isolate the term with x first:
2
𝑥
−
1
5
+
1
5
= 2.6 +
1
5
2
𝑥
= 2.6 +
1
5
• Combine like terms:
2
𝑥
= 2.8
• Cross multiply – multiply the denominators by the numerators on the opposite side:
2
𝑥
=
2.8
1
2 ∗ 1 = 2.8 ∗ 𝑥
2 = 2.8𝑥
• Solve for x:
2
2.8
=
2.8𝑥
2.8
0.714 = 𝑥
7. Misconception Alert!
• Be very careful not to accidentally make a variable in the denominator a
numerator.
• For example:
1
𝑥
= 10
– Do not just say x = 10!
– Cross multiply:
1 = 10𝑥
– Solve for x:
1
10
=
10𝑥
10
0.1 = 𝑥
8. Exponents
• When dealing with variables raised to a power, treat them the same and deal with the
exponent at the very end.
– Solve the following for x:
𝑥2
5
= 13
– Solution (isolate x2
, then solve for x):
• To get x as the only item on the left hand side of the equation, we multiply
both sides of the equation by 5 since the x was divided by 5:
5 ∗ 𝑥2
5
= 13 ∗ 5
• Combining like terms we get:
5 ∗ 𝑥2
5
= 13 ∗ 5
𝑥2
= 65
• Take the square root of both sides:
𝑥2 = 65
𝑥 = 8.1
9. Pause and Practice
• Solve each of the following for x:
– 2𝑥 + 5 = 9
–
5
𝑥
= 3
–
1
2𝑥−4
= 9
– 3𝑥2
= 27
–
1
𝑥2 + 4 = 13
10. Pause and Practice
• Solve each of the following for x:
– 2𝑥 + 5 = 9
• 𝑥 = 2
–
5
𝑥
= 3
• 𝑥 = 1.67
–
1
2𝑥−4
= 9
• 𝑥 = 2.06
– 3𝑥2
= 27
• 𝑥 = 3
–
1
𝑥2 + 4 = 13
• 𝑥 = 0.33
11. Try the exercises
• Please check your work before inputting the answers into Blackboard.
