Algebra Review

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.

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

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

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

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

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

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

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

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

Try the exercises

• Please check your work before inputting the answers into Blackboard.