1. Isolate an indicated variable in an equation.
Literal Equations
Isolate an indicated variable in an equation.

2. Solve Literal Equations
Our goal is to rearrange equations or formulas to isolate a desired variable by using inverse operations (just as you would use to solve an equation)
However you won’t end up with variable=constant number (like x=7), you will end with variable=expression (like r = 𝑑 𝑡 )
Use the properties of algebra to do this “legitimately.”

3. Solve for u. Give a property to justify each step
Solve for u. Give a property to justify each step. (Means isolate “u” in the following equations )
1/3u – 8 = y
1/3u = y + 8
u = 3(y + 8)
*That is the final answer. All you have to do is isolate the indicated variable. You will not end up with u=number.*
w = ux
w – 9 = 14ux
(w – 9) = u x
*This is the final answer. You are isolating the variable and will still have other variables in the equation*
Addition Property (add 8)
Multiplication Property (•3)
Subtraction Property (- 9)
Division Property (÷ 14x)

4. Isolate “w” in the following equations
Isolate “w” in the following equations. Give a property to justify each step.
y = 2x + w v
vy = 2x + w
vy – 2x = w
w + x = y 3
w = y – x 3
w = 3(y – x)
Multiplication Property (•v)
Subtraction Property (-2x)
Subtraction Property (-x)
Multiplication Property (•3)
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5. k = am + 3mx k = m(a + 3x) k = m a + 3x
Isolate “m” in the following equation. Give a property to justify each step.
The “m” is in both terms. It is a factor of both terms. You will need to “factor it out of them.” This is the distributive property backwards.
Distributive Property
Division Property (÷ [a + 3x])
k = am + 3mx
k = m(a + 3x)
k = m a + 3x
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6. Formulas Solve for h V = 𝟏 𝟑 𝝅 𝒓 𝟐 h
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There are many formulas that you have used
so far in your math career. Here are a few:
D = rt
A = bh
I = Prt
V = LWH
SA = 2LW + 2LH + 2WH
V = 1 3 𝜋 𝑟 2 h
Solve for h
V = 𝟏 𝟑 𝝅 𝒓 𝟐 h
3V = 𝜋 𝑟 2 h Mult. Prop of Eq. (By 3 to clear fraction)
3V = h Division Prop. of Eq.
𝝅 𝒓 𝟐