Solving literal equations

What is literal equation

What is the solving literal equation definition? It is when you have to solve a formula for a certain variable. Sometimes, you may need to solve for a variable that isn't the standard one. An example of this is the distance formula. The formula to finding distance is:

$D = rt$

$r$ stands for rate, and $t$ stands for time. The standard variable that you'd solve for is $D$, for distance.

Perhaps in the question you're faced with, you may need to find the rate rather than solving for distance. In that case, you may need to rearrange the formula so that you can solve the literal equation.

How to solve literal equations

Looking at literal equations, you may be worried about how you'll go about solving them. However, they're actually tackled in a way that's very similar to how you've been solving equations all along. The only difference is that since you're working with variables rather than numbers, you may not be able to simplify down your answer too much.

When you're given a question that asks you to solve for a literal equation, you'll be given a formula and then asked to solve for the indicated variable. That variable likely won't be isolated onto its own side, which is what you'll have to do. Take into account what you've done previously for equations where you've had to move terms to the other side of the equal sign. When you've got a positive number, take the negative of it from both sides. When you've got a number to multiply, take the division of it on both sides. When you're done isolating the variable to equalling the rest of the variables moved over to the other side of the equal sign, you've successfully solved your literal equation!

Example problems

Question 1:

Solve each of the formulas for the indicated variable.

i) $a = bc$ for $c$.

Solution:

We'll have to isolate $c$, which means $b$ has to be moved when we're solving for a variable ($c$ in this case).

$\frac{a}{b} = \frac{bc}{b}$

$b$ on the right side cancels out each other, and we are left with:

$\frac{a}{b} = c$

Our final answer:

$c = \frac{a}{b}$

ii) $\frac{x}{y} = z$ for $x$.

Solution:

$\frac{x}{y} = z$

Multiply $y$ to both sides in order to move $y$ over to the right hand side.

$y \bullet \frac{x}{y} = z \bullet y$

$y$ on the left side cancels out each other, and we're left with:

$x = yz$

Question 2:

Solve each of the formulas for the indicated variable.

i) $p = 3q + 3r$ for $r$.

Solution:

$p = 3q + 3r$

Subtract $3q$ from both sides

$p - 3q = 3r$

Divide $3$ on both sides

$\frac{p - 3q}{3} = \frac{3r}{3}$

Final answer:

$r = \frac{p - 3q}{3}$

ii) $r = 2x + 3xy$ for $x$.

Solution:

$r = 2x + 3xy$

Factor out $x$ on the right side

$r = x(2 + 3y)$

Divide $(2+3y)$from both sides

$\frac{r}{(2 + 3y)} = \frac{x(2 + 3y)}{(2 + 3y)}$

Final answer:

$x = \frac{r}{(2 + 3y)}$

Question 3:

Solve each of the formulas for the indicated variable.

i) $3(4x - y) = 6$ for $x$.

Solution:

$3(4x - y) = 6$

Divide $3$ from both sides

$\frac{3(4x - y)}{3} = \frac{6}{3}$

$4x - y = 2$

Add $y$ to both sides

$4x - y + y = 2 + y$

$4x = 2 + y$

Divide $4$ from both sides

$\frac{4x}{4} = \frac{(2 + y)}{4}$

Final answer:

$x = \frac{2 + y}{4}$

You can also check your answers online with this online literal equation solver.

If you need to brush up on linear equations and how to move terms from one side of the equation to another, check out this lesson that solves linear equations dealing with addition and subtraction, or this lesson that uses multiplication and division. You may also want to review how to tackle distributive property.