Lesson 3: 2.3 Solve a Formula for a Specific Variable

New Concept

This lesson teaches you how to rearrange formulas to solve for any specific variable. By isolating a variable, you can use existing equations in new ways, a crucial skill for geometry, science, and even spreadsheet calculations.

What’s next

Next, you'll master this skill through interactive examples. Then, you'll apply what you've learned to solve geometry-based challenge problems.

Property

To solve a formula for a specific variable means to isolate that variable on one side of the equals sign with a coefficient of one. All other variables and constants are moved to the other side of the equal sign. For example, to solve the formula $V = \frac{1}{3}\pi r^2 h$ for $h$, you would perform the following steps:

1. Write the formula: $V = \frac{1}{3}\pi r^2 h$
2. Remove the fraction by multiplying both sides by 3: $3V = \pi r^2 h$
3. Isolate $h$ by dividing both sides by $\pi r^2$: $\frac{3V}{\pi r^2} = h$

Examples

• To solve the simple interest formula $I = Prt$ for the time, $t$, divide both sides by $Pr$. This gives the formula $t = \frac{I}{Pr}$.
• To solve the perimeter formula for a rectangle, $P = 2L + 2W$, for the length, $L$, first subtract $2W$ from both sides to get $P - 2W = 2L$. Then, divide by 2 to get $L = \frac{P - 2W}{2}$.
• To solve the area formula for a triangle, $A = \frac{1}{2}bh$, for the base, $b$, first multiply by 2 to get $2A = bh$. Then, divide by the height $h$ to get $b = \frac{2A}{h}$.

Explanation

Think of this as rearranging a tool for a specific task. By isolating one variable, you create a new, direct formula to find its value, which is very efficient when you need to perform the same calculation multiple times.

Property

To solve geometry applications, follow this structured approach:

• Step 1. Read the problem to ensure all words and ideas are understood.
• Step 2. Identify what you are looking for.
• Step 3. Name what you are looking for by choosing a variable. Draw and label a figure.
• Step 4. Translate into an equation by writing the appropriate formula and substituting the given information.
• Step 5. Solve the equation using good algebra techniques.
• Step 6. Check the answer to ensure it makes sense.
• Step 7. Answer the question with a complete sentence.

Examples

• The area of a rectangular patio is 200 square feet and its width is 10 feet. To find the length, use $A=LW$. Substitute the values: $200 = L \cdot 10$. Solving for $L$ gives a length of 20 feet.
• A square garden has a perimeter of 64 meters. To find the length of a side $s$, use $P=4s$. Substitute the values: $64 = 4s$. Solving for $s$ shows that each side is 16 meters long.
• A triangular banner has an area of 40 square feet and a base of 8 feet. To find its height, use $A = \frac{1}{2}bh$. Substitute the values: $40 = \frac{1}{2}(8)h$. This simplifies to $40=4h$, so the height is 10 feet.

Explanation

This 7-step guide is a roadmap for tackling geometry word problems.
It turns a complex problem into a series of simple, manageable tasks, guiding you from understanding the question to finding and verifying your final answer.

Property

In any right triangle, where $a$ and $b$ are the lengths of the legs, and $c$ is the length of the hypotenuse, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse.

Examples

• A right triangle has legs of length 6 cm and 8 cm. To find the hypotenuse $c$, use the formula $6^2 + 8^2 = c^2$. This becomes $36 + 64 = 100$, so $c = \sqrt{100} = 10$ cm.
• A right triangle has a hypotenuse of 13 inches and one leg of 5 inches. To find the other leg $b$, use $5^2 + b^2 = 13^2$. This becomes $25 + b^2 = 169$. Solving for $b^2$ gives $144$, so $b = 12$ inches.
• A right triangle's hypotenuse measures 17 meters and one leg is 15 meters. To find the other leg $a$, use $a^2 + 15^2 = 17^2$. This becomes $a^2 + 225 = 289$. Solving for $a^2$ gives $64$, so $a = 8$ meters.

Explanation

This famous theorem is a special rule exclusively for right triangles.
It provides a powerful connection between the lengths of the two shorter sides (legs) and the longest side (hypotenuse), allowing you to find any missing side length.