Formulas: Videos & Practice Problems

Using formulas like $d=st$ for distance, speed, and time involves identifying known and unknown variables, substituting values, and solving for the unknown. Isolating variables in multivariable polynomials or linear equations requires simplifying, collecting terms, and isolating the target variable, enabling quick calculations for different inputs. Understanding units and operations like division and multiplication ensures accurate results. Mastery of these steps enhances problem-solving skills with formulas, supporting comprehension of terms, coefficients, and exponents in polynomial expressions and equations.

Solve Applied Problems Using Formulas

Solving word problems often involves using formulas, which are equations containing multiple variables designed for specific applications. A formula provides a structured way to relate known and unknown quantities, making problem-solving more efficient. For example, the formula for distance is expressed as \(d = s \times t\), where \(d\) represents distance, \(s\) is speed, and \(t\) is time.

To effectively use a formula, begin by identifying the known and unknown variables. Next, substitute the known values into the formula. Finally, solve for the unknown variable, applying algebraic techniques if necessary. For instance, if a train travels at a speed of 60 kilometers per hour for 2.5 hours, the distance traveled can be calculated by plugging these values into the distance formula: \(d = 60 \times 2.5 = 150\). Since speed is in kilometers per hour and time is in hours, the units of hours cancel out, leaving the distance in kilometers, confirming that \(d = 150\) kilometers.

When solving for a variable that is not isolated, such as time \(t\) in the equation \(d = s \times t\), algebraic manipulation is required. For example, if the distance \(d\) is 357 miles and speed \(s\) is 85 miles per hour, solving for time involves dividing both sides by speed: \(t = \frac{d}{s} = \frac{357}{85} = 4.2\). Analyzing the units, miles divided by miles per hour simplifies to hours, so \(t = 4.2\) hours, which aligns with the expected unit for time.

Understanding how to apply formulas in word problems enhances problem-solving skills by connecting real-world contexts with mathematical expressions. Whether the solution requires straightforward substitution or algebraic rearrangement, recognizing the relationship between variables and units ensures accurate and meaningful results.

A giant water tank takes the shape of a cylinder. It has a radius of $2m$ and a height of $5m$. How much water can this tank hold? Round to the nearest tenth of a .
Cylinder: $V=\pi r^{2}h$

• A. $125.7\operatorname{m}^3$
• B. $157.1\operatorname{m}^3$
• C. $62.8\operatorname{m}^3$
• D. $251.3\operatorname{m}^3$

Answer: (C)

The weather forecast says it will be $95\text{℉}$ today. What is this temperature in $\text{℃}$ ?
$C=\frac{5}{9}(F-32)$

• A. $63\text{℃}$
• B. $127\text{℃}$
• C. $35\text{℃}$
• D. $70.6\text{℃}$

Answer: (C)

Solve the formula for $P$ given $r=-5$, $t=1.5$, and $A=21$.
$A=P(1+rt)$

• A. $P=19.5$
• B. $P=12.0$
• C. $P=22.7$
• D. $P=19.4$

Answer: (C)

Solve Applied Problems Using Formulas Example 1

To calculate the pressure on sediment at a specific depth, the formula used relates pressure p in megapascals (MPa) to the depth h in meters. Given a sediment depth of 120 meters, the goal is to determine the corresponding pressure.

First, identify the known and unknown variables: the depth h is 120 meters, and the pressure p is the unknown value to be found. Substituting the known depth into the formula, the calculation proceeds as follows:

Calculate the difference between the depth and 50 meters: \$120 - 50 = 70$. Then, divide this result by 200:

\[\frac{70}{200} = 0.35\]

Finally, add 0.5 to this quotient to find the pressure:

\[p = 0.5 + 0.35 = 0.85 \text{ MPa}\]

This means the pressure on the sediment layer at 120 meters depth is 0.85 megapascals. Understanding how pressure varies with sediment depth is crucial in geology for assessing sediment compaction and related subsurface conditions.

Solve a Formula for a Specific Variable

Understanding how to manipulate formulas to solve for a specific variable is a fundamental skill in algebra and problem-solving. Variables act as placeholders for numbers, allowing us to treat them like numerical values and apply consistent steps to isolate the desired variable. When given a formula, you can either plug in known values immediately or first rearrange the formula to solve for the target variable before substituting values.

For example, consider a formula involving variables h, l, and w. If you know the values of h and l, you can substitute them directly into the formula and then simplify. The process involves simplifying the equation, collecting all terms containing the target variable on one side, and isolating that variable by performing inverse operations such as addition, subtraction, multiplication, or division. For instance, if the equation is \$3w = l + h\(, and you know \)h = 8\( and \)l = 4\(, substituting these values gives \)3w = 4 + 8\(, which simplifies to \)3w = 12\(. Dividing both sides by 3 isolates \)w\(, resulting in \)w = 4\(.

Alternatively, you can rearrange the formula to solve for the target variable first. Starting with the equation \)3w = l + h\(, you isolate \)w\( by dividing both sides by 3, yielding the formula \)w = \frac{l + h}{3}\(. This rearranged formula allows you to quickly calculate \)w\( for any values of \)l\( and \)h\( without repeating the entire solving process. For example, if \)h = 4\( and \)l = 5\(, substituting these values gives \)w = \frac{5 + 4}{3} = \frac{9}{3} = 3$.

This approach highlights the importance of simplifying expressions, collecting like terms, and isolating variables to solve equations efficiently. Mastering these steps enables you to handle a wide range of algebraic problems, whether you are plugging in values immediately or rearranging formulas to find a general solution. By practicing these techniques, you develop flexibility in problem-solving and deepen your understanding of algebraic relationships.

Solve a Formula for a Specific Variable Example 2

To solve for the variable b in the given formula, there are two effective methods that demonstrate different approaches to algebraic manipulation. The original formula is:

\(a = \frac{1}{2} B h + \frac{1}{2} b h\)

In the first method, the process begins by distributing the constants and variables, which simplifies the expression. After distributing, the equation remains the same but is prepared for collecting like terms. The goal is to isolate the variable b, so terms containing b are moved to one side, and all other terms to the opposite side. This is done by subtracting \(\frac{1}{2} B h\) from both sides:

\(a - \frac{1}{2} B h = \frac{1}{2} b h\)

Next, to isolate b, both sides are divided by h and multiplied by 2, effectively canceling the fractions and the variable h on the right side:

\(\frac{2a}{h} - B = b\)

This results in the formula rewritten with b isolated:

\(b = \frac{2a}{h} - B\)

The second method approaches the problem by first dividing both sides of the original equation by h and multiplying by 2, which simplifies the equation before isolating b:

\(\frac{2a}{h} = B + b\)

Then, subtracting B from both sides isolates b:

\(b = \frac{2a}{h} - B\)

Both methods arrive at the same solution, demonstrating that algebraic equations can be manipulated in different sequences to isolate a variable effectively. This flexibility in problem-solving is crucial for mastering algebraic techniques and applying them confidently in various contexts.

Rearrange the following formula to solve for the indicated variable.
$S=2(x+xz+yz)$
Solve for $y$.

• A. $y=\frac{\frac{S}{2}-x-xz}{z}$
• B. $y=\frac{\frac{S}{2}-x+xz}{z}$
• C. $y=\frac{\frac{S}{2}-x-xz}{2z}$
• D. $y=\frac{\frac{S}{2}+x-xz}{2z}$

Answer: (A)

Rearrange the following formula to solve for the indicated variable.
$T=50+\frac{N-40}{4}$
Solve for $N$.

• A. $N=4\left(T-50\right)-40$
• B. $N=4\left(T+50\right)+40$
• C. $N=4\left(50-T\right)+40$
• D. $N=4\left(T-50\right)+40$

Answer: (D)

Solve a Formula for a Specific Variable Example 3

To determine the temperature recorded on the fifth day given the average temperature over five days, we start by identifying the known and unknown values. The average temperature for the five days is 79 degrees. The temperatures for the first four days are 75, 82, 80, and 79 degrees, respectively. The unknown temperature for the fifth day is represented as e.

The formula for the average temperature over five days is:

\[\frac{a + b + c + d + e}{5} = \text{average}\]

Substituting the known values, we have:

\[\frac{75 + 82 + 80 + 79 + e}{5} = 79\]

To solve for e, multiply both sides of the equation by 5 to eliminate the denominator:

\[75 + 82 + 80 + 79 + e = 5 \times 79\]

Calculating the right side:

\[75 + 82 + 80 + 79 + e = 395\]

Next, sum the known temperatures on the left side:

\[316 + e = 395\]

Isolate e by subtracting 316 from both sides:

\[e = 395 - 316\]

Which simplifies to:

\[e = 79\]

Therefore, the temperature recorded on the fifth day was 79 degrees. This approach demonstrates how to use the average formula and algebraic manipulation to find an unknown value when given the average and other data points.

Here’s what students ask on this topic:

How do you use the formula $d=st$ to solve distance, speed, and time problems?

To use the formula $d=st$, first identify the known and unknown variables: distance ($d$), speed ($s$), and time ($t$). Plug in the known values for speed and time into the formula. Then, multiply speed by time to find the distance. For example, if a train travels at 60 km/h for 2.5 hours, substitute $s=60$ and $t=2.5$ to get $d=60\; imes\; 2.5=150$ kilometers. If solving for time or speed, rearrange the formula by isolating the unknown variable using algebraic operations like division. Always check units to ensure your answer makes sense, such as hours for time or kilometers for distance.

What are the steps to isolate a variable in a formula with multiple variables?

To isolate a variable in a formula with multiple variables, follow these steps: First, simplify both sides of the equation by combining like terms or distributing if needed. Second, collect all terms containing the target variable on one side and move all other terms to the opposite side using addition or subtraction. Third, isolate the variable by dividing or multiplying both sides by the coefficient attached to the variable. For example, in the formula $\mathrm{3w}=l+h$, to isolate $w$, divide both sides by 3 to get $w=rac\{l+h\}\{3\}$. This process allows you to solve for the variable and then substitute known values if needed.

How can you solve a formula for one variable before plugging in values?

Solving a formula for one variable before plugging in values involves algebraic manipulation. Start by simplifying the equation if possible. Then, use addition or subtraction to move all terms not containing the target variable to the other side. Next, isolate the variable by dividing or multiplying both sides by the coefficient of the variable. For example, given $\mathrm{3w}-l=h$, add $l$ to both sides to get $3w\; =\; l\; +\; h$. Then divide both sides by 3 to isolate $w$: $w\; =\; rac\{l\; +\; h\}\{3\}$. After isolating the variable, you can plug in any values for the other variables to quickly find the solution.

Why is it important to check units when solving formulas like distance = speed × time?

Checking units is crucial when solving formulas like $d=st$ because it ensures the answer is meaningful and correctly interpreted. For example, if speed is in kilometers per hour and time is in hours, multiplying them cancels the hours unit, leaving the distance in kilometers. Without verifying units, you might misinterpret the result or mix incompatible units, leading to incorrect answers. Always confirm that units align properly and cancel as expected to maintain accuracy and clarity in your solutions.

How do you solve word problems using formulas in algebra?

To solve word problems using formulas in algebra, first identify the known and unknown quantities described in the problem. Write down the formula that relates these quantities. Next, substitute the known values into the formula. If the unknown variable is not isolated, use algebraic steps such as simplifying, collecting like terms, and isolating the variable by division or multiplication. Finally, solve for the unknown and interpret the result in the context of the problem, including units. This systematic approach helps translate real-world situations into mathematical expressions and find accurate solutions.