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 requires simplifying, collecting terms, and dividing to solve linear equations. This process enhances understanding of terms, coefficients, and exponents within polynomial expressions. Mastery of these steps allows efficient problem-solving in standard form equations, supporting skills in manipulating expressions and applying formulas to real-world scenarios.

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 distance formula, expressed as \(d = s \times t\), relates distance (\(d\)), speed (\(s\)), and time (\(t\)). To effectively use a formula, begin by identifying the known and unknown variables, substitute the known values into the formula, and then solve for the unknown variable.

Consider a scenario where a train travels at a speed of 60 kilometers per hour for 2.5 hours. By substituting \(s = 60\) km/h and \(t = 2.5\) hours into the distance formula, the distance traveled is calculated as \(d = 60 \times 2.5 = 150\) kilometers. The units are crucial here; multiplying kilometers per hour by hours results in kilometers, confirming the distance measurement.

In another example, if the distance traveled is 357 miles and the speed is 85 miles per hour, but the time is unknown, the formula \(d = s \times t\) can be rearranged to solve for time: \(t = \frac{d}{s}\). Substituting the known values gives \(t = \frac{357}{85} = 4.2\) hours. Understanding unit cancellation is essential—miles divided by miles per hour simplifies to hours, which aligns with the time measurement.

Using formulas in word problems integrates algebraic manipulation and unit analysis, reinforcing problem-solving skills. Whether isolating variables directly or applying linear equation techniques, the process remains systematic: identify, substitute, and solve. Mastery of this approach enhances the ability to tackle diverse real-world problems involving distance, speed, time, and beyond.

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 $m^3$.
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. $63℃$
• B. $127℃$
• C. $35℃$
• D. $70.6℃$

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 unknown. Substituting the known depth into the formula, the calculation proceeds as follows:

The expression involves subtracting 50 from the depth, dividing the result by 200, and then adding 0.5:

\(p = \frac{h - 50}{200} + 0.5\)

Plugging in h = 120 meters:

\(p = \frac{120 - 50}{200} + 0.5 = \frac{70}{200} + 0.5 = 0.35 + 0.5 = 0.85 \text{ MPa}\)

This calculation shows that the pressure on the sediment at 120 meters depth is 0.85 megapascals. Understanding how pressure varies with sediment depth is crucial in geology for assessing sediment compaction and stability.

Solve Applied Problems Using Formulas Example 2

To determine the temperature at which the Fahrenheit value is numerically twice the Celsius value, we start with the relationship between Fahrenheit (F) and Celsius (C) given by the formula:

\[ C = \frac{5}{9} (F - 32) \]

Since we want to find when Fahrenheit is twice Celsius, we substitute \(F = 2C\) into the equation:

\[ C = \frac{5}{9} (2C - 32) \]

Distributing the fraction on the right side gives:

\[ C = \frac{5}{9} \times 2C - \frac{5}{9} \times 32 \]

Simplifying the terms:

\[ C = \frac{10}{9} C - \frac{160}{9} \]

To isolate the variable terms on one side, add \(\frac{160}{9}\) to both sides:

\[ C + \frac{160}{9} = \frac{10}{9} C \]

Then subtract \(C\) from both sides to collect like terms:

\[ \frac{160}{9} = \frac{10}{9} C - C \]

Since \(C\) is equivalent to \(\frac{9}{9} C\), the right side simplifies to:

\[ \frac{160}{9} = \frac{10}{9} C - \frac{9}{9} C = \frac{1}{9} C \]

Multiplying both sides by 9 to solve for \(C\):

\[ 160 = C \]

This means the temperature in Celsius is 160 degrees when Fahrenheit is twice Celsius.

To verify, substitute \(C = 160\) back into the original conversion formula to find Fahrenheit:

\[ 160 = \frac{5}{9} (F - 32) \]

Multiply both sides by the reciprocal \(\frac{9}{5}\):

\[ 160 \times \frac{9}{5} = F - 32 \]\[ \frac{1440}{5} = F - 32 \]\[ 288 = F - 32 \]

Add 32 to both sides:

\[ 288 + 32 = F \]\[ 320 = F \]

Since 320 is exactly twice 160, the solution is confirmed. Therefore, the temperature at which the Fahrenheit reading is twice the Celsius reading is 160°C and 320°F.

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. This involves combining like terms, moving all variable terms to one side and constants to the other, and isolating the target variable by performing inverse operations such as division or multiplication. 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, which is especially useful when you need to calculate the variable for multiple sets of values. Starting with the equation \)3w = l + h\(, you isolate \)w\( by dividing both sides by 3, yielding the formula \)w = \frac{l + h}{3}\(. This rearranged formula allows you to quickly substitute different values for \)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$.

The key steps in solving for a variable within a formula include simplifying both sides of the equation, collecting all terms containing the target variable on one side, and isolating that variable through inverse operations. Mastering these steps enhances your ability to manipulate equations efficiently and apply formulas flexibly across various problems.

Solve a Formula for a Specific Variable Example 3

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 to simplify the expression. After distributing, the equation remains the same but is prepared for collecting like terms. The goal is to isolate the lowercase b term. This is done by subtracting \(\frac{1}{2} B h\) from both sides, resulting in:

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

Next, to isolate b, divide both sides by h and multiply by 2, which cancels the fractions and the variable h on the right side:

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

This final expression shows b isolated, expressed in terms of a, B, and h.

The second method rearranges the steps by first dividing both sides of the original equation by h and multiplying by 2, simplifying the equation to:

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

Then, subtracting B from both sides isolates b:

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

Both methods yield the same result, demonstrating that algebraic equations can be manipulated in various sequences to isolate a variable effectively. Understanding these techniques enhances problem-solving flexibility and deepens comprehension of algebraic principles.

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 4

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:

\[75 + 82 + 80 + 79 + e = 5 \times 79\]\[75 + 82 + 80 + 79 + e = 395\]

Next, sum the known temperatures:

\[75 + 82 + 80 + 79 = 316\]

Isolate e by subtracting 316 from both sides:

\[e = 395 - 316\]\[e = 79\]

Therefore, the temperature recorded on the fifth day was 79 degrees. This problem demonstrates how to apply the concept of averages and algebraic manipulation to find an unknown value when given a set of data points and their average.

Here’s what students ask on this topic:

How do you use the formula $d=st$ to find distance, speed, or time?

To use the formula $d=st$, first identify the known and unknown variables. For example, if you know speed ($s$) and time ($t$), you can find distance ($d$) by multiplying speed by time. If distance and speed are known, solve for time by dividing distance by speed: $t=\frac{d}{s}$. Similarly, if distance and time are known, solve for speed by dividing distance by time: $s=\frac{d}{t}$. Always check units to ensure your answer makes sense, such as kilometers for distance or hours for time. This process involves substituting known values into the formula and isolating the unknown variable.

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

Isolating a variable in a formula with multiple variables involves three main 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 other terms to the opposite side using addition or subtraction. Third, isolate the target variable by dividing or multiplying both sides by the coefficient attached to it. For example, to isolate $w$ in $3w\; =\; l\; +\; h$, divide both sides by 3 to get $w=\frac{l}{h}3$. This method allows you to rewrite formulas to solve for any variable efficiently.

How do you solve word problems using formulas in algebra?

Solving word problems using formulas involves three key steps. First, identify the known and unknown quantities from the problem. Second, substitute the known values into the formula. Third, solve for the unknown variable by simplifying and isolating it using algebraic operations like addition, subtraction, multiplication, or division. For example, if a train travels at 60 km/h for 2.5 hours, use $d=st$ to find distance: $d=\; 60\; imes\; 2.5\; =\; 150$ kilometers. Always pay attention to units to ensure your answer is meaningful and correct.

Why is it important to understand units when working with formulas like distance = speed × time?

Understanding units is crucial when working with formulas because units tell you what the numbers represent and ensure your calculations make sense. For example, in the formula $d=st$, if speed is in kilometers per hour and time is in hours, multiplying them gives distance in kilometers. Units help you verify that your answer matches the quantity you are solving for. Ignoring units can lead to incorrect answers or confusion. Always include units in your calculations and check that they cancel or combine properly to give the correct unit for the unknown variable.

How can you rewrite a formula to solve for a different variable?

To rewrite a formula to solve for a different variable, follow these steps: simplify the equation if needed, collect all terms containing the target variable on one side, and isolate that variable by performing inverse operations. For example, starting with $d=st$, to solve for $t$, divide both sides by $s$: $t=\frac{d}{s}$. This process allows you to create a new formula that directly gives the value of the variable you want, making it easier to plug in values and solve problems efficiently.