Formulas: Videos & Practice Problems

Solving word problems often involves using formulas, which are equations with multiple variables like $d=st$ for distance. The process includes identifying known and unknown quantities, substituting values, and isolating the target variable by simplifying and collecting terms. This approach applies to multivariable polynomials and linear equations alike. Isolating variables before plugging in values allows quick evaluation for different inputs. Understanding how to manipulate formulas enhances problem-solving skills, especially when dealing with terms, coefficients, and exponents in polynomial expressions.

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. Thus, the train travels 150 kilometers.

When solving for a different variable, such as time, rearranging the formula may be required. For example, if the distance \(d\) is 357 miles and the speed \(s\) is 85 miles per hour, solving for time \(t\) involves dividing both sides of the equation by speed: \(t = \frac{d}{s} = \frac{357}{85} = 4.2\). Analyzing the units confirms that miles cancel out, leaving time measured in hours. This confirms that the travel time is 4.2 hours.

Understanding how to manipulate formulas and interpret units is crucial for solving real-world problems involving distance, speed, and time. Whether the solution requires straightforward substitution or algebraic manipulation, the process remains consistent: identify knowns and unknowns, substitute values, and solve while ensuring units are correctly handled. Mastery of these steps enhances problem-solving skills and builds confidence in applying mathematical concepts to everyday situations.

A giant water tank takes the shape of a cylinder. It has a radius of $2$ $m$ 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. $125.7m^3$
• B. $157.1m^3$
• C. $62.8m^3$
• D. $251.3m^3$

Answer: (C)

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

• A. $63°C$
• B. $127°C$
• C. $35°C$
• D. $70.6°C$

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 depth of 120 meters, we use the given formula where p represents pressure in megapascals (MPa) and h represents depth in meters. The problem provides the depth h = 120 meters, and we need to find the corresponding pressure p.

The formula involves subtracting 50 from the depth, dividing the result by 200, and then adding 0.5 to find the pressure:

\[ p = \frac{h - 50}{200} + 0.5 \]

Substituting the known depth:

\[ p = \frac{120 - 50}{200} + 0.5 \]

First, subtract 50 from 120:

\[ 120 - 50 = 70 \]

Next, divide 70 by 200:

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

Finally, add 0.5 to 0.35 to find the pressure:

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

Therefore, the pressure on the sediment layer at a depth of 120 meters is 0.85 megapascals. This calculation demonstrates how pressure increases with depth in sediment layers, an important concept in geology and sedimentology for understanding 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 these directly into the formula and then simplify to find w. This involves simplifying the equation, collecting variable terms on one side and constants on the other, and isolating the target variable by performing inverse operations such as division or multiplication.

Alternatively, you can first rearrange the formula to solve for w in terms of h and l. This process includes simplifying both sides of the equation, collecting all terms containing w on one side, and isolating w by dividing both sides by the coefficient of w. For instance, if the original formula is:

\[3w = l + h\]

then solving for w gives:

\[w = \frac{l + h}{3}\]

Once w is isolated, you can easily substitute different values for l and h to find corresponding values of w. This method is especially useful when you need to calculate w multiple times with varying inputs.

Whether you choose to plug in values first or rearrange the formula before substituting, the key steps remain consistent: simplify the equation, collect like terms to isolate the target variable, and then solve for that variable. Mastering this approach enhances your ability to work flexibly with formulas and solve a wide range of algebraic problems efficiently.

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 lowercase b, so terms containing b are moved to one side, and all other terms to the opposite side. Subtracting \(\frac{1}{2} B h\) from both sides yields:

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

Next, to isolate b, divide both sides by \(h\) and multiply 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 rearranges the steps by first dividing both sides of the original equation by \(h\) and multiplying by 2, which simplifies the equation before collecting like terms:

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

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 various sequences to isolate a variable. This flexibility is crucial for solving equations efficiently and understanding the underlying relationships between variables. Practicing these techniques enhances problem-solving skills 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. $N=4(T-50)-40$
• B. $N=4(T+50)+40$
• C. $N=4(50-T)+40$
• D. $N=4(T-50)+40$

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:

What are the steps to solve word problems using formulas in algebra?

To solve word problems using formulas in algebra, follow three main steps. First, identify the known and unknown quantities in the problem. This means recognizing which variables have given values and which you need to find. Second, substitute the known values into the formula. This involves plugging in the numbers for the variables you know. Third, solve for the unknown variable by simplifying the equation, collecting like terms, and isolating the variable. Sometimes this requires basic arithmetic, and other times you may need to use algebraic techniques like dividing both sides of the equation. Always check the units of measurement to ensure your answer makes sense in context. This process helps you systematically approach problems involving multiple variables and formulas.

How do you isolate a variable in a formula before plugging in values?

Isolating a variable in a formula means rewriting the equation so that the variable you want to solve for is alone on one side. To do this, first simplify both sides of the equation by combining like terms or distributing if needed. Next, collect all terms containing the target variable on one side and move all other terms to the opposite side by adding or subtracting. Finally, isolate the variable by dividing or multiplying both sides by the coefficient attached to it. For example, if you have $\mathrm{3w}=l+h$, divide both sides by 3 to get $w=\frac{l+h}{3}$. This method allows you to plug in different values for other variables quickly without rearranging the formula each time.

How can formulas be used to solve problems involving distance, speed, and time?

Formulas like $d=st$ relate distance ($d$), speed ($s$), and time ($t$). To solve problems, first identify which quantities are known and which is unknown. For example, if speed and time are given, plug those values into the formula to find distance by multiplying speed and time. If distance and speed are known but time is unknown, rearrange the formula to isolate time: $t=\frac{d}{s}$. Then substitute the known values and solve. Always pay attention to units (e.g., kilometers per hour, hours) to ensure the answer is in the correct measurement. This approach helps solve real-world problems involving travel and motion efficiently.

What is the advantage of isolating a variable in a formula before substituting values?

Isolating a variable in a formula before substituting values offers flexibility and efficiency. Once the formula is rearranged to solve for the target variable, you can quickly plug in different values for the other variables without repeating the algebraic steps. This is especially useful when you need to evaluate the formula multiple times with varying inputs. For example, if you isolate $w$ in a formula like $w=\frac{l+h}{3}$, you can easily calculate $w$ for different values of $l$ and $h$ without rearranging the equation each time. This saves time and reduces errors in calculations.

How do you check units when solving formulas in word problems?

Checking units is crucial to ensure your answer makes sense in context. When you multiply or divide quantities with units, apply unit cancellation rules. For example, in the distance formula $d=st$, if speed ($s$) is in kilometers per hour and time ($t$) is in hours, multiplying them cancels the hours unit, leaving kilometers for distance. Similarly, when dividing distance by speed to find time, units cancel appropriately to give time in hours. Always write units alongside your calculations and verify that the final unit matches the quantity you are solving for, such as distance in meters or miles, or time in seconds or hours. This practice helps avoid mistakes and confirms the solution's validity.