02/14 2. Linear Equations and Applications

Solving Linear Equations

An equation is a statement that two expressions have the same value. The central idea is balance: any operation used to change one side must also be used on the other side.

A linear equation in one variable has the general form

where $a$, $b$, and $c$ are constants and $a\ne0$. The variable has exponent $1$ and is not multiplied by another variable or placed in a denominator or radical.

The Properties of Equality allow you to preserve a true equation:

• Add the same quantity to both sides.

• Subtract the same quantity from both sides.

• Multiply both sides by the same quantity.

• Divide both sides by the same nonzero quantity.

A dependable solution process is:

1. Simplify each side by distributing and combining like terms.

2. Collect variable terms on one side.

3. Collect constants on the other side.

4. Divide or multiply to isolate the variable.

5. Substitute the result into the original equation to check it.

For example, solve

Subtract $2x$ from both sides, add $7$ to both sides, and divide by $3$:

Checking gives $5(6)-7=23$ and $2(6)+11=23$, so the solution is correct.

To remove fractions, multiply every term by the least common denominator. For

multiplying every term by $6$ gives

so $x=1$.

An equation can have one solution, no solution, or infinitely many solutions. For example, $3x+2=11$ has one solution, $2x+5=2x+9$ reduces to the false statement $5=9$ and has no solution, and $4x+8=4(x+2)$ reduces to the true statement $0=0$ and has infinitely many solutions.

Takeaway: Simplify first, preserve balance by doing the same operation on both sides, isolate the variable, and check the result.

Rearranging Formulas

A literal equation contains several variables, and the goal is to isolate the variable named in the question. Treat the other variables as constants while applying the Properties of Equality.

For the area formula

solve for $h$ by multiplying both sides by $2$ and then dividing by $b$, assuming $b\ne0$:

The same approach works with the distance formula. From

solving for $t$ gives

assuming $r\ne0$.

After rearranging a formula, substitute known values and retain the units. If a rectangle has area $84\text{ cm}^2$ and length $12\text{ cm}$, then using $A=lw$ gives

The width is $7\text{ cm}$.

Takeaway: Identify the requested variable, undo operations in reverse order, state any nonzero assumptions, and substitute values only after the formula has been rearranged.

Using Proportions

A proportion is an equation in which two ratios are equal:

with nonzero denominators. Cross-multiplication produces the equivalent equation

For example, solve

Cross-multiplying gives

so $120=5x$ and $x=24$.

In applications, corresponding quantities must occupy corresponding positions, and their units must be compatible. If $4$ notebooks cost $\$7.20$, the cost of $9$ notebooks can be represented by

Cross-multiplication gives

so

Thus, $9$ notebooks cost $\$16.20$ at the same unit price.

Takeaway: Set up matching quantities in the same relative positions, verify compatible units, and cross-multiply only when the denominators are nonzero.

Solving Percent Problems

A percent represents a ratio per $100$. Convert it to a decimal by dividing by $100$:

The percent equation is

Use the wording of the problem to identify the unknown:

• “What number is $20\%$ of $80$?” becomes $n=0.20(80)$.

• “$15$ is what percent of $60$?” becomes $15=p(60)$.

• “$30$ is $25\%$ of what number?” becomes $30=0.25n$.

For example, $35\%$ of $240$ is

To find the percent, solve $18=p(72)$:

To find the whole, solve $45=0.30n$:

For a percent increase or decrease, use

where $r$ is the decimal rate. Use $+$ for an increase and $-$ for a decrease. An $\$80$ item discounted by $15\%$ costs

so the sale price is $\$68$. The discount itself is $0.15(80)=\$12$.

For a tax, tip, or commission added to a base amount, calculate the percentage and add it. A $20\%$ tip on a $\$52$ bill is $0.20(52)=\$10.40$, making the total $\$62.40$.

Takeaway: Translate percent language carefully, use a decimal multiplier, and distinguish the part, the percent, and the whole.

Modeling Applications

Applications become manageable when a verbal situation is translated into a mathematical relationship.

1. Identify what is known and what must be found.

2. Define a variable with a clear meaning.

3. Write an equation using a formula or relationship.

4. Solve the equation.

5. Check the result in the original situation.

6. State the answer with appropriate units and meaning.

For number problems, translate phrases directly. If the sum of a number and $17$ is $42$, let $n$ represent the number:

so $n=25$.

Consecutive integers differ by $1$. If the first is $n$, the next two are $n+1$ and $n+2$. A sum of $72$ gives

so $n=23$. The integers are $23$, $24$, and $25$.

For geometry, define the dimensions before using the formula. If a rectangle has perimeter $50$ meters and its length is $3$ meters more than its width, let the width be $w$ and the length be $w+3$. Then

which gives $w=11$ and length $14$. The dimensions are $11\text{ m}$ by $14\text{ m}$.

For rate problems, use

A cyclist traveling $42$ miles in $3$ hours has average speed

or $14$ miles per hour. Units must be consistent before solving.

For money applications, total cost often equals quantity times unit price, while earnings may equal hourly rate times hours. If a worker earns $\$18$ per hour and receives $\$72$ for overtime, then

so $h=4$ overtime hours.

Common checks include distributing negative signs correctly, combining only like terms, applying multiplication or division to every term, delaying rounding until the final step, and asking whether the result is reasonable in context.

Takeaway: Define the variable, translate the relationships, solve with units, and interpret the answer in the original context.

• rate, distance, and time