What are linear relationships?

A linear relationship is any relationship between two variables that creates a line when graphed in the $xy$-plane. Linear relationships are very common in everyday life.

Recognizing Linear Relationships

1. Maya is $3$ inches taller than Geoff. How can we express the relationship between Maya's height in inches, $y$, and Geoff's height in inches, $x$?

Solution: Maya's height equals Geoff's height plus $3$:

$$\begin{aligned} \text{Maya's height} &= \text{Geoff's height} + 3 \\ y &= x+3 \end{aligned}$$

2. Tai runs $2$ miles every day. How can we express the relationship between the distance Tai runs in miles, $y$, and the number of days they run, $x$?

Solution: The distance equals $2$ miles times the number of days:

$$\begin{aligned} \text{distance} &= 2 \cdot \text{days} \\ y &= 2x \end{aligned}$$

Linear relationships

Linear equations can be used to represent the relationship between two variables, most commonly $x$ and $y$. To form the simplest linear relationship, we can make our two variables equal:

By plugging numbers into the equation, we can find some relative values of $x$ and $y$.

If we plot those points in the $xy$-plane, we create a line. The equation $y=x$ is graphed in the $xy$-plane, with the points $(0,0)$, $(1,1)$, $(2,2)$, and $(3,3)$

Every possible linear relationship is just a modification of this simple equation. We might multiply one of the variables by a coefficient or add a constant to one side of the equation, but we'll still be creating a linear relationship.

How do we translate word problems into linear equations?

Modeling real-world scenarios

For a walkthrough of modeling real-world scenarios with linear equations, see this Khan Academy video.

Translating word problems

It may not be hard to translate "Maya is $3$ inches taller than Geoff" into a linear equation, but some word problems are several sentences long, and the information we need to build an equation may be scattered around.

Translating Word Problems into Equations

1. A car with a price of $17,000 is to be purchased with an initial payment of $5,000 and monthly payments of $240. Which equation can be used to find the number of monthly payments, $m$, required to complete the purchase, assuming there are no taxes or fees?

Solution: We're given three important values here:

• $17,000 is the total price, so that's what everything else needs to add up to.
• $5,000 is a one-time payment.
• $240 is a constant amount paid every month, so it needs to be multiplied by $m$, the number of months.

The total price, $17,000, equals the sum of the other payments: the initial $5,000 payment and the $240 paid each month ($m$):

$$17{,}000 = 5{,}000+240m$$

2. The width of a rectangular vegetable garden is $w$ feet. The length of the garden is $8$ feet longer than its width. Which expression gives the perimeter, in feet, of the vegetable garden in terms of $w$?

Solution: We're only given one value here, but the context provides the rest of the information we need:

• The width of the garden is $w$ feet.
• The length of the garden ($\ell$) is $8$ feet more than the width, so $w+8$ feet.
• The perimeter of a rectangle, $P$, is the sum of its two lengths and two widths: $P=2\ell+2w$.

(The garden's four sides are labeled $w$, $w+8$, $w$, and $w+8$ going around the rectangle.)

We can write the equation for the perimeter and substitute $w+8$ for $\ell$:

$$\begin{aligned} P &= 2\ell+2w \\ &= 2(w+8)+2w \\ &= 2w+16+2w \\ &= 4w+16 \end{aligned}$$

The perimeter is equal to the sum of $16$ and $4$ times the width: $P=4w+16$.

3. The concession stand at a high school baseball game sold bags of peanuts for $2.50 each and hot dogs for $3.00 each. If the concession stand brought in $196 and sold $42$ hot dogs, how many bags of peanuts did the concession stand sell?

Solution: We have a number of important values to keep track of here:

• $196 is the total amount of money, so that's what everything else needs to add up to.
• $3.00 is the money from each hot dog sold.
• $42$ is the number of hot dogs sold, so we multiply that by $3.00 to find the money from selling all of them.
• $2.50 is the money from each bag of peanuts sold.
• The only unknown is how many bags of peanuts were sold, so that's our variable: $p$.

We can write an equation that represents this relationship — total money equals hot dog money plus peanut money:

$$196 = 3.00(42)+2.50(p)$$

Simplifying and solving for $p$:

$$\begin{aligned} 196 &= 126+2.50p \\ 70 &= 2.50p \\ 28 &= p \end{aligned}$$

The concession stand sold $28$ bags of peanuts.

Linear functions

Any linear equation with two variables is technically a function. Linear functions are usually written in either slope-intercept form or standard form. We need a thorough and flexible understanding of these forms in order to approach many questions about linear relationships.

Slope-intercept form

The slope-intercept form of a linear function, $y=mx+b$, where $m$ and $b$ are constants, tells us both the slope and the $y$-intercept of the line:

• The slope is equal to $m$.

  • Slope describes the rate of change in a linear relationship.
  • On a graph, slope is the direction and steepness of the line.
  • In a word problem, slope is the value that's applied multiple times, for instance a monthly payment on a large purchase.
  • Words like "rate", "per", and "each" are clues that we're looking at a value that's the slope.

  If we have any two points from a linear relationship, we can calculate slope by dividing the change in $y$ by the change in $x$:

  $$\text{slope}=\frac{\text{change in }y}{\text{change in }x}=\frac{y_2-y_1}{x_2-x_1}$$

• The $y$-intercept is equal to $b$.

  • The $y$-intercept of a line is the $y$-value when $x=0$.
  • On a graph, the $y$-intercept is the point where the line crosses the $y$-axis.
  • In a word problem, the $y$-intercept represents a constant value applied one time, for instance the initial down payment on a large purchase.
  • Words like "initial", "starting", and "one-time" are clues that we're looking at a value that's the $y$-intercept.

  If we know the slope of a linear relationship, we can plug any point into the linear equation and solve for the $y$-intercept.

Standard form

The standard form of a linear function, $Ay+Bx=C$, where $A$, $B$, and $C$ are constants, will often be used in word problem scenarios that have two inputs, instead of an input and an output. To find the slope or $y$-intercept of a line in standard form, it's often most convenient to convert the equation to slope-intercept form by isolating $y$.

More Word-Problem Patterns

Beyond straightforward linear relationships, a few recurring patterns show up again and again in algebra word problems: work-rate, mixture, and percent-change problems. Each has its own setup, but all three are still just linear equations once you translate the words correctly.

Work-Rate Problems

In a work-rate problem, an agent (a pipe, a worker, a machine) completes a job at a steady rate. If something takes $t$ hours to finish a job alone, its rate is $\frac{1}{t}$ jobs per hour. When two agents work together, their rates add, and the combined time to finish is $1$ divided by the combined rate.

1. A tank is filled by two pipes. Pipe A fills the tank in $6$ hours, and pipe B fills it in $10$ hours. How long does it take to fill the tank if both pipes run together?

Solution: Each pipe's rate is the reciprocal of the time it takes alone, in tanks per hour:

$$\begin{aligned} \text{combined rate} &= \frac{1}{6}+\frac{1}{10} \\ &= \frac{5}{30}+\frac{3}{30} \\ &= \frac{8}{30} \\ &= \frac{4}{15} \text{ tanks per hour} \end{aligned}$$

The time to fill one tank is the reciprocal of the combined rate:

$$\text{time} = \frac{1}{4/15} = \frac{15}{4} \text{ hours} = 3\tfrac{3}{4} \text{ hours}$$

Mixture Problems

In a mixture problem, a percentage describes a fraction of an evolving total. The key is to track the actual amount of the ingredient you care about — not just its percentage — both before and after the mixture changes.

1. (2007 AMC 8 #17) A $30$-liter paint mixture is $25\%$ red, $30\%$ yellow, and $45\%$ water. Five liters of yellow tint are added to the mixture. What percent of the new mixture is yellow?

Solution: First, find the actual amount of yellow paint in the original mixture:

$$0.30 \times 30 = 9 \text{ liters of yellow}$$

After adding $5$ liters of yellow tint, the new yellow amount and new total volume are:

$$\begin{aligned} \text{new yellow} &= 9+5 = 14 \text{ liters} \\ \text{new total} &= 30+5 = 35 \text{ liters} \end{aligned}$$

The new percentage of yellow is:

$$\frac{14}{35} = \frac{2}{5} = 40\%$$

Percent-Change Problems

A common trap: successive percent changes multiply, they don't add or cancel. A $p\%$ increase multiplies a value by $\left(1+\frac{p}{100}\right)$, and a $p\%$ decrease multiplies it by $\left(1-\frac{p}{100}\right)$.

1. A price increases by $20\%$ and then decreases by $20\%$. What is the net percent change?

Solution: Multiply the two scale factors together:

$$1.2 \times 0.8 = 0.96$$

A result of $0.96$ means the final price is $96\%$ of the original — a net $4\%$ decrease, not $0\%$. This happens because the second decrease of $20\%$ is taken from the already-increased (larger) price, so it removes more than the first increase added.

Things to Remember

The slope-intercept form of a linear equation, $y=mx+b$, tells us both the slope and the $y$-intercept of the line:

• The slope is equal to $m$.
• The $y$-intercept is equal to $b$.

We can write the equation of a line as long as we know either of the following:

• The slope of the line and a point on the line
• Two points on the line

For the additional word-problem patterns above:

• Work-rate: combined rate = sum of individual rates ($\frac{1}{t_1}+\frac{1}{t_2}$); total time = $1 \div$ combined rate.
• Mixture: track the actual quantity of the ingredient of interest, not just its percentage, before and after the change.
• Percent change: successive percent changes multiply as scale factors $\left(1\pm\frac{p}{100}\right)$ — they do not simply add or cancel out.

Practice Problems