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2. Equations & Inequalities in Two Variables
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Integrated Math 1

2.01 Slope-intercept form

Lesson
Practice
Lesson

Concept summary

The slope-intercept form of a line is:

\displaystyle y=mx+b
\bm{m}
The slope of the line
\bm{b}
The y-intercept of the line

A benefit of slope-intercept form is that we can easily identify two key features from the equation.

Rate of change

The ratio of change in one quantity to the corresponding change in another quantity

Slope

The ratio of change in the vertical direction (y-direction) to change in the horizontal direction (x-direction).

To interpret the slope, it can be helpful to look at the units.

x
y
y-intercept

A point where a line or graph intersects the y-axis. The value of x is 0 at this point, so it often represents the initial value or flat fee.

x
y
x-intercept

A point where a line or graph intersects the x-axis. The value of y is 0 at this point.

x
y

Based on the context, some values might be calculated algebraically, but are not reasonable based on the restrictions of the scenario. For example, time and lengths generally cannot be negative, which can create restrictions on the possible values x and y can take on.

Viable solutions

A valid solution that makes sense within the context of the question or problem

Non-viable solutions

An algebraically valid solution that does not make sense within the context of the question or problem

A special case of slope-intercept form is y=b, when m=0. This will be a horizontal line since the rise will be 0.

We can also have vertical lines, but their equations will not be in slope intercept form as they have an undefined slope.

x
y
A horizontal line of the form y=b
x
y
A vertical line of the form x=a

Worked examples

Example 1

Draw the graph of the line 4x+2y=10 by first converting to slope-intercept form.

Approach

We will:

  1. Rearrange the equation to the form y=mx+b
  2. Identify the slope and y-intercept from the equation
  3. Plot the y-intercept
  4. Use the rise and run to plot another point
  5. Draw the graph of the line

Solution

First we need to convert to slope-intercept form by solving for y.

\displaystyle 4x+2y\displaystyle =\displaystyle 10Given
\displaystyle 2y\displaystyle =\displaystyle -4x+10Subtraction property of equality
\displaystyle y\displaystyle =\displaystyle -2x+5Division property of equality
-4
-3
-2
-1
1
2
3
4
x
-1
1
2
3
4
5
6
7
y

The y-intercept is \left(0,5\right).

The slope is m=-2, so we can have:

\text{rise}=-2 and \text{run}=1

This means that going down 2 units and right by 1 unit gives another point on the line.

Reflection

Sometimes we will see this equation written as y=5-2x, but this does not change the graph.

Example 2

Imogen is a cyclist. She typically bikes at 15 \, \text{mph}. She is doing a 50 mile bike ride for charity.

a

Draw a graph that shows her distance remaining throughout the 50 mile ride if she bikes her typical speed.

Approach

We need to decide what our axes should represent before we can graph anything.

Since we are given her speed in miles per hour and speed typically is the slope of our graph we know that \dfrac{\text{miles}}{\text{hour}} \to \dfrac{\text{Rise}}{\text{Run}}, so we can use the units of "miles" for our y-axis and "hours" for our x-axis.

We also need to consider what scale would be appropriate. We know the distance goes from 50 down to 0, so we can use the speed to determine and appropriate scale for the x-axis.

Solution

We can use a table of values to see when the distance gets down below 0 as we don't need to go beyond that.

Time (hours)01234
Distance remaining (miles)5035205-10
1
2
3
4
\text{Time (hours)}
5
10
15
20
25
30
35
40
45
50
55
\text{Distance remaining (miles)}

This would be an appropriate scale and axes.

1
2
3
4
\text{Time (hours)}
5
10
15
20
25
30
35
40
45
50
55
\text{Distance remaining (miles)}

Initially, when \text{Time}=0, the distance remaining would be 50, this means we have a y-intercept of (0,50). Then using our labeled axes, we can use the rate of change, -15 \, \text{mph}, as the slope of the line to graph or use our table of values.

b

Write the linear equation that represents the graph in part (a).

Approach

We can use x and y as variables, or any letters of our choice, as long as we state what they represent.

Since this is a linear function, we can write the equation in slope-intercept form.

Solution

Let t represent the time in hours that Imogen has been biking for.

Let d represent the distance in miles remaining after t hours.

Slope: -15

y-intercept: 50

Equation: d=-15t+50

Reflection

We can also write the equation as d=50-15t.

c

Explain whether or not we can predict the distance remaining after 5 hours.

Approach

We can use the equation from part (b) or the graph from part (a), but either way, we need to consider whether or not the solution we find is viable.

Solution

1
2
3
4
5
\text{Time (hours)}
-20
-10
10
20
30
40
50
\text{Distance remaining (miles)}

Looking at the graph and extending it beyond the x-intercept, we can see that for any x-value greater than 3 \frac{1}{3} the distance remaining would be negative. Since 5>3 \frac{1}{3}, we would get a negative distance remaining in our prediction.

We can conclude that using this model, we cannot have a viable prediction for the distance remaining after 5 hours.

Reflection

If we had used the equation from part (b), we would get a negative answer which is not a viable distance remaining as she stops biking after 50 miles.

\displaystyle d\displaystyle =\displaystyle -15t+50Equation from part (b)
\displaystyle d\displaystyle =\displaystyle -15(5)+50Substitute t=5
\displaystyle d\displaystyle =\displaystyle -75+50Evaluate the product
\displaystyle d\displaystyle =\displaystyle -25Evaluate the sum

Example 3

A bathtub has a clogged drain, so it needs to be pumped out. It currently contains 30 gallons of water.

The table of values shows the linear relationship of the amount of water remaining in the tub, y, after x minutes.

\text{Time in minutes } (x)0123
\text{Water remaining in gallons } (y)30282624
a

Determine the linear equation in slope-intercept form that represents this situation.

Approach

We can pick two points to calculate the rate of change for the slope. Then we can recognize that the y-intercept is given in the table of values.

Solution

Find the slope using the values \left(0,30\right) and \left(1,28\right):

\displaystyle m\displaystyle =\displaystyle \dfrac{y_2-y_1}{x_2-x_1}
\displaystyle =\displaystyle \dfrac{28-30}{1-0}
\displaystyle =\displaystyle -2

Notice that the initial value, or y-intercept is given in the table as \left(0,30\right).

The equation that represents this situation is y=-2x+30.

Reflection

If we had not noticed that the y-intercept was given, then we could have substituted in any pair of values for x and y, and solved for b.

b

Draw the graph of this linear relationship with a clearly labeled scale. Only show the viable solutions.

Approach

We can't have a negative time (x \geq 0) and we should end the graph when the tub is empty (y=0). We need to find the time when the tub is empty.

Solution

To find when the tub is empty:

\displaystyle y\displaystyle =\displaystyle -2x+30Given equation
\displaystyle 0\displaystyle =\displaystyle -2x+30Substitute y=0
\displaystyle -30\displaystyle =\displaystyle -2xSubtraction property of equality
\displaystyle 15\displaystyle =\displaystyle xDivision property of equality

So we have the restriction that 0 \leq x\leq 15

The point at x=0 is \left(0,30\right).

The point at x=15 is \left(15,0\right).

5
10
15
\text{Time }(x)
5
10
15
20
25
30
\text{Water }(y)
c

Describe how the graph would change if instead there was initially 40 gallons of water in the tub and it emptied at 2.5 gallons per minute.

Approach

The initial value is represented by the y-intercept on the graph. The rate of change is represented by the slope of the graph. We should consider how these are changing from the original question.

Solution

The original question had an initial value of 30 gallons and the new scenario has an initial value of 40 gallons. This means the new y-intercept will be higher on the y-axis.

The original question had a rate of change of -2 as it was decreasing at 2 gallons per minute. The new scenario has a rate of change of -2.5 as it is emptying at 2.5 gallons per minute. This means the slope will be steeper in the new scenario.

We can see this on the graph:

5
10
15
20
\text{Time }(x)
5
10
15
20
25
30
35
40
\text{Water }(y)

Reflection

Notice that despite starting with 10 extra gallons of water, the tub with 40 gallons of water only takes 1 more minute to empty than the 30 gallon tub due to the steeper slope.

Outcomes

M1.N.Q.A.1

Use units as a way to understand real-world problems.*

M1.N.Q.A.1.A

Choose and interpret the scale and the origin in graphs and data displays.*

M1.A.SSE.A.2

Use the structure of an algebraic expression to identify ways to rewrite it.

M1.A.CED.A.2

Create equations in two variables to represent relationships between quantities and use them to solve problems in a real-world context. Graph equations with two variables on coordinate axes with labels and scales, and use the graphs to make predictions.*

M1.A.CED.A.3

Create individual and systems of equations and/or inequalities to represent constraints in a contextual situation, and interpret solutions as viable or non-viable.*

M1.A.CED.A.4

Rearrange formulas to isolate a quantity of interest using algebraic reasoning.

M1.A.REI.D.4

Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line).

M1.F.LE.B.3

Interpret the parameters in a linear or exponential function in terms of a context.*

M1.MP2

Reason abstractly and quantitatively.

M1.MP3

Construct viable arguments and critique the reasoning of others.

M1.MP4

Model with mathematics.

M1.MP5

Use appropriate tools strategically.

M1.MP6

Attend to precision.

M1.MP7

Look for and make use of structure.

M1.MP8

Look for and express regularity in repeated reasoning.

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