**Solving a Linear Equation**

The key to solving equations symbolically is to

understand equality. Many students think of

equality as a signal “to do something.” For

example, students encounter the following:

Students think of an equality as calculating a

set of numbers to get an answer. This can be a

source of many misconceptions. Instead, equality

is a statement that states two quantities are equal.

In this unit, students develop understanding of

equality. Equality can be thought of as a

“balance.” To solve an equation means to

maintain the equality between two quantities.

There are several ways to find the value of a

variable in a linear relationship if the value of the

other variable is known. For instance, in

C = 150 + 10n, where C is the rental cost and n

is the number of bikes, we might want to know

the cost to rent 75 bikes or how many bikes we

can rent for $300. In either case, we are solving for

one of the variables or we are solving a linear

equation with one variable or “one unknown.”

Students can find the value of the variable by

using these methods.

• Solving the equation using symbolic methods.

• Interpreting the information from a table or

graph.

• Reasoning about the situation in verbal

form—“if it costs $10 per bike, then 75 bikes

will cost $750 plus the fixed costs of $150 for a

total of $900.”

Investigation 3 develops symbolic methods for

solving equations. To solve an equation

symbolically, we write a series of equivalent

equations until we have one in which it is easy to

read the value of the variable. Equivalent

equations have the same solutions. Equality or

equivalence can be maintained by adding,

subtracting, multiplying by, or dividing by the

same quantity on both sides of the equation. For

multiplication and division, the quantity must be

nonzero. These procedures are called the

**properties of equality.**

To help students develop their understanding

of equality, Investigation 3 begins by making the

connection to tables and graphs. Linear equations

can be solved using tables or graphs. For example,

the equation 750 = 150 + 10n is associated with

the related linear function y = 150 + 10x. If

y = 750, then we are looking for a corresponding

value of x. Or, conversely, we can find a

corresponding value of y if we are given a value of

x. At first we expect that students will use graphs

or tables and, in some cases, simply substitute

values into the equation and calculate the missing

value.

Once the concept and properties of equality

have been explored, though, we move to a

pictorial situation to develop a symbolic method

for solving linear equations. First, students explore

adding, subtracting, multiplying, or dividing the

same number to both sides of a numeric sentence:

85 = 70 + 15.

Next, students explore equations like the

following with coins and pouches.

Each pouch in an equation must have the same

number of coins and the number of coins on both

sides of the equality sign are equal.

Students can solve this by taking 1 coin from

each side. This leaves 4 coins on the left and

2 pouches on the right. Because each pouch must

have the same number of coins, students can

intuitively divide both sides by 2 to find that each

pouch contains 2 coins. This provides a transition

to the more abstract situations of solving linear

equations in one unknown. Students first find the

number of coins using the pictures. Then they

translate each picture into a symbolic statement.

For example, if x represents the number of coins

in a pouch, then the preceding pictorial statement

can be represented as 5 = 2x + 1. Next, students

apply the properties of equality to isolate the

variable—that is, they solve the equation for x.

The set of equations is selected carefully to allow

students an opportunity to look at what each

symbolic statement means.

In this unit, we solve the following types of

equations:

6 - 3x = 10

5 + 17x = 12x – 9

2(x + 3) = 10

Integers and the distributive property were

developed in Accentuate the Negative. Review of

integers and the distributive property are

provided in the Connections section of each ACE.

**Solving a System
of Two Linear Equations
**

Students informally solve systems of linear

equations throughout the unit. They use graphs

and tables to find the point of intersection of two

lines. For example, a problem in Investigation 1

has students comparing pledge plans. Students are

asked to determine if any of the two pledge plans

will have the same amount of money.

C represents the amount of money collected

for x kilometers.

To find the number of kilometers that make

two plans equal, students start by using tables or

graphs. Later, students can solve two of the

preceding equations symbolically. To find when

Gilberto’s plan equals Alana’s plan, they write

2x = 5 + 0.5x and solve for x.

This can be done for either of the two plans.

From the graphs of the three plans, students note

that all three plans will never have the same costs

for a given amount of kilometers. In Investigation

3, students learn that situations like this one

represent a system of linear equations. Without

calling attention to it, students solve a linear

system by finding the point of intersection of the

two lines symbolically. They find the x value when

the y values are equal or by setting the two

equations equal. Students can also use a table or

graph to find the point of intersection.

It is important that students understand what a

solution is, whether they are dealing with a

symbolic solution or a graphical solution, and that

they can connect these two views of a solution.

Thus, is a solution for 2x = 5 + 0.5x. It

means that if Gilberto and Alana each walk

kilometers, they earn the same amount.

Graphically, the lines y = 2x and y = 5 + 0.5x

intersect at .This solution means that

when they each walk kilometers, they each

earn about $6.67.

Ideas about inequality are informally explored

by asking questions like: “if x = 4, does Gilberto

earn more or less than Alana?” Students can

answer this question by finding the missing

coordinate in for each equation, y = 2x

and y = 5 + 0.5x and then comparing the

coordinates. In the Shapes of Algebra unit,

students explore linear inequalities and systems of

linear equations in depth.

**Finding the Equation of a Line
**In this unit, most of the linear equations are of the

form y = mx + b. In some situations, equations

can be obtained by translating the verbal situation

directly into symbols.

To find the equation of a line symbolically, we

find the slope of the line (coefficient of x) and the

y-intercept (the constant term b) and substitute

these values into the equation y = mx + b.

Students are given one of the following pairs of

information about a line:

• The slope and y-intercept

• The slope and a point on the line

• Two points on a line

In the first case, the values are substituted

directly into the equation y = mx + b. In the

second case, the slope is substituted into the

equation y = mx + b and the coordinates of the

point are substituted into the equation to find the

y-intercept. In the third case, the slope is

determined using two points and then another

point is used to find the y-intercept.The following

methods can be used to find the y-intercept of a

line.

Finding the y-Intercept Symbolically

First, find the slope of the line.

Given the two points, (1, 4) and (3, 10), the

slope is 3.

Then substitute the slope into the equation

y = mx + b to obtain y = 3x + b. Since both

points lie on the line, the x and y values of the

points satisfy the equation. Choose one point

[(1, 4)] and substitute it into the equation.

4 = 3(1) + b

Solving for b, b = 1.

Now substitute for b.

y = 3x + 1

This is the equation of the line that passes

through the points (1, 4) and (3, 10).

**Method 2:
Finding the y-Intercept Using a Table**

Some students prefer to find the y-intercept using

a table and working backwards. Consider the

following table.

First, students note that this is a linear

relationship because as x increases by 1 unit, y

increases by 3 units. So, the slope is 3.To find the

y-intercept, they work backwards using the slope.

That is, as x decreases by 1 unit, y decreases by

3 units. This is repeated until x = 0.

The y-intercept is -2 and the equation of the

line is y = 3x - 2. Some students have a strong

sense of the y-intercept as a starting point, to

which you repeatedly add a constant number to

generate the table.

**Method 3:
Finding the y-Intercept Using a Graph
**

Students also use a graph to find the y-intercept.

They extend the line to intercept the y-axis. Some

use the ratio definition of slope to work from a

point on the graph backward or forward until they

hit the y-axis.