Graphing Linear Relationships

a

We will graph this equation by finding and plotting its intercepts, then connecting them with a line. To find the

x

- and

y

-intercepts, we will need to substitute

0

for one variable, solve, then repeat for the other variable.

Finding the $x$ -intercept

Think of the point where the graph of an equation crosses the

x

-axis. The

y

-value of that coordinate pair is equal to

0,

and the

x

-value is the

x

-intercept. To find the

x

-intercept of the given equation, we should substitute

0

for

y

and solve for

x .

9 x + 5 y = 45

Substitute

y = 0

9 x + 5 \cdot 0 = 45

MultiplyMultiply

9 x = 45

DivEqn

LHS / 9 = RHS / 9

x = 5

An

x

-intercept of

5

means that the graph passes through the

x

-axis at the point

(5, 0) .

Finding the $y$ -intercept

Let's use the same concept to find the

y

-intercept. Consider the point where the graph of the equation crosses the

y

-axis. The

x

-value of the coordinate pair at the

y

-intercept is

0 .

Therefore, substituting

0

for

x

will give us the

y

-intercept.

9 x + 5 y = 45

Substitute

x = 0

9 \cdot 0 + 5 y = 45

MultiplyMultiply

5 y = 45

DivEqn

LHS / 5 = RHS / 5

y = 9

A

y

-intercept of

9

means that the graph passes through the

y

-axis at the point

(0, 9) .

Graphing the equation

We can now graph the equation by plotting the intercepts and connecting them with a line.

b

Let's start by identifying the intercepts.

Finding the $x$ -intercept

To find the

x

-intercept we'll substitute

0

for

y

and then solve for

x .

- 2 x + 6.4 y = 16

Substitute

y = 0

- 2 x + 6.4 \cdot 0 = 16

MultiplyMultiply

- 2 x = 16

DivEqn

LHS / - 2 = RHS / - 2

x = - 8

An

x

-intercept of

- 8

means that the graph passes through the

x

-axis at the point

(- 8, 0) .

Finding the $y$ -intercept

To find the

y

-intercept we do the opposite, substituting

x

for

0 .

- 2 x + 6.4 y = 16

Substitute

x = 0

- 2 \cdot 0 + 6.4 y = 18

MultiplyMultiply

6.4 y = 16

DivEqn

LHS / 6.4 = RHS / 6.4

y = \frac{1 6}{2 . 4}

CalcQuotCalculate quotient

y = 2.5

A

y

-intercept of

2.5

means that the graph passes through the

y

-axis at the point

(0, 2.5) .

Graphing the equation

We can now graph the equation by plotting the intercepts and connecting them with a line.

c

One again, we start with the intercepts.

Finding the $x$ -intercept

Here,

y

is substituted with

0

and then solve the equation for

x .

8 x - 2 y = - 8

Substitute

y = 0

8 x + 2 \cdot 0 = - 8

MultiplyMultiply

8 x = - 8

DivEqn

LHS / 8 = RHS / 8

x = - 1

The

x

-intercept is

(- 1, 0) .

Finding the $y$ -intercept

For the

y

-intercept we'll substitute

x

with

0

instead.

8 x - 2 y = - 8

Substitute

x = 0

8 \cdot 0 - 2 y = - 8

MultiplyMultiply

- 2 y = - 8

DivEqn

LHS / - 2 = RHS / - 2

y = 4

The

y

-intercept is

(0, 4) .

Graphing the equation

We can now graph the equation by plotting the intercepts and connecting them with a line.