Master Rewriting Equations in Standard Form and Understand the Standard Form of a Linear Equation

In this lesson, a new form of linear equations will be introduced. Given a linear function in this form, its key components such as

x -

and

y -

intercepts will be identified and used to draw its graph. Additionally, converting between the forms of linear equations will be discussed.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Coordinate Plane
Least Common Denominator
Slope-Intercept Form
Point-Slope Form

Discussion

Standard Form of a Line

In the standard form of a line all $x -$ and $y -$ terms are on one side of the linear equation or function and the constant is on the other side.

$A x + B y = C$

In this form, $A,$ $B,$ and $C$ are real numbers. It is important to know that $A$ and $B$ cannot both be $0 .$ Different combinations of $A,$ $B,$ and $C$ can represent the same line on a graph. It is preferred to use the smallest possible whole numbers for $A,$ $B,$ and $C$ and it is also better if $A$ is a positive number.

Pop Quiz

Identifying the Standard Form of a Linear Equation

Consider the given linear equation that shows the relationship between the variables $x$ and $y .$ Determine whether the equation is written in standard form or not.

Linear equation written in different forms

Discussion

Graphing a Linear Function in Standard Form

A linear function written in standard form has quickly identifiable

x -

and

y -

intercepts. Since two points determine a line, this provides enough information to graph the function. Consider the following linear equation written in standard form.

3 x + 5 y = 30

The graph of this function can be drawn in two steps.

Find the Intercepts

Begin by substituting

y = 0

to find the

x -

intercept of the equation.

3 x + 5 y = 30

Substitute

$y = 0$

3 x + 5 (0) = 30

▼

Solve for

x

ZeroPropMult

Zero Property of Multiplication

3 x + 0 = 30

IdPropAdd

Identity Property of Addition

3 x = 30

DivEqn

$LHS / 3 = RHS / 3$

x = 10

The

x -

intercept is

(10, 0) .

The

y -

intercept can be found in a similar way. Substitute

x = 0

into the equation and solve for

y .

3 x + 5 y = 30

Substitute

$x = 0$

3 (0) + 5 y = 30

▼

Solve for

y

ZeroPropMult

Zero Property of Multiplication

0 + 5 y = 30

IdPropAdd

Identity Property of Addition

5 y = 30

DivEqn

$LHS / 5 = RHS / 5$

y = 6

The

y -

intercept is

(0, 6) .

Equation : x - intercept : y - intercept : 3 x + 5 y = 30 (10, 0) (0, 6)

Plot the Intercepts

Now it is time to plot the intercepts in a coordinate plane.

Only intercepts of 3x+5y=30 are depicted as points: y-intercept at (0, 6) and x-intercept at (10, 0).

Draw the Line Passing Through the Intercepts

Lastly, draw a line passing through these points.

Graph of the function 3x+5y=30 with y-intercept at (0,6) and x-intercept at (10,0).

Extra

General Formulas for the Intercepts of an Equation in Standard Form

Note that general formulas for the intercepts can be derived for any linear function written in standard form $A x + B y = C .$

Assumption	$x -$ intercept	$y -$ intercept
$A \neq = 0,$ $B \neq = 0$	$(\frac{C}{A}, 0)$	$(0, \frac{C}{B})$
$A = 0,$ $B \neq = 0$	The line is horizontal, $y = \frac{C}{B},$ so it does not cross the $x -$ axis.	$(0, \frac{C}{B})$
$A \neq = 0,$ $B = 0$	$(\frac{C}{A}, 0)$	The line is vertical, $x = \frac{C}{A},$ so it does not cross the $y -$ axis.

Example

Using Intercepts to Graph a Linear Equation

On his way home from school, Ignacio stops to buy fruit at a market in his neighborhood. Oranges cost

$ 3

per kilogram and apples cost

$ 4

per kilogram. He has

$ 24

to spend. The following linear equation models this situation.

3 x + 4 y = 24

Here,

x

represents the number of kilograms of oranges and

y

represents the number of kilograms of apples.

a Find and interpret the intercepts of the linear equation.

b Graph the equation.

Answer

x - intercept :

(8, 0)

$y - intercept :$ $(0, 6)$

Interpretation: See solution.

Hint

a To find the intercepts, substitute

0

for one of the variables and solve the equation for the other variable.

b Plot the intercepts and connect them with a line segment. Recall that the number of kilograms cannot be negative.

Solution

a The given equation models the total cost of the fruit, where

x

is the number of kilograms of oranges purchased and

y

is the number of kilograms of apples purchased.

3 x + 4 y = 24

The

x -

intercept will be found first. To do so, substitute

0

for

y

and solve the equation for

x .

3 x + 4 y = 24

Substitute

$y = 0$

3 x + 4 (0) = 24

▼

Solve for

x

ZeroPropMult

Zero Property of Multiplication

3 x + 0 = 24

IdPropAdd

Identity Property of Addition

3 x = 24

DivEqn

$LHS / 3 = RHS / 3$

x = 8

The point

(8, 0)

is the

x -

intercept, which means that if Ignacio does not buy any apples, he can buy

8

kilograms of oranges. Next, the

y -

intercept will be found. To do so, substitute

0

for

x

and solve for

y .

3 x + 4 y = 24

Substitute

$x = 0$

3 (0) + 4 y = 24

▼

Solve for

y

ZeroPropMult

Zero Property of Multiplication

0 + 4 y = 24

IdPropAdd

Identity Property of Addition

4 y = 24

DivEqn

$LHS / 4 = RHS / 4$

y = 6

The

y -

intercept is

(0, 6) .

This means that if Ignacio does not buy any oranges, he can buy

6

kilograms of apples.

b In order to graph the equation, the intercepts found in the previous part can be used.

Equation : x - intercept : y - intercept : 3 x + 4 y = 24 (8, 0) (0, 6)

Plot them on a coordinate plane and connect them with a line.

Since the number of kilograms of fruit purchased cannot be negative, only positive values of $x$ and $y$ make sense in this context.

Example

Using the Standard Form of a Linear Equation to Model a Real-Life Situation

LaShay has one part-time job that she works at after school and a second part-time job that she works at on weekends. One pays

$ 7

per hour and the other pays

$ 10

per hour. She wants to make

$ 350

per week.

Amount Paid Per Hour Job I : $ 7 Job II : $ 10

a Write a linear equation in standard form that describes this situation and draw its graph.

b If LaShay is allowed to work only

21

hours per week at the

$ 10

per hour job, how many hours does she have to work per week at the other job in order to make

$ 350 ?

Answer

a Equation:

7 x + 10 y = 350

Graph:

20

hours per week

Hint

a To find the intercepts, substitute

0

for one of the variables and solve the equation for the other variable.

b Substitute the given value into the equation from Part A.

Solution

a Let

x

be the number of hours worked at Job I and

y

be the number of hours worked at Job II. Then, the amount of money LaShay can make from each part-time job can be written in terms of

x

and

y .

Job	Amount Paid Per Hour $($)$	Amount LaShay Makes $($)$
I	$7$	$7 x$
II	$10$	$10 y$

Since LaShay wants to make

$ 350

per week, the sum of

7 x

and

10 y

should be equal to

350 .

7 x + 10 y = 350

To graph this equation, its intercepts will be found. Substitute

y = 0

to find the

x -

intercept and

x = 0

to find the

y -

intercept.

	$7 x + 10 y = 350$
Operation	$x -$ intercept	$y -$ intercept
Substitution	$7 x + 10 (0) = 350$	$7 (0) + 10 y = 350$
Calculation	$x = 50$	$y = 35$
Point	$(50, 0)$	$(0, 35)$

Now, plot the intercepts on a coordinate plane and connect them with a line segment. Since the number of hours worked cannot be negative, only positive values of $x$ and $y$ make sense.

b Recall that the variable

y

in the equation written in Part A represents the number of hours worked at Job II. Therefore, by substituting

21

for

y

into the equation, the number of hours that LaShay needs to work per week at Job I can be found.

7 x + 10 y = 350

Substitute

$y = 21$

7 x + 10 (21) = 350

▼

Solve for

x

Multiply

7 x + 210 = 350

SubEqn

$LHS - 210 = RHS - 210$

7 x = 140

DivEqn

$LHS / 7 = RHS / 7$

x = 20

LaShay needs to work

20

hours per week at Job I in order to achieve her goal.

Discussion

Writing a Linear Equation in Standard Form

Any linear equation can be rewritten in standard form. Consider the following linear equation that is written in slope-intercept form.

y = \frac{4}{5} x + \frac{2}{3}

Using the Properties of Equality, the equation can be rewritten in standard form.

A x + B y = C

Here,

A,

B,

and

C

are real numbers and

A

and

B

cannot both be equal to

0 .

It can be noted that representing

A,

B,

and

C

with the smallest possible integers is preferred, as well as

A

being positive.

Remove All Fractions

When a linear equation contains fractions, the first step is to remove all fractions. The equation is multiplied by the least common denominator of the fractions. In this case, it is the product of

5

and

3,

which is

15 .

Using the Multiplication Property of Equality, the equation can be written as follows.

y = \frac{4}{5} x + \frac{2}{3}

MultEqn

$LHS \cdot 15 = RHS \cdot 15$

15 y = (\frac{4}{5} x + \frac{2}{3}) 15

▼

Simplify right-hand side

Distr

Distribute $15$

15 y = \frac{4}{5} x \cdot 15 + \frac{2}{3} \cdot 15

CommutativePropMult

Commutative Property of Multiplication

15 y = \frac{4}{5} \cdot 15 \cdot x + \frac{2}{3} \cdot 15

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

15 y = \frac{4 \cdot 1 5}{5} x + \frac{2 \cdot 1 5}{3}

ReduceFrac

$\frac{a}{b} = \frac{a / 5}{b / 5}$

15 y = \frac{4 \cdot 3}{1} x + \frac{2 \cdot 1 5}{3}

ReduceFrac

$\frac{a}{b} = \frac{a / 3}{b / 3}$

15 y = \frac{4 \cdot 3}{1} x + \frac{2 \cdot 5}{1}

DivByOne

$\frac{a}{1} = a$

15 y = 4 \cdot 3 x + 2 \cdot 5

Multiply

15 y = 12 x + 10

Rearrange the Terms of the Equation

Now, move the terms containing variables to the left-hand side of the equation. In addition to this, move the constant to the right-hand side using the Subtraction Property of Equality.

15 y = 12 x + 10

SubEqn

$LHS - 12 x = RHS - 12 x$

- 12 x + 15 y = 10

Since a positive coefficient for

x

is preferable, the equation can be multiplied by

- 1 .

- 12 x + 15 y = 10

MultEqn

$LHS \cdot (- 1) = RHS \cdot (- 1)$

12 x - 15 y = - 10

This equation is now in the standard form. Note that the values of

A,

B,

and

C

are in their smallest possible integer forms.

Example

Writing a Linear Equation in Standard Form From a Graph

Jordan wants to buy some songs and movies online to enjoy after school. She can buy songs for $$ 0.75$ each and movies for $$ 5$ each. The graph represents the relationship between the number of songs purchased $x$ and the number of movies purchased $y .$

Write an equation in standard form that describes the relationship between

x

and

y .

Give the answer such that

A,

B,

and

C

are the smallest possible integers and

A

is positive.

Hint

Start by writing the equation of the line in point-slope form. Then, convert it into the standard form.

Solution

From the given graph, the $x -$ and $y -$ intercepts can be identified.

The intercepts are

(60, 0)

and

(9, 0) .

When two points on a line are known, the point-slope form can be used to write the equation of the line. Recall that an equation in point-slope form follows a specific format.

y - y_{1} = m (x - x_{1})

In this form,

m

is the slope and

(x_{1}, y_{1})

is a point on the graph of the line. To find the slope of the line, the intercepts can be used. Substitute them into the Slope Formula.

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

SubstitutePoints

Substitute $(60, 0)$ & $(0, 9)$

m = \frac{9 - 0}{0 - 6 0}

▼

Simplify

SubTerms

Subtract terms

m = \frac{9}{- 6 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 3}{b / 3}$

m = \frac{3}{- 2 0}

MoveNegDenomToFrac

Put minus sign in front of fraction

m = - \frac{3}{2 0}

Using the

y -

intercept

(0, 9)

— or any other point on the line — and the slope

- \frac{3}{2 0},

the equation of the line can be written.

y - 9 = - \frac{3}{2 0} (x - 0) \Leftrightarrow y - 9 = - \frac{3}{2 0} x

Finally, this equation needs to be converted into the standard form. To do so, all fractions will be removed and the variable terms will be on the left-hand side of the equation.

y - 9 = - \frac{3}{2 0} x

MultEqn

$LHS \cdot 20 = RHS \cdot 20$

20 y - 180 = - 3 x

AddEqn

$LHS + 3 x = RHS + 3 x$

3 x + 20 y - 180 = 0

AddEqn

$LHS + 180 = RHS + 180$

3 x + 20 y = 180

This equation is in standard form.

Alternative Solution

Recall the standard form of a linear equation.

A x + B y = C

In the context of the problem,

x

is the number of songs Jordan can purchase and

y

is the number of movies she can purchase. If the values of

A

and

B

are considered as the costs of a song and a movie, respectively, then the equation shows the amount of money Jordan spends.

A x + B y = C ⇓ 0.75 x + 5 y = C

Now, the value of

C

can be calculated by using one of the intercepts.

The graph intercepts the axes at

(60, 0)

and

(0, 9) .

For simplicity, the

y -

intercept

(0, 9)

will be used.

0.75 x + 5 y = C

SubstituteII

$x = 0$ , $y = 9$

0.75 (0) + 5 (9) = C

▼

Solve for

C

ZeroPropMult

Zero Property of Multiplication

0 + 5 (9) = C

Multiply

0 + 45 = C

IdPropAdd

Identity Property of Addition

45 = C

RearrangeEqn

Rearrange equation

C = 45

Therefore, the relationship between

x

and

y

can be expressed by the following equation.

0.75 x + 5 y = 45

The number on the right hand-side

45

can be interpreted as Jordan's budget for her multimedia purchases. Finally, since it is preferred to rewrite the coefficients as the smallest possible integers, multiply the equation by

4 .

0.75 x + 5 y = 45

MultEqn

$LHS \cdot 4 = RHS \cdot 4$

3 x + 20 y = 180

Example

Converting Between the Forms of a Linear Equation

Dominika and Ali are working on an extra credit assignment after school. They have been given a linear equation written in standard form to solve.

3 x + 4 y = - 33

They converted the equation into alternative forms of a linear equation.

a What linear equation forms did Dominika and Ali write?

b Are their calculations correct? If not, describe and correct any mistakes.

Answer

a Dominika: Point-Slope Form
Ali: Slope-Intercept Form

b Are Ali's calculations right? Yes.
Are Dominika's calculations right? No, see solution

Hint

a A linear equation in slope-intercept form has the form

y = m x + b .

A linear equation in point-slope form has the form

y - y_{1} = m (x - x_{1}) .

b Pay close attention when factoring out a negative number.

Solution

a Consider the last lines of the steps.

The equation found by Dominika is written in point-slope form and the other equation is written in slope-intercept form.

Point-Slope Form	Slope-Intercept Form
$y - y_{1} = m (x - x_{1})$	$y = m x + b$
$Dominika ’ s Equation y - \frac{3}{4} = - \frac{3}{4} (x - 12)$	$Ali ’ s Equation y = - \frac{3}{4} x + (- \frac{3 3}{4})$

b It can be seen that Ali wrote the given equation in slope-intercept form correctly. He started by subtracting

3 x

from both sides of the equation. Then, he applied the Division Property of Equality without making a mistake.

However, Dominika's calculations are not entirely correct. Until the last step everything is correct. However, she made a mistake when factoring out $- \frac{3}{4} .$

The given equation can be written correctly in point-slope form as follows.

y - \frac{3}{4} = - \frac{3}{4} x - \frac{3 6}{4}

▼

Factor out

- \frac{3}{4}

Rewrite

Rewrite $36$ as $3 \cdot 12$

y - \frac{3}{4} = - \frac{3}{4} x - \frac{3 \cdot 1 2}{4}

MovePartNumRight

$\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

y - \frac{3}{4} = - \frac{3}{4} x - \frac{3}{4} \cdot 12

FactorOut

Factor out $- \frac{3}{4}$

y - \frac{3}{4} = - \frac{3}{4} (x + 12)

Pop Quiz

Finding the Values of $A,$ $B,$ and $C$

Consider the standard form of a linear equation.

A x + B y = C

In general,

A,

B,

and

C

are real numbers. However, it is preferred for

A,

B,

and

C

to be the smallest possible integers and for

A

to be positive. With this information in mind, write the values of

A,

B,

and

C

by finding the equation of the given line in standard form. Give the answers such that

A,

B,

and

C

are the smallest possible integers and

A

is positive.

Closure

Exploring the Standard Form of a Linear Equation

Throughout the lesson, the standard form of a linear equation has been discussed.

A x + B y = C

In general,

A,

B,

and

C

are real numbers. However, it was previously noted that there are preferred properties for these numbers.

$A,$ $B,$ and $C$ are the smallest possible integers. In other words, the greatest common factor of these three numbers is $1 .$
$A$ is positive or $0 .$
$A$ and $B$ cannot both equal $0 .$

For each line there is exactly one equation in standard form that meets these properties. However, infinitely many equivalent linear equations in standard form can be obtained by using the Multiplication Property of Equality. Linear equations are equivalent if they describe the same line.

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Catch-Up and Review

Extra

Answer

Hint

Solution

Answer

Hint

Solution

Hint

Solution

Alternative Solution

Answer

Hint

Solution