6. Writing and Graphing Equations in Standard Form
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Chapter 2
6.
Writing and Graphing Equations in Standard Form
This lesson delves into the intricacies of two key mathematical topics: rewriting equations into standard form and understanding what the standard form of a linear equation means. It serves as a step-by-step guide, enriched with practical examples, to help you navigate these complex subjects. Ideal for students aiming to excel in algebra or geometry, as well as teachers looking for instructional lessons, this content is a valuable educational asset. The use cases extend beyond academics to fields like engineering, computer science, and economics, where understanding linear equations is essential.
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Lesson Settings & Tools
| | 11 Theory slides |
| | 13 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Writing and Graphing Equations in Standard Form
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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.
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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.
Ax+By=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.
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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.
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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. 3x+5y=30 The graph of this function can be drawn in two steps.
1
Find the Intercepts
expand_more Begin by substituting y= 0 to find the x-intercept of the equation.
3x+5y=30
Substitute
y= 0
3x+5( 0)=30
▼
Solve for x
ZeroPropMult
Zero Property of Multiplication
3x+0=30
IdPropAdd
Identity Property of Addition
3x = 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.
3x+5y=30
Substitute
x= 0
3( 0)+5y=30
▼
Solve for y
ZeroPropMult
Zero Property of Multiplication
0+5y=30
IdPropAdd
Identity Property of Addition
5y = 30
DivEqn
.LHS /5.=.RHS /5.
y = 6
The y-intercept is (0,6). rc Equation: & 3x+5y = 30 x-intercept: & (10,0) y-intercept: & (0,6)
2
Plot the Intercepts
expand_more Now it is time to plot the intercepts in a coordinate plane.
3
Draw the Line Passing Through the Intercepts
expand_more Lastly, draw a line passing through these points.
Extra
General Formulas for the Intercepts of an Equation in Standard FormNote that general formulas for the intercepts can be derived for any linear function written in standard form Ax+ By= C.
| Assumption | x-intercept | y-intercept |
|---|---|---|
| A≠ 0, B≠ 0 | (C/A,0) | (0,C/B) |
| A= 0, B≠ 0 | The line is horizontal, y= C B, so it does not cross the x-axis. | (0,C/B) |
| A≠ 0, B= 0 | (C/A,0) | The line is vertical, x= C A, so it does not cross the y-axis. |
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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. 3x+4y = 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
a x-intercept: (8,0)
y-intercept: (0,6)
b
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.
3x+4y=24 The x-intercept will be found first. To do so, substitute 0 for y and solve the equation for x.
3x+4y=24
Substitute
y= 0
3x+4( 0)=24
▼
Solve for x
ZeroPropMult
Zero Property of Multiplication
3x+0=24
IdPropAdd
Identity Property of Addition
3x=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.
3x+4y=24
Substitute
x= 0
3( 0)+4y=24
▼
Solve for y
ZeroPropMult
Zero Property of Multiplication
0+4y=24
IdPropAdd
Identity Property of Addition
4y=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.
rc Equation: & 3x+4y = 24 x-intercept: & (8,0) y-intercept: & (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.
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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 [-0.9em] 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 $350?
Answer
a Equation: 7x+10y=350
Graph:
Graph:
b 20 hours per week
Hint
a To find the intercepts, substitute 0 for one of the variables and solve the equation for the other variable.
b Substitute the given value into the equation from Part A.
Solution
a Let x be the number of hours worked at Job I and y be the number of hours worked at Job II. Then, the amount of money LaShay can make from each part-time job can be written in terms of x and y.
| Job | Amount Paid Per Hour ($) | Amount LaShay Makes ($) |
|---|---|---|
| I | 7 | 7 x |
| II | 10 | 10 y |
Since LaShay wants to make $350 per week, the sum of 7x and 10y 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.
| 7x+10y=350 | ||
|---|---|---|
| Operation | x-intercept | y-intercept |
| Substitution | 7x+10( 0)=350 | 7( 0)+10y=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.
LaShay needs to work 20 hours per week at Job I in order to achieve her goal.
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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 = 4/5x + 2/3 Using the Properties of Equality, the equation can be rewritten in standard form. Ax+By=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.
1
Remove All Fractions
expand_more 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= 4/5x + 2/3
MultEqn
LHS * 15=RHS* 15
15y=(4/5x+2/3)15
▼
Simplify right-hand side
Distr
Distribute 15
15y=4/5x* 15+2/3* 15
CommutativePropMult
Commutative Property of Multiplication
15y=4/5* 15* x+2/3* 15
MoveRightFacToNum
a/c* b = a* b/c
15y=4* 15/5x+2* 15/3
ReduceFrac
a/b=.a /5./.b /5.
15y=4* 3/1x+2* 15/3
ReduceFrac
a/b=.a /3./.b /3.
15y=4* 3/1x+2* 5/1
DivByOne
a/1=a
15y=4* 3x+2* 5
Multiply
Multiply
15 y = 12x + 10
2
Rearrange the Terms of the Equation
expand_more 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.
Since a positive coefficient for x is preferable, the equation can be multiplied by - 1.
This equation is now in the standard form. Note that the values of A, B, and C are in their smallest possible integer forms.
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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.
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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 = y_2-y_1/x_2-x_1
SubstitutePoints
Substitute ( 60,0) & ( 0,9)
m=9- 0/0- 60
▼
Simplify
SubTerms
Subtract terms
m=9/- 60
ReduceFrac
a/b=.a /3./.b /3.
m=3/- 20
MoveNegDenomToFrac
Put minus sign in front of fraction
m=- 3/20
Using the y-intercept ( 0, 9) — or any other point on the line — and the slope - 320, the equation of the line can be written. y- 9= -3/20(x- 0)⇔ y-9=- 3/20x 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=- 3/20x
MultEqn
LHS * 20=RHS* 20
20y-180 = - 3x
AddEqn
LHS+3x=RHS+3x
3x+20y -180= 0
AddEqn
LHS+180=RHS+180
3x+20y = 180
This equation is in standard form.
Alternative Solution
Recall the standard form of a linear equation.
Ax+By=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.
Ax+By=C ⇓ 0.75x+5y = 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.
0.75x+5y = 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
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. 0.75x+5y = 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.
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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. 3x+4y=- 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
Ali: Slope-Intercept Form
b Are Ali's calculations right? Yes.
Are Dominika's calculations right? No, see solution
Are Dominika's calculations right? No, see solution
Hint
a A linear equation in slope-intercept form has the form y=mx+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 = mx+ b |
| Dominika's Equation [-0.7em] y- 3/4= -3/4(x- 12) | Ali's Equation [-0.7em] y = -3/4x+ ( -33/4) |
b It can be seen that Ali wrote the given equation in slope-intercept form correctly. He started by subtracting 3x 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 - 34.
The given equation can be written correctly in point-slope form as follows.
y-3/4=-3/4x-36/4
▼
Factor out -3/4
Rewrite
Rewrite 36 as 3* 12
y-3/4=-3/4x-3* 12/4
MovePartNumRight
a* b/c=a/c* b
y-3/4=-3/4x-3/4* 12
FactorOut
Factor out -3/4
y-3/4=-3/4(x+12)
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Pop Quiz
Finding the Values of A, B, and C
Consider the standard form of a linear equation. Ax+By= 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.
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Closure
Exploring the Standard Form of a Linear Equation
Throughout the lesson, the standard form of a linear equation has been discussed. Ax+By=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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Writing and Graphing Equations in Standard Form
Exercise 1.1
Determine the form in which each linear equation is written.
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We will begin by reviewing the three forms in which a linear equation can be written.
| Form | Equations | Features |
|---|---|---|
| Slope-Intercept Form | y=mx+n | m is the slope and b is the y-intercept. |
| Point-Slope Form | (y-y_1)=m(x-x_1) | m is the slope and (x_1,y_1) is the coordinate of any point on the line. |
| Standard Form | Ax+By=C | A, B, and C are real numbers and A and B are not both zero. |
We can now determine the form of the given linear equations. Let's analyze the first equation. y+7 = - (x-21) ⇕ y-( - 7)= -1(x- 21) We can identify the point ( 21, - 7) and the slope m= - 1. This means that this equation is in point-slope form. Next, we will consider the second equation. 7x-y= 21 ⇕ 7x+(- 1)y=21 This one is in standard form, where 7, - 1, and 21 are real numbers. Finally, we can identify the form of our last equation. y= 21x + 7 This is in slope-intercept form, where -2 is the slope and 7 is the y-intercept.
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Solution
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Think of the point where the graph of an equation crosses the x-axis. The y-value of that ( x, y) coordinate pair is 0, and the x-value is the x-intercept. To find the x-intercept of the equation, substitute 0 for y and solve for x.
An x-intercept of - 6 means that the graph passes through the x-axis at the point ( - 6,0).
Find the y-intercept following a similar fashion. Consider the point where the graph of the equation crosses the y-axis. The x-value of the ( x, y) coordinate pair at the y-intercept is 0. Therefore, substituting 0 for x will give the y-intercept.
The y-intercept of the equation is 24, which means that the graph passes through the y-axis at the point (0, 24).
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Solution
Ignacio's school organizes a benefit performance to cover some school expenses. An adult ticket costs $5 and a student ticket costs $2.50. Ignacio expects to make $600 from the performance. The linear equation below models the situation. 2.5x+5y = 600 In this equation, x represents the number of students and y represents the number of adults. Which graph represents the solution set of the equation?
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The given equation models the earnings from the performance, where x is the number of students and y is the number of adults. 2.5x+5y=600 To graph this equation, we will first find its intercepts. This can be done by substituting 0 for one variable and solving the equation for the other variable. Let's first substitute 0 for y and solve the equation for x.
The point (240,0) is the x-intercept, which means that 240 students can attend the performance if no adults attend. Next, we will find the y-intercept. To do so, we substitute 0 for x and solve for y.
The y-intercept is (0,120), which means that 120 adults can attend the performance if no students attend. Finally, we can use the intercepts to draw the graph. rc Equation: & 2.5x+5y = 600 x-intercept: & (240,0) y-intercept: & (0,120) Let's plot these points on a coordinate plane and connect them with a line.
We can only consider the first quadrant because a negative number of attendees makes no sense in this context.
This corresponds to option C.
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Solution
Write the equation in standard form using integers. y = 2/5x+3/4
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Equations in standard form are written in a specific format. Ax+ By= C Here, A, B, and C are integers and A and B cannot both be zero simultaneously. We will use the Properties of Equality to rewrite the given equation in this form.
Now we can see that our equation follows the standard form. Ax+ B y&= C 8x+ (- 20)y&= - 15
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Solution
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Let's examine Ali's solution.
In the first step, Ali subtracts 9x from both sides of the equation. Here, we can say that the Subtraction Property of Equality is correctly applied. Then, he multiplies the left-hand side of the equation by - 1, but forgets to multiply the same number on the other side of the equation.
Therefore, we can say that the Multiplication Property of Equality is not applied correctly. The answer is C.
Consider the standard form of a linear equation.
Ax+By=C
In this equation A, B, and C are real numbers, and A and B cannot be zero simultaneously. In this form, the variable terms should be on the left-hand side and the constant term should be on the right-hand side. Let's write the given equation in standard form using the Properties of Equality.
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Solution
The diagram shows the graph of a linear equation.
Which equation has the graph shown in the diagram? rr I: & - 3x + 2y = 2 II: & - 3x -2 y = - 2 III:& 3x + 2y = - 2 IV: & 3x - 2y = 2 V: & - 3x - 2y = 2
{"type":"choice","form":{"alts":["I","II","III","IV","V"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":3}
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We will first write the equation of the line in slope-intercept form. y = mx + b In this form, m is the slope of the line and b is the y-intercept. We can directly identify the slope and the y-intercept on the given graph.
The y-intercept b is - 1 and the slope m is 32. Let's substitute these values into the slope-intercept form. y = mx + b ⇓ y = 3/2x + ( - 1) Now, we need to rewrite this equation in standard form. To do so, we remove the fraction and then move the variable terms to the left-hand side of the equation.
The equation is in standard form, which corresponds to the equation in IV.
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What is y= - 73x-5 written in standard form using integers? A& 7/3x+y= 5 [1em] B& 7x+3y= - 5 [1em] C& - 7x+3y= 15 [1em] D& 7x+3y= - 15
{"type":"choice","form":{"alts":["A","B","C","D"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":3}
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Equations in standard form are written in a specific format. Ax+ By= C In this form, A, B, and C are integers and A and B cannot both be zero. We will use the Properties of Equality to rewrite the given equation in this form. To get rid of the fraction, we need to multiply both sides by its denominator.
Now we can see that our equation follows the standard form. Ax+ By&= C 7x+ 3y&= - 15 Therefore, the correct answer is D.
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Writing and Graphing Equations in Standard Form
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