8.EE.C.7Common CoreMathExpressions and EquationsGrade 8
8.EE.C.7: Solving Linear Equations with One, No or Infinitely Many Solutions
In plain English: 8.EE.C.7 is the Common Core grade 8 math standard that asks students to solve linear equations in one variable, including equations with fractions, decimals, parentheses and variables on both sides. Students also show whether an equation has one solution, no solution or infinitely many by simplifying it to x = a, a = b or a = a.
Solve linear equations in one variable.
a.Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers).
b.Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
Common Core State Standards for Mathematics · Domain: Expressions and Equations (EE) · Cluster: Analyze and solve linear equations and pairs of simultaneous linear equations. Also written as 8.EE.7 · Official standard
Students solve linear equations in one variable that need several steps: distributing, collecting like terms (terms with the same variable part), moving variable terms from one side to the other, and working with fraction and decimal coefficients (the numbers multiplied by the variable). Every step makes a simpler equation with the same solutions, and students learn to name the step they used.
The second idea is that not every equation has exactly one answer. When the variable cancels, what is left decides the case: a true statement such as 12 = 12 means infinitely many solutions, and a false statement such as 3 = -4 means no solution. Students give their own examples of each type and prove the type by transforming the equation.
Learning Objectives
By the end of this lesson, students will be able to:
Solve linear equations in one variable with variables on both sides
Solve equations that need the distributive property and collecting like terms
Solve equations with fraction, decimal and negative coefficients
Decide whether an equation has one solution, no solution or infinitely many solutions by transforming it to x = a, a = b or a = a
Write their own examples of equations with one, no and infinitely many solutions
Prior Knowledge Required
Students should already be comfortable with:
Using the distributive property and collecting like terms in expressions 7.EE.A.1
Solving equations of the forms px + q = r and p(x + q) = r 7.EE.B.4
Adding, subtracting, multiplying and dividing fractions, decimals and negative numbers 7.NS.A.3
Checking whether a number makes an equation true by substituting it 6.EE.B.5
Read the number trick aloud. Every student picks a different starting number and follows the steps on paper.
Warm-Up Prompt
"Pick any number. Double it. Add 10. Take half of the result. Subtract the number you picked. What do you get?" Then: "Can you find a number n that makes n + 2 = n + 3 true?"
Everyone gets 5, whatever number they picked. Write the trick as an equation, (2n + 10) ÷ 2 - n = 5, and ask: "Which numbers make this true?" All of them. Then look at n + 2 = n + 3. A number plus 2 can never equal the same number plus 3, so no number works. Tell students that today they will learn to tell these cases apart from equations that have exactly one answer.
Direct Instruction20 minutes
Define the key words and write each one on an anchor chart:
Linear equation in one variable: an equation with one letter, such as x, where the letter is never squared, multiplied by itself or in a denominator. Example: 3x - 4 = x + 6.
Solution: a number that makes the equation true when you substitute it for the variable.
Coefficient: the number multiplied by the variable. In -2.5x the coefficient is -2.5. A constant is a number with no variable. Rational number coefficients are coefficients that are fractions, decimals or negative numbers, such as (2/3)x or 0.4x.
Recall from grade 7: the distributive property says a(b + c) = ab + ac, so 3(x - 2) = 3x - 6. Like terms are terms with the same variable part, such as 4x and -x, or 5 and -2. You collect like terms by adding them.
Equivalent equations have exactly the same solutions. Each legal step (distributing, collecting like terms, adding or subtracting the same thing on both sides, multiplying or dividing both sides by the same number that is not 0) makes a simpler equivalent equation. Transforming an equation means rewriting it with legal steps like these.
Three possible endings: x = a means one solution, the number a. A true statement a = a, such as 7 = 7, means every number is a solution (infinitely many solutions). A false statement a = b, such as 3 = 5, means no number is a solution.
Work through the examples below. After each step, ask: "Is this new equation equivalent to the one before it? Which legal step did we use?"
One solution, variables on both sides
Solve 7x - 4 = 3x + 20.
Equation: Subtract 3x from both sides: 4x - 4 = 20. Add 4: 4x = 24. Divide by 4: x = 6. Check: 7(6) - 4 = 38 and 3(6) + 20 = 38. The equation ends in the form x = a, so it has one solution.
How many solutions does 4(x + 3) - x = 3(x + 4) have?
Equation: Distribute: 4x + 12 - x = 3x + 12. Collect like terms: 3x + 12 = 3x + 12. Subtract 3x: 12 = 12. This is always true, so every number is a solution. Try x = 10: 4(13) - 10 = 42 and 3(14) = 42.
No solution (a = b)
How many solutions does 2(3x - 1) + 5 = 6x - 4 have?
Equation: Distribute: 6x - 2 + 5 = 6x - 4. Collect like terms: 6x + 3 = 6x - 4. Subtract 6x: 3 = -4. This is never true, so no number is a solution.
Fraction coefficients
Solve (2/3)(x - 6) = (1/4)x + 1.
Equation: Multiply every term on both sides by 12, the least common denominator (the smallest number that both denominators divide into): 8(x - 6) = 3x + 12. Distribute: 8x - 48 = 3x + 12. Subtract 3x and add 48: 5x = 60, so x = 12. Check: (2/3)(6) = 4 and (1/4)(12) + 1 = 4.
Diagram 1 puts Examples 1, 3 and 4 side by side. In all three, the transforming steps are the same kind; only the ending is different. Stress that when the x-terms cancel, the work is not over: read what is left. For Example 5, show why multiplying by the least common denominator (the smallest number that every denominator divides into, here 12) is allowed: it is one legal step, done to every term on both sides. Decimals can be cleared the same way, by multiplying by 10 or 100.
Diagram 2 gives a picture of the three cases. Graph y = (left side) and y = (right side) as two lines. Where the lines cross, both sides are equal, so the x-value there is the solution. Parallel lines never meet (no solution), and two sides that give the same line are equal for every x (infinitely many solutions). This picture is a look ahead to systems of equations in 8.EE.C.8; students do not need to graph to solve today.
Guided Practice15 minutes
Pairs solve four equations, one at a time. After each one, a pair shows its steps at the board and names the final form (x = a, a = a or a = b).
9 - 2(x + 4) = 3x - 14. (1 - 2x = 3x - 14, so 15 = 5x and x = 3.)
Write your own equation that uses the distributive property and has no solution. Trade with your partner and transform each other's equation to prove it.
Closure5-10 minutes
Exit ticket: (1) Solve 4(x + 0.5) = 2x + 11. (4x + 2 = 2x + 11, so 2x = 9 and x = 4.5.) (2) How many solutions does 3(x + 2) = 4x + 6 have? (One: 3x + 6 = 4x + 6 gives x = 0. Zero is a number, so this is not "no solution.") (3) Finish the sentence: "When the variable disappears, I look at what is left. If it is true, ...; if it is false, ..."
Differentiation Strategies
For Struggling Students
Give a step-labeling template with two columns: the new equation on the left and the legal step used on the right
Start with equations that need only one kind of step, then add parentheses, then fractions
Keep a card on the desk: "Variable gone? True statement: all numbers. False statement: no number."
For Advanced Students
Ask for an equation with fractions on both sides that has infinitely many solutions, and one with decimals that has no solution
Ask: for which values of a does ax + 3 = 5x + 3 have exactly one solution? (every a except 5)
Extension (beyond this standard): preview HSA.REI.B.3 by solving ax + b = c for x, with letters as coefficients
Assessment Guidance
What to Look For
Ask students to say the step they are using, not only to write it. Watch for four errors: multiplying only the first term inside parentheses, losing a negative sign when distributing a negative number, multiplying only some terms by the common denominator, and stopping when the variable cancels. When the variable cancels, a strong answer names the final statement (for example "3 = -4 is false") and then the number of solutions.
02
Classroom Activities
3 Activities
1
One, None or Infinitely Many? Card Sort
20 minPairs
Pairs sort 12 equation cards into three piles: "One solution," "No solution" and "Infinitely many solutions." They must transform every equation on a whiteboard until it has the form x = a, a = a or a = b, and write that final form on the back of the card before placing it.
Shuffle the cards and place them face down. Partners take turns drawing a card
The partner who draws transforms the equation aloud; the other partner checks each step
For every "One solution" card, substitute the answer into the original equation to check it
When all 12 cards are placed, compare piles with another pair and settle any disagreement by redoing the steps
Discussion Questions
Each pile has four cards. Which card did you first place in the wrong pile, and what made you change it?
Cards 6 and 9 both have a 3 in front of parentheses, but they are in different piles. What did you look at to decide between "no solution" and "infinitely many"?
Can you tell the pile of Card 4 without solving it? (Clue: compare the x-coefficients on the two sides after distributing.)
Modification for Distance Learning
Put the cards on a shared slide with three labeled boxes. Pairs drag each card into a box and type the final form (x = a, a = a or a = b) in a text box next to it.
2
Build an Equation Challenge
15 minPairs
This activity asks students to create equations of each type, which is the first thing the standard asks for. Each pair gets one starting card with a left side that uses the distributive property. They write three right sides: one that gives exactly one solution, one that gives no solution, and one that gives infinitely many solutions.
Starting Cards (left sides)
Card A: 2(x + 3) - x = ?
Card B: 3(x - 2) + 2x = ?
Card C: 0.5(6x + 8) = ?
Card D: (1/3)(9x - 6) + x = ?
Sample Record
A pair with Card A first simplified the left side: 2(x + 3) - x = x + 6. They wrote 2(x + 3) - x = x + 6 (it becomes 6 = 6, infinitely many solutions), 2(x + 3) - x = x + 1 (it becomes 6 = 1, no solution) and 2(x + 3) - x = 3x (it becomes x + 6 = 3x, so x = 3).
Procedure
Simplify your left side first, by distributing and collecting like terms
Write your three right sides on a strip of paper, in a random order, without labels
Trade strips with another pair. They must transform each equation and label it
Trade back and check the labels together
Discussion Questions
To get infinitely many solutions, what must the two sides have in common after simplifying?
To get no solution, what must be the same and what must be different?
Any other right side gives exactly one solution. Why are "one solution" equations the easiest to write?
Challenge Variation
Write a right side that uses the distributive property too, with a fraction or decimal coefficient, for each of the three types.
3
Error Analysis Gallery Walk
20 minGroups of 3
Hang four posters around the room. Each poster shows an equation and a student's work with one mistake. Groups rotate every 4 minutes, find the mistake, write the corrected solution on a sticky note and add it to the poster.
Poster 1: The -2 was not multiplied through: -2 times 3 is -6, not +6. Correct: 5 - 2x - 6 = 9, so -2x - 1 = 9, -2x = 10 and x = -5.
Poster 2: 4x and 3 are not like terms, so they cannot be added. Correct: 3x + 3 = 2x + 10, so x = 7.
Poster 3: Only the fraction terms were multiplied by 6. Every term must be: 2x + 12 = 3x - 6, so x = 18.
Poster 4: The steps are right, but 0 = 0 is a true statement. Every number is a solution, so there are infinitely many solutions.
Procedure
At each poster, one student reads the work aloud, one finds the first wrong step, and one writes the corrected solution
Before moving on, the group checks its answer by substituting into the original equation (or, for Poster 4, by trying two different numbers)
After the last rotation, each group reads the sticky notes on its first poster and agrees or disagrees
Discussion Questions
Which two posters show a mistake with the distributive property or with like terms?
On Poster 4, the student's steps were correct. Why was the answer still wrong?
What check would have caught each mistake?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Three Equations, Three Endings
The same kind of steps (distribute, collect like terms, move the variable terms) leads to three different endings. x = 6 means one solution. 3 = -4 is false for every x, so there is no solution. 12 = 12 is true for every x, so there are infinitely many solutions.
Diagram 2: Each Side of the Equation as a Line
Each side of an equation is graphed as a line, drawn to scale for x from 0 to 4: the solid line is the left side and the dashed line is the right side. Left: y = 2x + 1 and y = x + 3 cross at (2, 5), so x = 2 is the only solution of 2x + 1 = x + 3. Middle: y = x + 3 and y = x + 1 are parallel and never meet, so x + 3 = x + 1 has no solution. Right: both sides of 2(x + 1) = 2x + 2 give the same line, so every x is a solution.
04
Homework Assignment
~30 min
8.EE.C.7 Homework: Solving Linear Equations
Directions: Show every step. Name the step when it helps (distribute, collect like terms, add or subtract on both sides). Check each "one solution" answer by substituting it into the original equation. When the variable disappears, write the final statement and what it means.
Solve each equation. Clear the fractions or decimals first if it helps: (a) (3/4)x + 2 = (1/2)x + 5 (b) 0.25(x + 8) = 0.75x - 3 (c) (1/2)(x - 3) = (1/4)(x + 1)
A school garden is a rectangle. Its length is 4 m less than 3 times its width, and its perimeter is 48 m. Let w be the width. Write an equation that uses the distributive property, solve it, and give the width and the length.
Part 2: How Many Solutions? (Problems 4-6)
For each equation, transform it until you reach the form x = a, a = a or a = b. Then state how many solutions it has: (a) 5(x - 2) + 3 = 5x - 7 (b) 2(4x + 1) - 3x = 5x - 2 (c) 3(x + 2) - x = 4x - 2
Complete the equation 6x - 4 = ____ in three different ways, so that it has (a) exactly one solution, (b) no solution and (c) infinitely many solutions. Use the distributive property in at least one of your right sides, and show the steps that prove each answer.
Priya says the equation 0.2(10x - 5) = 2x - 1 has no solution, "because the x-terms cancel." (a) Transform the equation and decide whether she is right. (b) Explain in one sentence what she should look at after the x-terms cancel. (c) Change one number in the equation so that it really has no solution.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Distributing and Like Terms
Every expansion and combination is correct, including signs
One sign or distribution error
Several errors
Solutions
All answers correct and checked
Most answers correct, or checks missing
Most answers wrong
Number of Solutions
Final form (x = a, a = a or a = b) written and read correctly every time
Final form written but misread once
Final forms missing or misread
Examples and Explanations
Own examples work and explanations use the final statement
Examples work but explanations are vague
Examples missing or incorrect
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Solve 9x + 4 = 5x + 32.
Answer: C
Subtract 5x: 4x + 4 = 32. Subtract 4: 4x = 28. Divide by 4: x = 7. Choice A adds 4 instead of subtracting it (4x = 36). Choice B moves 5x to the left side without changing its sign (14x = 28). Choice D stops at 4x = 28 and forgets to divide.
Question 2 of 20 · Multiple Choice
Solve 3(x - 5) = 2x + 1.
Answer: B
Distribute: 3x - 15 = 2x + 1. Subtract 2x: x - 15 = 1. Add 15: x = 16. Choice A multiplies only x by 3 (3x - 5 = 2x + 1). Choice C writes 3x + 15, a sign error. Choice D computes 15 - 1 instead of 1 + 15 when it adds 15 to both sides of x - 15 = 1.
Question 3 of 20 · Multiple Choice
Which equation has infinitely many solutions?
Answer: D
In choice D, the left side is 2x + 12 + x = 3x + 12, the same as the right side, so the equation becomes 12 = 12. Choices A and C become false statements (12 = 4 and 12 = -12), so they have no solution. Choice B has one solution, x = 0; a student who sees the same constant on both sides may think it has infinitely many.
Question 4 of 20 · Multiple Choice
How many solutions does 5x - 2(x + 3) = 3x + 4 have?
Answer: B
Distribute and collect like terms: 5x - 2x - 6 = 3x - 6, so 3x - 6 = 3x + 4. Subtract 3x: -6 = 4, which is false, so there is no solution. Choice C stops when the x-terms cancel and does not read the false statement. Choice D adds 5x and 2x instead of subtracting (7x - 6 = 3x + 4, so x = 2.5).
Question 5 of 20 · Multiple Choice
Solve (1/2)x - 4 = (1/3)x + 1.
Answer: D
Multiply every term by 6: 3x - 24 = 2x + 6, so x = 30. Check: 15 - 4 = 11 and 10 + 1 = 11. Choice B multiplies only the fraction terms by 6 (3x - 4 = 2x + 1). Choice A subtracts 4 instead of adding it, getting (1/6)x = -3. Choice C reaches (1/6)x = 5 and multiplies by 1/6 instead of dividing.
Question 6 of 20 · Multiple Choice
Solve 1.2x - 0.5 = 0.7x + 2.
Answer: A
Subtract 0.7x: 0.5x - 0.5 = 2. Add 0.5: 0.5x = 2.5. Divide by 0.5: x = 5. Choice C subtracts 0.5 instead of adding it (0.5x = 1.5). Choice B multiplies 2.5 by 0.5 instead of dividing. Choice D divides 0.5 by 2.5, the wrong way around.
Question 7 of 20 · Multiple Choice
Andre transforms an equation step by step, and the last line is 5 = 5. What does this tell him?
Answer: C
5 = 5 is a true statement with no variable left, so the original equation is true for every value of x: infinitely many solutions. Choice A reads the leftover number as the value of x. Choice B confuses a true statement (a = a) with a false one (a = b). Choice D assumes the variable must be 0 because it disappeared.
Question 8 of 20 · Multiple Choice
For which value of k does 4(x - 3) = 4x + k have infinitely many solutions?
Answer: A
The left side is 4x - 12. The two sides are the same expression only when k = -12, and then the equation becomes -12 = -12. Choice B drops the minus sign. Choice C multiplies only x by 4 (4x - 3). Choice D uses the coefficient 4, which is already the same on both sides. With any k other than -12, the equation has no solution.
Question 9 of 20 · Multiple Choice
Which equation is equivalent to 3(2x - 1) - (x - 4) = 14 after you expand and collect like terms?
Answer: A
Expand: 6x - 3 - x + 4 = 14. Collect like terms: 5x + 1 = 14. Choice B changes the sign of x but not the sign of -4 when it removes the parentheses (6x - 3 - x - 4). Choice C adds x instead of subtracting it. Choice D multiplies only 2x by 3 (6x - 1 - x + 4).
Question 10 of 20 · Multiple Choice
Ms. Lee buys 4 notebooks and 4 pens for $22.00. Each pen costs $1.50 less than each notebook. She writes 4(n + n - 1.50) = 22, where n is the price of one notebook. What does one notebook cost?
Answer: D
Collect like terms inside: 4(2n - 1.50) = 22. Distribute: 8n - 6 = 22, so 8n = 28 and n = 3.50. A notebook costs $3.50 and a pen $2.00, and 4($3.50 + $2.00) = $22.00. Choice B is the price of a pen. Choice C is 22 ÷ 4, the cost of one notebook and one pen together. Choice A leaves out the notebook and solves 4(n - 1.50) = 22.
Question 11 of 20 · Multiple Choice
Which equation has exactly one solution?
Answer: B
In choice B, 2x + 6 = 3x + 2 gives x = 4, one solution. The x-coefficients are different (2 and 3), which is the clue. Choice A has the same expression on both sides, so it has infinitely many solutions. Choices C and D keep the same x-coefficient but change the constant: they become 6 = 3 and 6 = 8, so they have no solution.
Question 12 of 20 · Multiple Choice
Kim solves 6(x + 2) = 4x + 20. Her first line is 6x + 2 = 4x + 20. Which statement is true?
Answer: A
The distributive property gives 6x + 12 = 4x + 20. Then 2x = 8 and x = 4. Check: 6(6) = 36 and 4(4) + 20 = 36. Choice D follows Kim's wrong line to the end (2x = 18). Choice C adds 6 and 2 instead of multiplying them. Choice B reaches 2x = 8 and multiplies by 2 instead of dividing.
Question 13 of 20 · Multiple Choice
Solve (3/4)(x + 8) = x + 1.
Answer: D
Distribute: (3/4)x + 6 = x + 1. Subtract (3/4)x and subtract 1: 5 = (1/4)x, so x = 20. Check: (3/4)(28) = 21 and 20 + 1 = 21. Choice A multiplies only x by 3/4 ((3/4)x + 8 = x + 1). Choice B multiplies 5 by 1/4 instead of dividing. Choice C loses a minus sign: a student who subtracts x from both sides gets -(1/4)x = -5 but writes -(1/4)x = 5.
Question 14 of 20 · Multiple Choice
One plumber charges $65 for a visit plus $40 per hour. Another charges $35 plus $50 per hour. For how many hours of work do they charge the same amount?
Answer: C
65 + 40h = 35 + 50h. Subtract 40h and subtract 35: 30 = 10h, so h = 3. Both plumbers charge $185 for 3 hours. Choice A adds 65 and 35 instead of subtracting (100 = 10h). Choice B divides 10 by 30. Choice D stops at 10h = 30.
Question 15 of 20 · Short Answer
Solve 4(2x - 1) - 3(x - 5) = 26. Show each step.
Distribute: 8x - 4 - 3x + 15 = 26. Collect like terms: 5x + 11 = 26. Subtract 11: 5x = 15. Divide by 5: x = 3. Check: 4(5) - 3(-2) = 20 + 6 = 26. Watch for -3 times -5: it is +15.
Question 16 of 20 · Short Answer
Transform each equation until you reach x = a, a = a or a = b, and say how many solutions it has. (a) 6x + 3(1 - 2x) = 3 (b) 7(x - 1) = 7x + 1
(a) 6x + 3 - 6x = 3, so 3 = 3: a true statement, so there are infinitely many solutions. (b) 7x - 7 = 7x + 1, so -7 = 1: a false statement, so there is no solution.
Question 17 of 20 · Short Answer
Write one linear equation that uses the distributive property and has no solution, and one that has infinitely many solutions. Show the final form of each.
Answers vary. Sample: 5(x + 2) = 5x + 7 becomes 5x + 10 = 5x + 7, then 10 = 7, which is false: no solution. 5(x + 2) = 5x + 10 becomes 10 = 10, which is true: infinitely many solutions. Full credit needs both final forms, not only the equations.
Question 18 of 20 · Short Answer
Solve 0.6(x - 5) = 0.2x + 1.4.
Distribute: 0.6x - 3 = 0.2x + 1.4. Subtract 0.2x: 0.4x - 3 = 1.4. Add 3: 0.4x = 4.4. Divide by 0.4: x = 11. Check: 0.6(6) = 3.6 and 0.2(11) + 1.4 = 3.6. Students may also multiply every term by 10 first: 6(x - 5) = 2x + 14.
Question 19 of 20 · Short Answer
Every ticket to a school play has the same full price. The Garcia family buys 5 tickets with a coupon for $2 off each ticket. The Kim family buys 3 tickets at full price and an $8 program. Both families spend the same amount. Write and solve an equation to find the full price of one ticket.
Let t be the full price. 5(t - 2) = 3t + 8. Distribute: 5t - 10 = 3t + 8. Subtract 3t and add 10: 2t = 18, so t = 9. A ticket costs $9, and each family spends $35: 5($7) = $35 and 3($9) + $8 = $35.
Question 20 of 20 · Short Answer
Nico transforms 2(5x - 3) = 10x - 6 and gets 0 = 0. He writes "x = 0." Explain his mistake, and give two different numbers that are solutions, with a check for each.
0 = 0 is a true statement, so the equation is true for every number: it has infinitely many solutions, not only x = 0. Sample checks: x = 1 gives 2(2) = 4 and 10 - 6 = 4; x = 2 gives 2(7) = 14 and 20 - 6 = 14. Any two numbers with correct checks earn full credit.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does 8.EE.C.7 mean?
8.EE.C.7 means students can solve any linear equation in one variable, including ones with fractions, decimals, parentheses and variables on both sides. They also learn that such an equation has exactly one solution, no solution or infinitely many solutions, and they show which case it is by simplifying until they reach x = a, a = a or a = b.
Is 8.EE.C.7 taught in grade 8 or in Algebra I?
8.EE.C.7 is a grade 8 standard in the Expressions and Equations domain. Schools that teach Algebra I in grade 8 cover it early in that course. High school standard HSA.REI.B.3 continues the work, adding inequalities and equations with letters as coefficients.
How can an equation have no solution?
An equation has no solution when simplifying it leads to a false statement such as 4 = 9. For example, x + 7 = x + 2 asks for a number that stays the same when you add 7 and when you add 2, and no such number exists. Subtracting x from both sides shows it: 7 = 2 is false.
What does "infinitely many solutions" mean?
It means every number makes the equation true. When you simplify both sides and they turn into the same expression, the variable cancels and you are left with a true statement such as 8 = 8. You can test it: pick any two numbers, substitute them, and both sides will match each time.
If I get x = 0, does that mean there is no solution?
No. x = 0 is one solution, the number zero. For example, 5x + 2 = 2x + 2 gives 3x = 0, so x = 0, and substituting 0 makes both sides equal 2. "No solution" only happens when you reach a false statement with no variable left.
What are rational number coefficients in 8.EE.C.7?
They are coefficients that are fractions, decimals or negative numbers, such as -3x, 0.75x or (5/8)x. Part b of the standard asks students to solve equations with these numbers, not only whole numbers. Students use the same steps; the arithmetic is the new challenge.
Should students clear fractions and decimals first?
It often helps, but it is a choice, not a rule. Multiplying every term on both sides by the least common denominator (the smallest number every denominator divides into) turns fractions into whole numbers, and multiplying by 10 or 100 clears decimals. A frequent slip is multiplying only some of the terms, so ask students to count the terms before and after.
What mistakes do students make with the distributive property?
A common mistake is multiplying only the first term inside the parentheses, for example writing 4(x + 5) as 4x + 5 instead of 4x + 20. Another is losing a negative sign: -3(x - 2) is -3x + 6, not -3x - 6. A third is combining unlike terms, such as adding 2x and 5 to get 7x.
How is 8.EE.C.7 different from 7.EE.B.4?
In grade 7, 7.EE.B.4 asks for equations of the forms px + q = r and p(x + q) = r, with the variable on one side. 8.EE.C.7 allows the variable on both sides, several sets of parentheses and like terms to collect, and it adds the idea that an equation can have no solution or infinitely many.
How does 8.EE.C.7 connect to systems of equations?
Solving a system in 8.EE.C.8 often ends with a one-variable equation, and 8.EE.C.7 skills solve it. The three cases match too: a system of two lines has one solution, no solution (parallel lines) or infinitely many solutions (the same line). Students who master this standard find systems much easier.
07
Related Standards
5 standards
These standards connect to 8.EE.C.7: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
7.EE.A.1Prerequisite
Add, subtract, factor and expand linear expressions with rational coefficients
