UNIT 4Linear equationsLesson 2 of 21 Equations with Variables on Both Sides2 Equations with One, No or Infinitely Many Solutions
Equations with One, No or Infinitely Many Solutions
Most equations have one solution. Some have none. Some are true for every number.
What you'll be able to do
- Tell whether an equation has one, no or infinitely many solutions.
- Simplify each side before you decide.
- Find the number that gives each type.
- Say what each type means in a story.
The big idea
Solve as usual. Get the x terms on one side. Sometimes the x terms cancel. Then look at what is left.
No solution. Solve 2x + 5 = 2x − 1. Subtract 2x: 5 = −1. That is never true. So no number works.
Infinitely many. Solve 4(x − 2) = 4x − 8. Distribute: 4x − 8 = 4x − 8. Subtract 4x: −8 = −8. That is always true. So every number works.
One solution. If the x terms don't cancel, there is exactly one solution. It can be 0. In 6x + 1 = 2x + 1, x = 0 is the one solution.
Simplify first. Distribute and combine each side before you decide. Two sides can look different but be the same.
A quick test. Write both sides as (number)x + (number). Different x parts: one solution. Same x parts but different numbers: no solution. Both the same: infinitely many.
Words you need
- Solution
- A number that makes an equation true.
- No solution
- No number makes the equation true, like 5 = −1.
- Infinitely many solutions
- Every number makes the equation true, like −8 = −8.
- Simplify each side
- Distribute and add like terms until each side is short.
What is left after the x terms cancel?
| What is left | How many solutions |
|---|---|
| x = a number | One |
| A false statement, like 5 = −1 | None |
| A true statement, like −8 = −8 | Infinitely many |
Worked examples
Example 1
Solve 2x + 5 = 2x − 1.
- Subtract 2x from both sides: 5 = −1.
- That is false for every x.
- So there is no solution.
Answer: No solution
Example 2
Solve 4(x − 2) = 4x − 8.
- Distribute: 4x − 8 = 4x − 8.
- Subtract 4x: −8 = −8. That is always true.
- So every number works.
Answer: Every value works (infinitely many solutions)
Example 3
Solve 6x + 1 = 2x + 1.
- Subtract 2x: 4x + 1 = 1.
- Subtract 1: 4x = 0.
- So x = 0. That is one solution.
Answer: x = 0
Common mistakes
Calling x = 0 "no solution". x = 0 is a number, so it is one solution.
Reading 8 = 8 as x = 8. A true statement with no x means every number works.
Deciding before you simplify. Distribute and combine each side first. 2(x + 3) and 2x + 6 are the same.
Dividing both sides by x. That can throw away the solution x = 0. Subtract instead.
Practice: 20 questions
11 of 20 are word problems. For a word problem, sketch it first and mark what you know.
Tier 1 Foundations Use the rule directly
- Ana has $40 and saves $5 a week. Ben has $25 and saves $5 a week. When will they have the same amount? Word problem
- Jo starts at the 5 m mark and runs 3 m each second. Kim starts at the 5 m mark and runs 4 m each second. When are they side by side? Word problem
- Solve 5x − 2 = 2x + 3x − 2.
- Chen solves 4x + 8 = 2x + 8 + 2x. He gets 8 = 8 and writes x = 8. Find his mistake.
- Lea has $40 and saves $5 a week. Omar has $25 and saves $8 a week. When will they have the same amount? Word problem
Tier 2 Intermediate One more step to think through
- How many solutions does 2(3x − 4) + 5 = 6x − 3 have?
- Which equation has no solution? A: 5(x + 2) − x = 4x + 10. B: (1/2)(4x − 6) = 2x + 3.
- Maya's bus pass costs $12 plus $2.50 a ride. Leo pays $2.50 a ride plus two $6 fees. For how many rides do they pay the same? Word problem
- Lia pays $4 a month, plus 3 months up front: 4(x + 3) dollars for x months. Max pays 4x + c dollars. Which c makes their costs always equal? Which c make them never equal? Word problem
- A square has sides of 2x + 3 cm. A rectangle is 3x + 1 cm long and x + 5 cm wide. For which x do they have the same perimeter? Word problem
Tier 3 Advanced Several steps - find the hidden structure
- Gym A costs $40 to join and $15 a month. Gym B costs $25 to join and $k a month. For which k does 40 + 15x = 25 + kx have no solution? Word problem
- Find a and b so that 2(x + a) = bx + 8 has infinitely many solutions.
- Dev says ax + 4 = 2x + 4 has exactly one solution for every value of a. Find his mistake.
- Rectangle A is x + 5 cm by x cm. Rectangle B is x + 2 cm by x + 1 cm. For which x do they have the same perimeter? Word problem
- Plan A costs $20 plus $5 per GB, plus a $10 fee: 20 + 5x + 10. Plan B costs $30 plus $k per GB. For which k do they cost the same for every number of GB? Word problem
Tier 4 Challenge Stretch problems, some above grade level
- Sara solves 3x = 7x by dividing both sides by x. She gets 3 = 7 and says there is no solution. Find her mistake.
- Think of a number n. Add 5, then double it. Subtract 4, then subtract twice your number. Kai says you always get 6. Write an equation to show why. Word problem
- In 2(ax − 3) = 6x + b, which a and b give infinitely many solutions? Which give none? Which give exactly one?
- Mia raises $24 plus $4 a lap. Noah raises $6 a lap plus a $b bonus. He gives $2 a lap to the water fund. Which bonus makes their totals always equal? Can any bonus make them equal at exactly one lap count? Word problem
- |a| is the distance from a to 0. Solve |2x − 1| = 7. How many solutions does |2x − 1| = −3 have? Above grade level
Answers
Tier 1 Foundations
- No solution. 40 + 5x = 25 + 5x. Subtract 5x: 40 = 25. That is never true, so they never have the same amount.
- x = 0. 5 + 3x = 5 + 4x gives x = 0. That is one solution: they are side by side only at the start. Check x = 0: left side 5, right side 5.
- Every value works (infinitely many solutions). Both sides simplify to the same expression, so 5x − 2 = 2x + 3x − 2 is true for every x.
- Every value works (infinitely many solutions). 8 = 8 has no x in it. It is always true, so every number works, not just 8. Both sides simplify to the same expression, so 4x + 8 = 2x + 8 + 2x is true for every x.
- x = 5. Solve 40 + 5x = 25 + 8x. Check x = 5: left side 65, right side 65.
Tier 2 Intermediate
- Every value works (infinitely many solutions). Both sides simplify to the same expression, so 2 · (3x − 4) + 5 = 6x − 3 is true for every x.
- B has no solution. Simplify each side. A becomes 4x + 10 = 4x + 10, which is always true. B becomes 2x − 3 = 2x + 3, which is never true. So B has no solution.
- Every value works (infinitely many solutions). Leo pays 2.5x + 6 + 6 = 2.5x + 12. That is the same as Maya's cost. So they pay the same for every number of rides.
- c = 12 gives infinitely many solutions. Any c except 12 gives no solution. Lia pays 4x + 12. If c = 12, the costs always match. Any other c gives 12 = c, which is false, so they never match.
- Every value works (infinitely many solutions). The square's perimeter is 4(2x + 3) = 8x + 12. The rectangle's is 2(3x + 1) + 2(x + 5) = 8x + 12. They match for every positive x.
Tier 3 Advanced
- k = 15 gives no solution. The x terms cancel only when k = 15. Then 40 = 25, which is false. So k = 15 gives no solution.
- a = 4 and b = 2 give infinitely many solutions. The left side is 2x + 2a. The x parts must match, so b = 2. The numbers must match too: 2a = 8, so a = 4.
- a = 2 gives infinitely many solutions. Any a except 2 gives exactly one solution. When a = 2, both sides are 2x + 4, so every number works. Any other a gives one solution, x = 0. Move every term of ax + 4 = 2x + 4 to one side: (a − 2)x = 0. No solution: the x part is 0 but the number part isn't. Infinitely many: both parts are 0. One solution: the x part isn't 0.
- No solution. A's perimeter is 4x + 10. B's is 4x + 6. Then 4x + 10 = 4x + 6 gives 10 = 6, which is false. So no x works.
- k = 5 gives infinitely many solutions. Plan A is 5x + 30. Plan B is kx + 30. They match for every number of GB when k = 5.
Tier 4 Challenge
- x = 0. Dividing by x is not allowed when x could be 0. Subtract 3x instead: 0 = 4x, so x = 0. x = 0 works: 3(0) = 7(0). Dividing by x hid this solution. Check x = 0: left side 0, right side 0.
- Every value works (infinitely many solutions). 2(n + 5) − 4 − 2n = 2n + 10 − 4 − 2n = 6. The n terms cancel and 6 = 6 is always true. So every starting number gives 6.
- a = 3 and b = −6 give infinitely many solutions. a = 3 and any b except −6 give no solution. Any a except 3 gives exactly one solution. The left side is 2ax − 6. The x parts match when 2a = 6, so a = 3. Then the numbers match only when b = −6. If a isn't 3, there is one solution.
- b = 24 gives infinitely many solutions. Exactly one solution: never. Noah's total is 6x + b − 2x = 4x + b. Both gain $4 a lap, so the x terms always cancel. With b = 24 they are always equal. Otherwise they are never equal.
- x = 4 or x = −3. |2x − 1| = −3 has no solution. 2x − 1 is 7 or −7. So x = 4 or x = −3. A distance can't be −3, so |2x − 1| = −3 has no solution.
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