Kenneth ChengMathematics · models · ideas
Search中文

PAPER · REASONING · PRACTICE

Equations: brackets and constant denominators — Teacher reference

Distribute to every term, clear constant denominators on the whole equality and verify in the original equation.

40 minutes · Year 7, upper volume · text pp. 63–66 · foundation lessons 32–33

Textbook reference: Year 7, upper volume · Chapter 3 · Linear equations

Chapter core practice: use after the referenced textbook lessons, or for chapter review. See the upper-volume plan

One worked example, one guided attempt, eight core questions, four follow-ups and two extensions. Write on paper; the teacher controls the reveal.

Student classroom pageArrange in My Classroom

  1. 4′Check prerequisites
  2. 5′Observe and predict
  3. 8′Worked example and guided attempt
  4. 15′Independent practice
  5. 5′Discuss and correct
  6. 3′Exit questions

Observe · explain

A question before practising

Compare clearing the denominator in (x + 2)/4 − 3 = 1 with multiplying only the fraction. Ask what must happen to the standalone −3 and to the right side.

Organise the key idea

A multiplier applies to every term of each side. These linear equations have constant nonzero denominators; multiplying by their common multiple preserves the original solutions.

Open the original chapterOpen the interactive inquiry

Use the lesson question here for paper discussion; the original activity opens separately.

Worked example1

Remove brackets, solve and check the original equation.

4(x−3)=2x+64(x-3)=2x+6
Check your work
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Guided attempt2

Multiply both sides by 3 first, then solve and check.

x+13=5\frac{x+1}{3}=5
  1. 3 × [(x + 1)/3] = 3 × ____.
  2. x + 1 = ____, so x = ____.
  3. Original check: (____ + 1)/3 = ____.
Check your work
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Prerequisite3

Expand the product; no equation needs to be solved.

2(x−4)2(x-4)
Check your work
  • Keep each sign and collect only terms with the same variable powers.
  • Give an equivalent simplified expression.
Prerequisite4

Calculate the product used when clearing a denominator of 2.

6×126\times\frac{1}{2}
Check your work
  • Apply the relevant sign and order rules to every operation.
  • Show an intermediate calculation and obtain the exact stated value.
Independent practice5

Solve and verify.

2(x+4)−3=x+92(x+4)-3=x+9
Check your work
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Independent practice6

Solve, taking care with the negative multiplier.

5−2(x−1)=3x−85-2(x-1)=3x-8
Check your work
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Independent practice7

Clear the denominator and solve. Do not leave the standalone 1 unchanged.

x2+1=4\frac{x}{2}+1=4
Check your work
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Independent practice8

Use a common multiplier for 3 and 5, then solve.

x−23=x+45\frac{x-2}{3}=\frac{x+4}{5}
Check your work
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Exit question9

Solve and check in the original fractional equation.

2x+13−1=4\frac{2x+1}{3}-1=4
Check your work
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Exit question10

A learner multiplies (x + 2)/4 − 3 = 1 by 4 but writes x + 2 − 3 = 4. Identify the first error and solve correctly.

x+24−3=1\frac{x+2}{4}-3=1
Check your work
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Further practice11

Solve after expanding both sides.

3(x+2)=2(x+5)3(x+2)=2(x+5)
Check your work
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Further practice12

Solve with the outside minus applied to both enclosed terms.

−(x−5)=2x−1-(x-5)=2x-1
Check your work
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Further practice13

Clear both denominators using 6, then solve.

x3+x6=5\frac{x}{3}+\frac{x}{6}=5
Check your work
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Further practice14

Multiply both sides by 4, then solve.

3x−24=x+22\frac{3x-2}{4}=\frac{x+2}{2}
Check your work
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Extension15

Solve, showing both distributive steps before collecting.

2(x−1)−3(x+2)=4−2x2(x-1)-3(x+2)=4-2x
Check your work
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Extension16

Use 6 as a common multiplier and retain the subtraction of the entire second numerator.

x+12−x−23=3\frac{x+1}{2}-\frac{x-2}{3}=3
Check your work
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Preparation and assessment notes

Prerequisites

The two existing collecting/transposition lessons, distributivity and equivalent fractions.

Look for in students’ work

Use the prerequisite checks to choose support, project the worked and guided items, then let learners work on paper. Select supplements or extensions as needed. No learner answers are saved by this presentation.

Collect the paper exit questions and note completion and common errors. This page does not grade or store paper answers.

All reference answers

Worked example1

Remove brackets, solve and check the original equation.

4(x−3)=2x+64(x-3)=2x+6
Reference answer

x = 9.

x=9x=9
  1. Distribute: 4x − 12 = 2x + 6.
  2. Subtract 2x and add 12 on both sides: 2x = 18, so x = 9.
  3. Check: 4(9 − 3) = 24 and 2 × 9 + 6 = 24.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Guided attempt2

Multiply both sides by 3 first, then solve and check.

x+13=5\frac{x+1}{3}=5
  1. 3 × [(x + 1)/3] = 3 × ____.
  2. x + 1 = ____, so x = ____.
  3. Original check: (____ + 1)/3 = ____.
Reference answer

x = 14.

x=14x=14
  1. Multiply both complete sides by 3: x + 1 = 15.
  2. Subtract 1: x = 14.
  3. Check: (14 + 1)/3 = 5.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Prerequisite3

Expand the product; no equation needs to be solved.

2(x−4)2(x-4)
Reference answer

2x−8.

2x−82x-8
  1. 2 × x = 2x and 2 × (−4) = −8.
  2. The result is 2x − 8.
  • Keep each sign and collect only terms with the same variable powers.
  • Give an equivalent simplified expression.
Prerequisite4

Calculate the product used when clearing a denominator of 2.

6×126\times\frac{1}{2}
Reference answer

The value is 3.

33
  1. 6 divided by 2 is 3.
  2. Thus multiplying 1/2 by 6 gives 3.
  • Apply the relevant sign and order rules to every operation.
  • Show an intermediate calculation and obtain the exact stated value.
Independent practice5

Solve and verify.

2(x+4)−3=x+92(x+4)-3=x+9
Reference answer

x = 4.

x=4x=4
  1. Expand: 2x + 8 − 3 = x + 9, or 2x + 5 = x + 9.
  2. Subtract x and 5 on both sides: x = 4.
  3. Check: 2(4 + 4) − 3 = 13 = 4 + 9.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Independent practice6

Solve, taking care with the negative multiplier.

5−2(x−1)=3x−85-2(x-1)=3x-8
Reference answer

x = 3.

x=3x=3
  1. Expand: 5 − 2x + 2 = 3x − 8.
  2. 7 − 2x = 3x − 8 gives 15 = 5x, so x = 3.
  3. Check: 5 − 2(3 − 1) = 1 and 3 × 3 − 8 = 1.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Independent practice7

Clear the denominator and solve. Do not leave the standalone 1 unchanged.

x2+1=4\frac{x}{2}+1=4
Reference answer

x = 6.

x=6x=6
  1. Multiply every term by 2: x + 2 = 8.
  2. x = 6; check 6/2 + 1 = 4.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Independent practice8

Use a common multiplier for 3 and 5, then solve.

x−23=x+45\frac{x-2}{3}=\frac{x+4}{5}
Reference answer

x = 11.

x=11x=11
  1. Multiply both sides by 15: 5(x − 2) = 3(x + 4).
  2. 5x − 10 = 3x + 12 gives 2x = 22 and x = 11.
  3. Check: (11 − 2)/3 = 3 and (11 + 4)/5 = 3.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Exit question9

Solve and check in the original fractional equation.

2x+13−1=4\frac{2x+1}{3}-1=4
Reference answer

x = 7.

x=7x=7
  1. Multiply all terms by 3: 2x + 1 − 3 = 12.
  2. 2x = 14, so x = 7.
  3. Check: (2 × 7 + 1)/3 − 1 = 5 − 1 = 4.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Exit question10

A learner multiplies (x + 2)/4 − 3 = 1 by 4 but writes x + 2 − 3 = 4. Identify the first error and solve correctly.

x+24−3=1\frac{x+2}{4}-3=1
Reference answer

The −3 must also be multiplied by 4. Correct: x + 2 − 12 = 4, so x = 14.

x=14x=14
  1. The multiplier applies to the complete left side, including −3.
  2. x + 2 − 12 = 4 gives x − 10 = 4 and x = 14.
  3. Check: (14 + 2)/4 − 3 = 1.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Further practice11

Solve after expanding both sides.

3(x+2)=2(x+5)3(x+2)=2(x+5)
Reference answer

x = 4.

x=4x=4
  1. 3x + 6 = 2x + 10.
  2. x = 4; both original sides are 18.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Further practice12

Solve with the outside minus applied to both enclosed terms.

−(x−5)=2x−1-(x-5)=2x-1
Reference answer

x = 2.

x=2x=2
  1. Expand to −x + 5 = 2x − 1.
  2. 6 = 3x, so x = 2; both sides are 3.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Further practice13

Clear both denominators using 6, then solve.

x3+x6=5\frac{x}{3}+\frac{x}{6}=5
Reference answer

x = 10.

x=10x=10
  1. Multiplying by 6 gives 2x + x = 30.
  2. 3x = 30, so x = 10. Check 10/3 + 10/6 = 5.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Further practice14

Multiply both sides by 4, then solve.

3x−24=x+22\frac{3x-2}{4}=\frac{x+2}{2}
Reference answer

x = 6.

x=6x=6
  1. 3x − 2 = 2(x + 2).
  2. 3x − 2 = 2x + 4, so x = 6.
  3. Both original sides equal 4.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Extension15

Solve, showing both distributive steps before collecting.

2(x−1)−3(x+2)=4−2x2(x-1)-3(x+2)=4-2x
Reference answer

x = 12.

x=12x=12
  1. Expand: 2x − 2 − 3x − 6 = 4 − 2x.
  2. −x − 8 = 4 − 2x. Add 2x and 8: x = 12.
  3. Check: 2 × 11 − 3 × 14 = −20 and 4 − 24 = −20.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Extension16

Use 6 as a common multiplier and retain the subtraction of the entire second numerator.

x+12−x−23=3\frac{x+1}{2}-\frac{x-2}{3}=3
Reference answer

x = 11.

x=11x=11
  1. Multiply by 6: 3(x + 1) − 2(x − 2) = 18.
  2. 3x + 3 − 2x + 4 = 18, so x + 7 = 18 and x = 11.
  3. Check: 12/2 − 9/3 = 6 − 3 = 3.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.

Paper for the class

Download all worksheets (ZIP · 318 PDFs · 47.2 MB)

All 53 Year 7–8 upper- and lower-volume packs in Traditional Chinese and English. Includes teacher answers; student core, further practice and teacher papers are in separate folders.

Core student worksheet: two A4 pages. Further practice is separate. Teacher copies include reference reasoning.

Textbook reference and authorship

A 40-minute selection of original core practice after studying the cited textbook lessons. It is not a replacement for every lesson in the chapter.

大湾区初中数学三维课堂（七年级上册）正文.pdf

Year 7, upper volume · text pp. 63–66 · foundation lessons 32–33

Original questions and explanations; the source textbook and answer PDFs are not reproduced here.

Equations: brackets and constant denominators

Year 7, upper volume · text pp. 63–66 · foundation lessons 32–33 · 1/2

Name: __________ Class: ______ Date: __________

Worked example1

Remove brackets, solve and check the original equation.

4(x−3)=2x+64(x-3)=2x+6
Guided attempt2

Multiply both sides by 3 first, then solve and check.

x+13=5\frac{x+1}{3}=5
  1. 3 × [(x + 1)/3] = 3 × ____.
  2. x + 1 = ____, so x = ____.
  3. Original check: (____ + 1)/3 = ____.
Prerequisite3

Expand the product; no equation needs to be solved.

2(x−4)2(x-4)
Prerequisite4

Calculate the product used when clearing a denominator of 2.

6×126\times\frac{1}{2}
g7-equations-brackets-fractions · 2026-10-14-upper-core-v1 · Conditions and reasons matter; measurement is not proof.

Equations: brackets and constant denominators

Year 7, upper volume · text pp. 63–66 · foundation lessons 32–33 · 2/2

Name: __________ Class: ______ Date: __________

Independent practice5

Solve and verify.

2(x+4)−3=x+92(x+4)-3=x+9
Independent practice6

Solve, taking care with the negative multiplier.

5−2(x−1)=3x−85-2(x-1)=3x-8
Independent practice7

Clear the denominator and solve. Do not leave the standalone 1 unchanged.

x2+1=4\frac{x}{2}+1=4
Independent practice8

Use a common multiplier for 3 and 5, then solve.

x−23=x+45\frac{x-2}{3}=\frac{x+4}{5}
Exit question9

Solve and check in the original fractional equation.

2x+13−1=4\frac{2x+1}{3}-1=4
Exit question10

A learner multiplies (x + 2)/4 − 3 = 1 by 4 but writes x + 2 − 3 = 4. Identify the first error and solve correctly.

x+24−3=1\frac{x+2}{4}-3=1
g7-equations-brackets-fractions · 2026-10-14-upper-core-v1 · Conditions and reasons matter; measurement is not proof.

Equations: brackets and constant denominators — Further practice

Year 7, upper volume · text pp. 63–66 · foundation lessons 32–33

Further practice11

Solve after expanding both sides.

3(x+2)=2(x+5)3(x+2)=2(x+5)
Further practice12

Solve with the outside minus applied to both enclosed terms.

−(x−5)=2x−1-(x-5)=2x-1
Further practice13

Clear both denominators using 6, then solve.

x3+x6=5\frac{x}{3}+\frac{x}{6}=5
Further practice14

Multiply both sides by 4, then solve.

3x−24=x+22\frac{3x-2}{4}=\frac{x+2}{2}
Extension15

Solve, showing both distributive steps before collecting.

2(x−1)−3(x+2)=4−2x2(x-1)-3(x+2)=4-2x
Extension16

Use 6 as a common multiplier and retain the subtraction of the entire second numerator.

x+12−x−23=3\frac{x+1}{2}-\frac{x-2}{3}=3

Equations: brackets and constant denominators — Teacher answers

Year 7, upper volume · text pp. 63–66 · foundation lessons 32–33

Use the prerequisite checks to choose support, project the worked and guided items, then let learners work on paper. Select supplements or extensions as needed. No learner answers are saved by this presentation.

Worked example1

Remove brackets, solve and check the original equation.

4(x−3)=2x+64(x-3)=2x+6
Reference answer

x = 9.

x=9x=9
  1. Distribute: 4x − 12 = 2x + 6.
  2. Subtract 2x and add 12 on both sides: 2x = 18, so x = 9.
  3. Check: 4(9 − 3) = 24 and 2 × 9 + 6 = 24.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Guided attempt2

Multiply both sides by 3 first, then solve and check.

x+13=5\frac{x+1}{3}=5
  1. 3 × [(x + 1)/3] = 3 × ____.
  2. x + 1 = ____, so x = ____.
  3. Original check: (____ + 1)/3 = ____.
Reference answer

x = 14.

x=14x=14
  1. Multiply both complete sides by 3: x + 1 = 15.
  2. Subtract 1: x = 14.
  3. Check: (14 + 1)/3 = 5.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Prerequisite3

Expand the product; no equation needs to be solved.

2(x−4)2(x-4)
Reference answer

2x−8.

2x−82x-8
  1. 2 × x = 2x and 2 × (−4) = −8.
  2. The result is 2x − 8.
  • Keep each sign and collect only terms with the same variable powers.
  • Give an equivalent simplified expression.
Prerequisite4

Calculate the product used when clearing a denominator of 2.

6×126\times\frac{1}{2}
Reference answer

The value is 3.

33
  1. 6 divided by 2 is 3.
  2. Thus multiplying 1/2 by 6 gives 3.
  • Apply the relevant sign and order rules to every operation.
  • Show an intermediate calculation and obtain the exact stated value.
Independent practice5

Solve and verify.

2(x+4)−3=x+92(x+4)-3=x+9
Reference answer

x = 4.

x=4x=4
  1. Expand: 2x + 8 − 3 = x + 9, or 2x + 5 = x + 9.
  2. Subtract x and 5 on both sides: x = 4.
  3. Check: 2(4 + 4) − 3 = 13 = 4 + 9.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Independent practice6

Solve, taking care with the negative multiplier.

5−2(x−1)=3x−85-2(x-1)=3x-8
Reference answer

x = 3.

x=3x=3
  1. Expand: 5 − 2x + 2 = 3x − 8.
  2. 7 − 2x = 3x − 8 gives 15 = 5x, so x = 3.
  3. Check: 5 − 2(3 − 1) = 1 and 3 × 3 − 8 = 1.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Independent practice7

Clear the denominator and solve. Do not leave the standalone 1 unchanged.

x2+1=4\frac{x}{2}+1=4
Reference answer

x = 6.

x=6x=6
  1. Multiply every term by 2: x + 2 = 8.
  2. x = 6; check 6/2 + 1 = 4.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Independent practice8

Use a common multiplier for 3 and 5, then solve.

x−23=x+45\frac{x-2}{3}=\frac{x+4}{5}
Reference answer

x = 11.

x=11x=11
  1. Multiply both sides by 15: 5(x − 2) = 3(x + 4).
  2. 5x − 10 = 3x + 12 gives 2x = 22 and x = 11.
  3. Check: (11 − 2)/3 = 3 and (11 + 4)/5 = 3.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Exit question9

Solve and check in the original fractional equation.

2x+13−1=4\frac{2x+1}{3}-1=4
Reference answer

x = 7.

x=7x=7
  1. Multiply all terms by 3: 2x + 1 − 3 = 12.
  2. 2x = 14, so x = 7.
  3. Check: (2 × 7 + 1)/3 − 1 = 5 − 1 = 4.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Exit question10

A learner multiplies (x + 2)/4 − 3 = 1 by 4 but writes x + 2 − 3 = 4. Identify the first error and solve correctly.

x+24−3=1\frac{x+2}{4}-3=1
Reference answer

The −3 must also be multiplied by 4. Correct: x + 2 − 12 = 4, so x = 14.

x=14x=14
  1. The multiplier applies to the complete left side, including −3.
  2. x + 2 − 12 = 4 gives x − 10 = 4 and x = 14.
  3. Check: (14 + 2)/4 − 3 = 1.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Further practice11

Solve after expanding both sides.

3(x+2)=2(x+5)3(x+2)=2(x+5)
Reference answer

x = 4.

x=4x=4
  1. 3x + 6 = 2x + 10.
  2. x = 4; both original sides are 18.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Further practice12

Solve with the outside minus applied to both enclosed terms.

−(x−5)=2x−1-(x-5)=2x-1
Reference answer

x = 2.

x=2x=2
  1. Expand to −x + 5 = 2x − 1.
  2. 6 = 3x, so x = 2; both sides are 3.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Further practice13

Clear both denominators using 6, then solve.

x3+x6=5\frac{x}{3}+\frac{x}{6}=5
Reference answer

x = 10.

x=10x=10
  1. Multiplying by 6 gives 2x + x = 30.
  2. 3x = 30, so x = 10. Check 10/3 + 10/6 = 5.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Further practice14

Multiply both sides by 4, then solve.

3x−24=x+22\frac{3x-2}{4}=\frac{x+2}{2}
Reference answer

x = 6.

x=6x=6
  1. 3x − 2 = 2(x + 2).
  2. 3x − 2 = 2x + 4, so x = 6.
  3. Both original sides equal 4.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Extension15

Solve, showing both distributive steps before collecting.

2(x−1)−3(x+2)=4−2x2(x-1)-3(x+2)=4-2x
Reference answer

x = 12.

x=12x=12
  1. Expand: 2x − 2 − 3x − 6 = 4 − 2x.
  2. −x − 8 = 4 − 2x. Add 2x and 8: x = 12.
  3. Check: 2 × 11 − 3 × 14 = −20 and 4 − 24 = −20.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.
Extension16

Use 6 as a common multiplier and retain the subtraction of the entire second numerator.

x+12−x−23=3\frac{x+1}{2}-\frac{x-2}{3}=3
Reference answer

x = 11.

x=11x=11
  1. Multiply by 6: 3(x + 1) − 2(x − 2) = 18.
  2. 3x + 3 − 2x + 4 = 18, so x + 7 = 18 and x = 11.
  3. Check: 12/2 − 9/3 = 6 − 3 = 3.
  • Use the original equality and apply each operation to both sides.
  • Obtain the solution and check it in the original equation or situation.