PAPER · REASONING · PRACTICE

Rational equations: checking and applications

Clear denominators completely, check extraneous candidates, and model work-rate and travel problems.

40 minutes · Year 8, upper volume · text pp. 97–104 · foundation lessons 49–52

Textbook reference: Year 8, upper · Chapter 15: rational expressions · lessons 49–52

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.

Teacher reference Arrange in My Classroom

Lesson routePhase

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

Observe · explain

A question before practising

Compare a valid solution with a candidate that makes an original denominator zero. Which line of the work exposes the difference?

Organise the key idea

Clearing denominators produces candidates. Only values that satisfy the original defined equation and the context are solutions.

Open the original chapter

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

Worked example1

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{x}{x-2}=\frac{2}{x-2}

Start with the forbidden values from the original denominators.
Multiply every term by the least common denominator, including standalone terms.
A candidate is a solution only when the original equation is defined and both sides agree.

The original equation requires x≠2.
Multiplying both sides by x −2 gives x=2.
That candidate makes both original denominators zero, so it is rejected.
The original equation therefore has no solution.

Check your work

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Guided attempt2

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{3}{x-1}=2

Restriction x≠___; cleared equation 3=___.
Candidate x=___; original denominator=___, so ___.

Start with the forbidden values from the original denominators.
Multiply every term by the least common denominator, including standalone terms.
A candidate is a solution only when the original equation is defined and both sides agree.

The restriction is x≠1.
Multiply every term by x −1:3=2(x−1).
3=2x−2, so x=5/2.
Check:3/(5/2−1)=3/(3/2)=2, and 5/2≠1.

Check your work

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Prerequisite3

Before solving any equation containing 1/(x−3), state its forbidden value.

\frac{1}{x-3}

Find values that make an original denominator or a divisor zero.
The denominator x−3 must be nonzero.
The simplified expression is (1)/(x-3); keep x≠3.

First record x≠3 from the original denominators and any divisor.
The denominator x−3 must be nonzero.
(1)/(x-3) = (1)/(x-3).
Therefore (1)/(x-3), with x≠3 still required.

Check your work

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Prerequisite4

A team completes one job in 6 days at a constant rate. What fraction of the job does it complete per day?

Identify the known quantities and the required quantity.
Write the relevant equality before calculating.
Substitute carefully, retain signs and units, and check the result.

Treat the whole job as 1.
Daily rate=whole job÷days=1/6.

Check your work

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Independent practice5

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{2}{x-1}=1

Start with the forbidden values from the original denominators.
Multiply every term by the least common denominator, including standalone terms.
A candidate is a solution only when the original equation is defined and both sides agree.

Record x≠1.
2=x−1 gives x=3.
Check 2/(3−1)=1 and the denominator 2 is nonzero.

Check your work

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Independent practice6

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{x+1}{x-2}=\frac{3}{x-2}

Start with the forbidden values from the original denominators.
Multiply every term by the least common denominator, including standalone terms.
A candidate is a solution only when the original equation is defined and both sides agree.

The original restriction is x≠2.
Clearing denominators gives x+1=3, so the candidate is x=2.
The candidate is forbidden; the original equation has no solution.

Check your work

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Independent practice7

Team A alone takes 12 days. A and B together take 4 days. Let x be B’s days alone. Form and solve a rate equation.

Express each rate or travel time using the same unknown and consistent units.
Add the completed work, or equate the sum of rates to the combined rate.
Solve, substitute into the original model, and reject forbidden or nonpositive contextual values.

A’s rate is 1/12 and B’s rate is 1/x; require x>0.
The rates add:1/12+1/x=1/4.
Thus 1/x=1/6, so x=6 days.
Check 1/12+1/6=1/4;6 is positive and nonzero.

Check your work

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Independent practice8

Two travellers take the same time. One covers 90 km at x km/h; the other covers 120 km at (x+10) km/h. Find x.

Express each rate or travel time using the same unknown and consistent units.
Use distance÷speed for each time and identify whether times are equal or differ.
Solve, substitute into the original model, and reject forbidden or nonpositive contextual values.

Times are 90/x and 120/(x+10); require x>0.
Equal times give 90(x+10)=120x after clearing denominators.
900=30x, so x=30 km/h.
Check 90/30=120/40=3 hours; both speeds are positive.

Check your work

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Exit question9

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{1}{x-1}=\frac{2}{x-1}

Start with the forbidden values from the original denominators.
Multiply every term by the least common denominator, including standalone terms.
A candidate is a solution only when the original equation is defined and both sides agree.

The original domain requires x≠1.
Clearing the common denominator gives 1=2, which is impossible.
Therefore there is no solution; x=1 cannot be used to evade the contradiction.

Check your work

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Exit question10

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{4}{x}=2

Start with the forbidden values from the original denominators.
Multiply every term by the least common denominator, including standalone terms.
A candidate is a solution only when the original equation is defined and both sides agree.

Record x≠0, then 4=2x.
x=2; check 4/2=2 in the original equation.

Check your work

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Further practice11

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{3}{x^2-4}=\frac{1}{x-2}

Start with the forbidden values from the original denominators.
Multiply every term by the least common denominator, including standalone terms.
A candidate is a solution only when the original equation is defined and both sides agree.

Factor x²−4=(x−2)(x+2), so x≠−2,2.
Multiply by the full denominator:3=x+2.
x=1; check 3/(1−4)=−1=1/(1−2).

Check your work

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Further practice12

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{1}{2x}=\frac{2}{x+3}

Start with the forbidden values from the original denominators.
Multiply every term by the least common denominator, including standalone terms.
A candidate is a solution only when the original equation is defined and both sides agree.

The restrictions are x≠0,−3.
Multiply by 2x(x+3):x+3=4x, so x=1.
Check 1/(2×1)=2/(1+3)=1/2.

Check your work

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Further practice13

A alone takes 6 days and B alone takes 3 days. Let x be their days working together. Form and solve a work equation.

Express each rate or travel time using the same unknown and consistent units.
Add the completed work, or equate the sum of rates to the combined rate.
Solve, substitute into the original model, and reject forbidden or nonpositive contextual values.

In x days the work fractions are x/6 and x/3.
x/6+x/3=1 gives x+2x=6, so x=2.
Check 2/6+2/3=1;2 days is positive.

Check your work

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Further practice14

Two trips take equal time:180 km at x km/h and 240 km at (x+20) km/h. Find x.

Express each rate or travel time using the same unknown and consistent units.
Use distance÷speed for each time and identify whether times are equal or differ.
Solve, substitute into the original model, and reject forbidden or nonpositive contextual values.

Use time=distance/speed and require x>0.
180(x+20)=240x gives 3600=60x, so x=60.
Check 180/60=240/80=3 hours.

Check your work

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Extension15

A alone takes 8 days, and A+B together take 3 days. How many days does B take alone? Let this number be x.

Express each rate or travel time using the same unknown and consistent units.
Add the completed work, or equate the sum of rates to the combined rate.
Solve, substitute into the original model, and reject forbidden or nonpositive contextual values.

Require x>0 and add the rates:1/8+1/x=1/3.
1/x=1/3−1/8=5/24, so x=24/5 days.
Check 1/8+5/24=3/24+5/24=1/3; the noninteger time is valid.

Check your work

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Extension16

A cyclist travels 12 km at x km/h. A bus travels the same route at 2x km/h but starts 1/2 hour later; they arrive together. Find x.

Express each rate or travel time using the same unknown and consistent units.
Use distance÷speed for each time and identify whether times are equal or differ.
Solve, substitute into the original model, and reject forbidden or nonpositive contextual values.

The cyclist’s time exceeds the bus time by 1/2 hour; require x>0.
12/x=6/x+1/2 gives 6/x=1/2, so x=12.
Check cyclist 1 hour, bus 1/2 hour, and the delayed starts yield the same arrival time.

Check your work

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Preparation and assessment notes

Prerequisites

Rational operations, linear equations, work fractions and distance÷speed.

Look for in students’ work

Use the worked model and guided attempt selectively, then the eight core questions. Keep further practice and extensions optional. Ask for a reason and an original-equation or expansion check.

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

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.

Print edition

Textbook reference and authorship

Chapter-core practice after the mapped textbook lessons; these tasks do not replace every first-teaching period.

大湾区初中数学三维课堂（八年级上册）正文.pdf

Year 8, upper volume · text pp. 97–104 · foundation lessons 49–52

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

Rational equations: checking and applications

Year 8, upper volume · text pp. 97–104 · foundation lessons 49–52 · 1/2

Name: __________ Class: ______ Date: __________

Worked example1

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{x}{x-2}=\frac{2}{x-2}

Guided attempt2

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{3}{x-1}=2

Restriction x≠___; cleared equation 3=___.
Candidate x=___; original denominator=___, so ___.

Prerequisite3

Before solving any equation containing 1/(x−3), state its forbidden value.

\frac{1}{x-3}

Prerequisite4

A team completes one job in 6 days at a constant rate. What fraction of the job does it complete per day?

Rational equations: checking and applications

Year 8, upper volume · text pp. 97–104 · foundation lessons 49–52 · 2/2

Name: __________ Class: ______ Date: __________

Independent practice5

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{2}{x-1}=1

Independent practice6

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{x+1}{x-2}=\frac{3}{x-2}

Independent practice7

Team A alone takes 12 days. A and B together take 4 days. Let x be B’s days alone. Form and solve a rate equation.

Independent practice8

Two travellers take the same time. One covers 90 km at x km/h; the other covers 120 km at (x+10) km/h. Find x.

Exit question9

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{1}{x-1}=\frac{2}{x-1}

Exit question10

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{4}{x}=2

Rational equations: checking and applications — Further practice

Year 8, upper volume · text pp. 97–104 · foundation lessons 49–52

Further practice11

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{3}{x^2-4}=\frac{1}{x-2}

Further practice12

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{1}{2x}=\frac{2}{x+3}

Further practice13

A alone takes 6 days and B alone takes 3 days. Let x be their days working together. Form and solve a work equation.

Further practice14

Two trips take equal time:180 km at x km/h and 240 km at (x+20) km/h. Find x.

Extension15

A alone takes 8 days, and A+B together take 3 days. How many days does B take alone? Let this number be x.

Extension16

A cyclist travels 12 km at x km/h. A bus travels the same route at 2x km/h but starts 1/2 hour later; they arrive together. Find x.

Rational equations: checking and applications — Teacher answers

Year 8, upper volume · text pp. 97–104 · foundation lessons 49–52

Use the worked model and guided attempt selectively, then the eight core questions. Keep further practice and extensions optional. Ask for a reason and an original-equation or expansion check.

Worked example1

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{x}{x-2}=\frac{2}{x-2}

Reference answer

No solution.

The original equation requires x≠2.
Multiplying both sides by x −2 gives x=2.
That candidate makes both original denominators zero, so it is rejected.
The original equation therefore has no solution.

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Guided attempt2

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{3}{x-1}=2

Restriction x≠___; cleared equation 3=___.
Candidate x=___; original denominator=___, so ___.

Reference answer

x=5/2.

The restriction is x≠1.
Multiply every term by x −1:3=2(x−1).
3=2x−2, so x=5/2.
Check:3/(5/2−1)=3/(3/2)=2, and 5/2≠1.

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Prerequisite3

Before solving any equation containing 1/(x−3), state its forbidden value.

\frac{1}{x-3}

Reference answer

(1)/(x-3); x≠3.

\frac{1}{x-3}

First record x≠3 from the original denominators and any divisor.
The denominator x−3 must be nonzero.
(1)/(x-3) = (1)/(x-3).
Therefore (1)/(x-3), with x≠3 still required.

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Prerequisite4

A team completes one job in 6 days at a constant rate. What fraction of the job does it complete per day?

Reference answer

1/6.

\frac{1}{6}

Treat the whole job as 1.
Daily rate=whole job÷days=1/6.

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Independent practice5

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{2}{x-1}=1

Reference answer

x=3.

Record x≠1.
2=x−1 gives x=3.
Check 2/(3−1)=1 and the denominator 2 is nonzero.

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Independent practice6

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{x+1}{x-2}=\frac{3}{x-2}

Reference answer

No solution.

The original restriction is x≠2.
Clearing denominators gives x+1=3, so the candidate is x=2.
The candidate is forbidden; the original equation has no solution.

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Independent practice7

Team A alone takes 12 days. A and B together take 4 days. Let x be B’s days alone. Form and solve a rate equation.

Reference answer

x=6 days; it is positive and satisfies the original model.

\frac{1}{12}+\frac{1}{x}=\frac{1}{4}

A’s rate is 1/12 and B’s rate is 1/x; require x>0.
The rates add:1/12+1/x=1/4.
Thus 1/x=1/6, so x=6 days.
Check 1/12+1/6=1/4;6 is positive and nonzero.

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Independent practice8

Two travellers take the same time. One covers 90 km at x km/h; the other covers 120 km at (x+10) km/h. Find x.

Reference answer

x=30 km/h; it is positive and satisfies the original model.

\frac{90}{x}=\frac{120}{x+10}

Times are 90/x and 120/(x+10); require x>0.
Equal times give 90(x+10)=120x after clearing denominators.
900=30x, so x=30 km/h.
Check 90/30=120/40=3 hours; both speeds are positive.

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Exit question9

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{1}{x-1}=\frac{2}{x-1}

Reference answer

No solution.

The original domain requires x≠1.
Clearing the common denominator gives 1=2, which is impossible.
Therefore there is no solution; x=1 cannot be used to evade the contradiction.

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Exit question10

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{4}{x}=2

Reference answer

x=2.

Record x≠0, then 4=2x.
x=2; check 4/2=2 in the original equation.

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Further practice11

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{3}{x^2-4}=\frac{1}{x-2}

Reference answer

x=1.

Factor x²−4=(x−2)(x+2), so x≠−2,2.
Multiply by the full denominator:3=x+2.
x=1; check 3/(1−4)=−1=1/(1−2).

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Further practice12

Solve the equation. State the restrictions and check every candidate in the original equation.

\frac{1}{2x}=\frac{2}{x+3}

Reference answer

x=1.

The restrictions are x≠0,−3.
Multiply by 2x(x+3):x+3=4x, so x=1.
Check 1/(2×1)=2/(1+3)=1/2.

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Further practice13

A alone takes 6 days and B alone takes 3 days. Let x be their days working together. Form and solve a work equation.

Reference answer

x=2 days; it is positive and satisfies the original model.

\frac{x}{6}+\frac{x}{3}=1

In x days the work fractions are x/6 and x/3.
x/6+x/3=1 gives x+2x=6, so x=2.
Check 2/6+2/3=1;2 days is positive.

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Further practice14

Two trips take equal time:180 km at x km/h and 240 km at (x+20) km/h. Find x.

Reference answer

x=60 km/h; it is positive and satisfies the original model.

\frac{180}{x}=\frac{240}{x+20}

Use time=distance/speed and require x>0.
180(x+20)=240x gives 3600=60x, so x=60.
Check 180/60=240/80=3 hours.

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Extension15

A alone takes 8 days, and A+B together take 3 days. How many days does B take alone? Let this number be x.

Reference answer

x=24/5 days; it is positive and satisfies the original model.

\frac{1}{8}+\frac{1}{x}=\frac{1}{3}

Require x>0 and add the rates:1/8+1/x=1/3.
1/x=1/3−1/8=5/24, so x=24/5 days.
Check 1/8+5/24=3/24+5/24=1/3; the noninteger time is valid.

Show a complete equation after clearing denominators.
Check original denominators and the original equality.

Extension16

A cyclist travels 12 km at x km/h. A bus travels the same route at 2x km/h but starts 1/2 hour later; they arrive together. Find x.

Reference answer

x=12 km/h; it is positive and satisfies the original model.

\frac{12}{x}=\frac{12}{2x}+\frac{1}{2}

The cyclist’s time exceeds the bus time by 1/2 hour; require x>0.
12/x=6/x+1/2 gives 6/x=1/2, so x=12.
Check cyclist 1 hour, bus 1/2 hour, and the delayed starts yield the same arrival time.

Show a complete equation after clearing denominators.
Check original denominators and the original equality.