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Two-variable systems: substitution and elimination

Reduce two unknowns to one, then recover the complete solution and check both equations.

40 minutes · Year 7, lower volume · text pp. 51–60 · textbook lessons 26–30

Textbook reference: Year 7 lower · Chapter 8 Simultaneous linear equations · text lessons 26–30

Chapter core practice: use after the referenced textbook lessons, or for chapter review. See the lower-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 referenceArrange 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

18 minutes using tickets: A=2 dollars, B=5 dollars, total 10, revenue 32. Predict a feasible pair (2 min), list possible integer pairs and the two equations (4 min), then have two groups solve the same model by substitution and by scaled subtraction (7 min). Compare how each method removes one unknown and check both original conditions (3 min). Change only revenue to 31 and explain why the real solution is not a feasible whole-ticket solution (2 min); do not round counts.

Organise the key idea

Reduce two unknowns to one, then recover the complete solution and check both equations.

Open the original chapterOpen the interactive inquiry

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

Ordinary route: use the first fixed case for the five-minute demonstration. Advanced route: follow the full 18-minute investigation; record predictions before checking.

Worked example1

Use substitution to solve the system, then check both equations.

{(1)x+(1)y=7(−2)x+(1)y=1\begin{cases}(1)x+(1)y=7\\(-2)x+(1)y=1\end{cases}
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Guided attempt2

Solve the system. Show elimination and check every original equation.

{(2)x+(1)y=11(1)x+(−1)y=1\begin{cases}(2)x+(1)y=11\\(1)x+(-1)y=1\end{cases}
  1. Add the original equations: (2x+y)+(x−y)=11+1(2x+y)+(x-y)=11+1.
  2. Collect: 3x=□3x=\square, so x=□x=\square.
  3. Back-substitute: y=11−2(□)=□y=11-2(\square)=\square; check both original equations.
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Prerequisite3

Does x=3,y=4x=3, y=4 solve the displayed system? Check every equation.

{(1)x+(1)y=7(2)x+(−1)y=2\begin{cases}(1)x+(1)y=7\\(2)x+(-1)y=2\end{cases}
  1. Yes, every original equation holds.
  2. No, checking one equation is enough to reject it.
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Prerequisite4

Does x=2,y=5x=2, y=5 solve the displayed system? Check every equation.

{(1)x+(1)y=7(2)x+(−1)y=2\begin{cases}(1)x+(1)y=7\\(2)x+(-1)y=2\end{cases}
  1. Yes, one matching equation is sufficient.
  2. No, at least one original equation fails.
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Independent practice5

Solve by substitution and check both equations.

{(1)x+(−1)y=2(2)x+(1)y=10\begin{cases}(1)x+(-1)y=2\\(2)x+(1)y=10\end{cases}
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Independent practice6

Solve the system. Show elimination and check every original equation.

{(2)x+(3)y=16(3)x+(−2)y=11\begin{cases}(2)x+(3)y=16\\(3)x+(-2)y=11\end{cases}
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Independent practice7

Solve the system. Show elimination and check every original equation.

{(3)x+(−2)y=0(2)x+(1)y=7\begin{cases}(3)x+(-2)y=0\\(2)x+(1)y=7\end{cases}
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Independent practice8

Solve the system. Show elimination and check every original equation.

{(1)x+(1)y=3(2)x+(−1)y=1\begin{cases}(1)x+(1)y=3\\(2)x+(-1)y=1\end{cases}
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Exit question9

Does x=2,y=3x=2, y=3 solve the displayed system? Check every equation.

{(1)x+(2)y=8(3)x+(−1)y=3\begin{cases}(1)x+(2)y=8\\(3)x+(-1)y=3\end{cases}
  1. Yes, every original equation holds.
  2. No, checking one equation is enough to reject it.
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Exit question10

To multiply 2x−y=52x-y=5 by 3, which equation is correct?

  1. 6x−3y=156x-3y=15
  2. 6x−y=156x-y=15
  3. 6x−3y=56x-3y=5
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Further practice11

Solve the system. Show elimination and check every original equation.

{(1)x+(1)y=9(1)x+(−1)y=3\begin{cases}(1)x+(1)y=9\\(1)x+(-1)y=3\end{cases}
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Further practice12

Solve the system. Show elimination and check every original equation.

{(2)x+(3)y=13(2)x+(−1)y=1\begin{cases}(2)x+(3)y=13\\(2)x+(-1)y=1\end{cases}
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Further practice13

Solve the system. Show elimination and check every original equation.

{(3)x+(2)y=19(5)x+(−3)y=0\begin{cases}(3)x+(2)y=19\\(5)x+(-3)y=0\end{cases}
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Further practice14

Solve the system. Show elimination and check every original equation.

{(1)x+(1)y=1(2)x+(−1)y=−7\begin{cases}(1)x+(1)y=1\\(2)x+(-1)y=-7\end{cases}
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Extension15

Describe every real solution of the system. Explain why two written equations do not guarantee a unique pair.

{(1)x+(1)y=6(2)x+(2)y=12\begin{cases}(1)x+(1)y=6\\(2)x+(2)y=12\end{cases}
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Extension16

Decide whether the system has a real solution. Explain by elimination.

{(1)x+(1)y=6(2)x+(2)y=13\begin{cases}(1)x+(1)y=6\\(2)x+(2)y=13\end{cases}
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Preparation and assessment notes

Prerequisites

Solve one-variable equations, distribute a multiplier and substitute a signed value into an expression.

Look for in students’ work

Use the worked and guided tasks selectively. The eight core paper tasks are prerequisites, independent tasks and exits; further practice and extensions are optional.

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.

Textbook reference and authorship

Original 40-minute core consolidation after the mapped textbook lessons; not a replacement for every first-teaching period. Only the supplied text has been mapped.

大湾区初中数学三维课堂（七年级下册）正文.pdf

Year 7, lower volume · text pp. 51–60 · textbook lessons 26–30

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

Two-variable systems: substitution and elimination

Year 7, lower volume · text pp. 51–60 · textbook lessons 26–30 · 1/2

Name: __________ Class: ______ Date: __________

Worked example1

Use substitution to solve the system, then check both equations.

{(1)x+(1)y=7(−2)x+(1)y=1\begin{cases}(1)x+(1)y=7\\(-2)x+(1)y=1\end{cases}
Guided attempt2

Solve the system. Show elimination and check every original equation.

{(2)x+(1)y=11(1)x+(−1)y=1\begin{cases}(2)x+(1)y=11\\(1)x+(-1)y=1\end{cases}
  1. Add the original equations: (2x+y)+(x−y)=11+1(2x+y)+(x-y)=11+1.
  2. Collect: 3x=□3x=\square, so x=□x=\square.
  3. Back-substitute: y=11−2(□)=□y=11-2(\square)=\square; check both original equations.
Prerequisite3

Does x=3,y=4x=3, y=4 solve the displayed system? Check every equation.

{(1)x+(1)y=7(2)x+(−1)y=2\begin{cases}(1)x+(1)y=7\\(2)x+(-1)y=2\end{cases}
  1. Yes, every original equation holds.
  2. No, checking one equation is enough to reject it.
Prerequisite4

Does x=2,y=5x=2, y=5 solve the displayed system? Check every equation.

{(1)x+(1)y=7(2)x+(−1)y=2\begin{cases}(1)x+(1)y=7\\(2)x+(-1)y=2\end{cases}
  1. Yes, one matching equation is sufficient.
  2. No, at least one original equation fails.
g7-systems-elimination · 2026-10-15-lower-core-v1 · Conditions and reasons matter; measurement is not proof.

Two-variable systems: substitution and elimination

Year 7, lower volume · text pp. 51–60 · textbook lessons 26–30 · 2/2

Name: __________ Class: ______ Date: __________

Independent practice5

Solve by substitution and check both equations.

{(1)x+(−1)y=2(2)x+(1)y=10\begin{cases}(1)x+(-1)y=2\\(2)x+(1)y=10\end{cases}
Independent practice6

Solve the system. Show elimination and check every original equation.

{(2)x+(3)y=16(3)x+(−2)y=11\begin{cases}(2)x+(3)y=16\\(3)x+(-2)y=11\end{cases}
Independent practice7

Solve the system. Show elimination and check every original equation.

{(3)x+(−2)y=0(2)x+(1)y=7\begin{cases}(3)x+(-2)y=0\\(2)x+(1)y=7\end{cases}
Independent practice8

Solve the system. Show elimination and check every original equation.

{(1)x+(1)y=3(2)x+(−1)y=1\begin{cases}(1)x+(1)y=3\\(2)x+(-1)y=1\end{cases}
Exit question9

Does x=2,y=3x=2, y=3 solve the displayed system? Check every equation.

{(1)x+(2)y=8(3)x+(−1)y=3\begin{cases}(1)x+(2)y=8\\(3)x+(-1)y=3\end{cases}
  1. Yes, every original equation holds.
  2. No, checking one equation is enough to reject it.
Exit question10

To multiply 2x−y=52x-y=5 by 3, which equation is correct?

  1. 6x−3y=156x-3y=15
  2. 6x−y=156x-y=15
  3. 6x−3y=56x-3y=5
g7-systems-elimination · 2026-10-15-lower-core-v1 · Conditions and reasons matter; measurement is not proof.

Two-variable systems: substitution and elimination — Further practice

Year 7, lower volume · text pp. 51–60 · textbook lessons 26–30

Further practice11

Solve the system. Show elimination and check every original equation.

{(1)x+(1)y=9(1)x+(−1)y=3\begin{cases}(1)x+(1)y=9\\(1)x+(-1)y=3\end{cases}
Further practice12

Solve the system. Show elimination and check every original equation.

{(2)x+(3)y=13(2)x+(−1)y=1\begin{cases}(2)x+(3)y=13\\(2)x+(-1)y=1\end{cases}
Further practice13

Solve the system. Show elimination and check every original equation.

{(3)x+(2)y=19(5)x+(−3)y=0\begin{cases}(3)x+(2)y=19\\(5)x+(-3)y=0\end{cases}
Further practice14

Solve the system. Show elimination and check every original equation.

{(1)x+(1)y=1(2)x+(−1)y=−7\begin{cases}(1)x+(1)y=1\\(2)x+(-1)y=-7\end{cases}
Extension15

Describe every real solution of the system. Explain why two written equations do not guarantee a unique pair.

{(1)x+(1)y=6(2)x+(2)y=12\begin{cases}(1)x+(1)y=6\\(2)x+(2)y=12\end{cases}
Extension16

Decide whether the system has a real solution. Explain by elimination.

{(1)x+(1)y=6(2)x+(2)y=13\begin{cases}(1)x+(1)y=6\\(2)x+(2)y=13\end{cases}

Two-variable systems: substitution and elimination — Teacher answers

Year 7, lower volume · text pp. 51–60 · textbook lessons 26–30

Use the worked and guided tasks selectively. The eight core paper tasks are prerequisites, independent tasks and exits; further practice and extensions are optional.

Ordinary route: prerequisites 4, short inquiry 5, worked/guided 8, independent 15, correction 5, exit 3 minutes. Advanced route: inquiry 18, organise 4, selected practice 13, exit 5 minutes.

Prerequisites: Solve one-variable equations, distribute a multiplier and substitute a signed value into an expression.

Ordinary route: use the first fixed case for the five-minute demonstration. Advanced route: complete the full investigation.

Inquiry and discussion: 18 minutes using tickets: A=2 dollars, B=5 dollars, total 10, revenue 32. Predict a feasible pair (2 min), list possible integer pairs and the two equations (4 min), then have two groups solve the same model by substitution and by scaled subtraction (7 min). Compare how each method removes one unknown and check both original conditions (3 min). Change only revenue to 31 and explain why the real solution is not a feasible whole-ticket solution (2 min); do not round counts.

Worked example1

Use substitution to solve the system, then check both equations.

{(1)x+(1)y=7(−2)x+(1)y=1\begin{cases}(1)x+(1)y=7\\(-2)x+(1)y=1\end{cases}
Reference answer

x=2,y=5x=2,\quad y=5.

x=2,y=5x=2,\quad y=5
  1. The second equation gives y=2x+1y=2x+1.
  2. Substitute in the first: x+(2x+1)=7x+(2x+1)=7, so x=2x=2.
  3. Then y=5y=5. Check 2+5=72+5=7 and −2(2)+5=1-2(2)+5=1.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Guided attempt2

Solve the system. Show elimination and check every original equation.

{(2)x+(1)y=11(1)x+(−1)y=1\begin{cases}(2)x+(1)y=11\\(1)x+(-1)y=1\end{cases}
  1. Add the original equations: (2x+y)+(x−y)=11+1(2x+y)+(x-y)=11+1.
  2. Collect: 3x=□3x=\square, so x=□x=\square.
  3. Back-substitute: y=11−2(□)=□y=11-2(\square)=\square; check both original equations.
Reference answer

x=4,y=3x=4,\quad y=3.

x=4,y=3x=4,\quad y=3
  1. Equation 2 minus 12\frac{1}{2} times equation 1 gives (0)x+(−32)y=−92(0)x+(\frac{-3}{2})y=\frac{-9}{2}.
  2. Back-substitute to obtain y=3y=3.
  3. Back-substitute to obtain x=4x=4.
  4. Substitute the complete tuple into every original equation; each left side equals its stated right side.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Prerequisite3

Does x=3,y=4x=3, y=4 solve the displayed system? Check every equation.

{(1)x+(1)y=7(2)x+(−1)y=2\begin{cases}(1)x+(1)y=7\\(2)x+(-1)y=2\end{cases}
  1. Yes, every original equation holds.
  2. No, checking one equation is enough to reject it.
Reference answer

Yes, every original equation holds.

  1. Equation 1 gives left side 7 and right side 7.
  2. Equation 2 gives left side 2 and right side 2.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Prerequisite4

Does x=2,y=5x=2, y=5 solve the displayed system? Check every equation.

{(1)x+(1)y=7(2)x+(−1)y=2\begin{cases}(1)x+(1)y=7\\(2)x+(-1)y=2\end{cases}
  1. Yes, one matching equation is sufficient.
  2. No, at least one original equation fails.
Reference answer

No, at least one original equation fails.

  1. Equation 1 gives left side 7 and right side 7.
  2. Equation 2 gives left side -1 and right side 2.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Independent practice5

Solve by substitution and check both equations.

{(1)x+(−1)y=2(2)x+(1)y=10\begin{cases}(1)x+(-1)y=2\\(2)x+(1)y=10\end{cases}
Reference answer

x=4,y=2x=4,\quad y=2.

x=4,y=2x=4,\quad y=2
  1. From the first equation, x=y+2x=y+2.
  2. Substitute: 2(y+2)+y=102(y+2)+y=10, so y=2y=2.
  3. Then x=4x=4. Both equations give their original right sides.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Independent practice6

Solve the system. Show elimination and check every original equation.

{(2)x+(3)y=16(3)x+(−2)y=11\begin{cases}(2)x+(3)y=16\\(3)x+(-2)y=11\end{cases}
Reference answer

x=5,y=2x=5,\quad y=2.

x=5,y=2x=5,\quad y=2
  1. Equation 2 minus 32\frac{3}{2} times equation 1 gives (0)x+(−132)y=−13(0)x+(\frac{-13}{2})y=-13.
  2. Back-substitute to obtain y=2y=2.
  3. Back-substitute to obtain x=5x=5.
  4. Substitute the complete tuple into every original equation; each left side equals its stated right side.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Independent practice7

Solve the system. Show elimination and check every original equation.

{(3)x+(−2)y=0(2)x+(1)y=7\begin{cases}(3)x+(-2)y=0\\(2)x+(1)y=7\end{cases}
Reference answer

x=2,y=3x=2,\quad y=3.

x=2,y=3x=2,\quad y=3
  1. Equation 2 minus 23\frac{2}{3} times equation 1 gives (0)x+(73)y=7(0)x+(\frac{7}{3})y=7.
  2. Back-substitute to obtain y=3y=3.
  3. Back-substitute to obtain x=2x=2.
  4. Substitute the complete tuple into every original equation; each left side equals its stated right side.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Independent practice8

Solve the system. Show elimination and check every original equation.

{(1)x+(1)y=3(2)x+(−1)y=1\begin{cases}(1)x+(1)y=3\\(2)x+(-1)y=1\end{cases}
Reference answer

x=43,y=53x=\frac{4}{3},\quad y=\frac{5}{3}.

x=43,y=53x=\frac{4}{3},\quad y=\frac{5}{3}
  1. Equation 2 minus 22 times equation 1 gives (0)x+(−3)y=−5(0)x+(-3)y=-5.
  2. Back-substitute to obtain y=53y=\frac{5}{3}.
  3. Back-substitute to obtain x=43x=\frac{4}{3}.
  4. Substitute the complete tuple into every original equation; each left side equals its stated right side.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Exit question9

Does x=2,y=3x=2, y=3 solve the displayed system? Check every equation.

{(1)x+(2)y=8(3)x+(−1)y=3\begin{cases}(1)x+(2)y=8\\(3)x+(-1)y=3\end{cases}
  1. Yes, every original equation holds.
  2. No, checking one equation is enough to reject it.
Reference answer

Yes, every original equation holds.

  1. Equation 1 gives left side 8 and right side 8.
  2. Equation 2 gives left side 3 and right side 3.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Exit question10

To multiply 2x−y=52x-y=5 by 3, which equation is correct?

  1. 6x−3y=156x-3y=15
  2. 6x−y=156x-y=15
  3. 6x−3y=56x-3y=5
Reference answer

6x−3y=156x-3y=15

  1. Multiply each term on both sides by 3.
  2. The coefficients and right side all scale.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Further practice11

Solve the system. Show elimination and check every original equation.

{(1)x+(1)y=9(1)x+(−1)y=3\begin{cases}(1)x+(1)y=9\\(1)x+(-1)y=3\end{cases}
Reference answer

x=6,y=3x=6,\quad y=3.

x=6,y=3x=6,\quad y=3
  1. Equation 2 minus 11 times equation 1 gives (0)x+(−2)y=−6(0)x+(-2)y=-6.
  2. Back-substitute to obtain y=3y=3.
  3. Back-substitute to obtain x=6x=6.
  4. Substitute the complete tuple into every original equation; each left side equals its stated right side.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Further practice12

Solve the system. Show elimination and check every original equation.

{(2)x+(3)y=13(2)x+(−1)y=1\begin{cases}(2)x+(3)y=13\\(2)x+(-1)y=1\end{cases}
Reference answer

x=2,y=3x=2,\quad y=3.

x=2,y=3x=2,\quad y=3
  1. Equation 2 minus 11 times equation 1 gives (0)x+(−4)y=−12(0)x+(-4)y=-12.
  2. Back-substitute to obtain y=3y=3.
  3. Back-substitute to obtain x=2x=2.
  4. Substitute the complete tuple into every original equation; each left side equals its stated right side.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Further practice13

Solve the system. Show elimination and check every original equation.

{(3)x+(2)y=19(5)x+(−3)y=0\begin{cases}(3)x+(2)y=19\\(5)x+(-3)y=0\end{cases}
Reference answer

x=3,y=5x=3,\quad y=5.

x=3,y=5x=3,\quad y=5
  1. Equation 2 minus 53\frac{5}{3} times equation 1 gives (0)x+(−193)y=−953(0)x+(\frac{-19}{3})y=\frac{-95}{3}.
  2. Back-substitute to obtain y=5y=5.
  3. Back-substitute to obtain x=3x=3.
  4. Substitute the complete tuple into every original equation; each left side equals its stated right side.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Further practice14

Solve the system. Show elimination and check every original equation.

{(1)x+(1)y=1(2)x+(−1)y=−7\begin{cases}(1)x+(1)y=1\\(2)x+(-1)y=-7\end{cases}
Reference answer

x=−2,y=3x=-2,\quad y=3.

x=−2,y=3x=-2,\quad y=3
  1. Equation 2 minus 22 times equation 1 gives (0)x+(−3)y=−9(0)x+(-3)y=-9.
  2. Back-substitute to obtain y=3y=3.
  3. Back-substitute to obtain x=−2x=-2.
  4. Substitute the complete tuple into every original equation; each left side equals its stated right side.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Extension15

Describe every real solution of the system. Explain why two written equations do not guarantee a unique pair.

{(1)x+(1)y=6(2)x+(2)y=12\begin{cases}(1)x+(1)y=6\\(2)x+(2)y=12\end{cases}
Reference answer

(x,y)=(t,6−t), t∈R(x,y)=(t,6-t),\ t\in\mathbb{R}; infinitely many solutions.

(x,y)=(t,6−t), t∈R(x,y)=(t,6-t),\ t\in\mathbb{R}
  1. The second equation is twice the first, so it adds no new restriction.
  2. Choose any real x=t and set y=6−t; both equations hold.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Extension16

Decide whether the system has a real solution. Explain by elimination.

{(1)x+(1)y=6(2)x+(2)y=13\begin{cases}(1)x+(1)y=6\\(2)x+(2)y=13\end{cases}
Reference answer

No solution. Subtracting twice the first from the second gives 0=1.

  1. Twice the first equation is 2x+2y=12.
  2. The second requires the same left side to equal 13, a contradiction.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.