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Three-variable systems: eliminate, solve and check

Eliminate the same unknown to obtain a two-variable system, then back-substitute all three values.

40 minutes · Year 7, lower volume · text pp. 67–70 · textbook lessons 34–35

Textbook reference: Year 7 lower · Chapter 8 Simultaneous linear equations · text lessons 34–35

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 on paper: start with x+y+z=9, x−y=−1, y−z=−1. Predict the order of the three values (3 min). Two groups eliminate a different chosen unknown consistently, each producing a two-variable system and then the complete triple (7 min). Compare results and check all three original equations (4 min). Change only the first total from 9 to 12; predict and verify how every value changes (3 min). State why obtaining only two values does not complete a three-variable solution (1 min). No three-variable interactive engine is claimed.

Organise the key idea

Eliminate the same unknown to obtain a two-variable system, then back-substitute all three values.

Open the original chapter

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

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

{(1)x+(1)y+(1)z=9(1)x+(−1)y+(0)z=−1(0)x+(1)y+(−1)z=−1\begin{cases}(1)x+(1)y+(1)z=9\\(1)x+(-1)y+(0)z=-1\\(0)x+(1)y+(-1)z=-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.

{(1)x+(1)y+(1)z=6(1)x+(−1)y+(0)z=0(0)x+(1)y+(−1)z=0\begin{cases}(1)x+(1)y+(1)z=6\\(1)x+(-1)y+(0)z=0\\(0)x+(1)y+(-1)z=0\end{cases}
  1. From x−y=0x-y=0, write y=□y=\square.
  2. From y−z=0y-z=0, write z=y=□z=y=\square.
  3. Substitute in the first equation: x+x+x=6x+x+x=6, so x=□x=\square; recover y,z and check all three.
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Prerequisite3

Does x=1,y=2,z=3x=1, y=2, z=3 solve the displayed system? Check every equation.

{(1)x+(1)y+(1)z=6(1)x+(−1)y+(0)z=−1(0)x+(1)y+(−1)z=−1\begin{cases}(1)x+(1)y+(1)z=6\\(1)x+(-1)y+(0)z=-1\\(0)x+(1)y+(-1)z=-1\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

To reduce three equations in x,y,z to a two-variable system, what should the two derived equations have in common?

  1. They both eliminate the same unknown and retain the same other two.
  2. One eliminates x and the other only y, leaving three unknowns together.
  3. Their right sides must both be zero.
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Independent practice5

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

{(1)x+(1)y+(1)z=12(1)x+(−1)y+(0)z=2(0)x+(1)y+(−1)z=−1\begin{cases}(1)x+(1)y+(1)z=12\\(1)x+(-1)y+(0)z=2\\(0)x+(1)y+(-1)z=-1\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.

{(1)x+(1)y+(1)z=10(2)x+(−1)y+(0)z=−1(0)x+(1)y+(−1)z=2\begin{cases}(1)x+(1)y+(1)z=10\\(2)x+(-1)y+(0)z=-1\\(0)x+(1)y+(-1)z=2\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.

{(1)x+(1)y+(0)z=5(0)x+(1)y+(1)z=7(1)x+(0)y+(1)z=6\begin{cases}(1)x+(1)y+(0)z=5\\(0)x+(1)y+(1)z=7\\(1)x+(0)y+(1)z=6\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+(1)z=3(1)x+(−1)y+(0)z=1(0)x+(1)y+(−1)z=1\begin{cases}(1)x+(1)y+(1)z=3\\(1)x+(-1)y+(0)z=1\\(0)x+(1)y+(-1)z=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=3,z=4x=2, y=3, z=4 solve the displayed system? Check every equation.

{(1)x+(1)y+(1)z=9(1)x+(−1)y+(0)z=−1(0)x+(1)y+(−1)z=−1\begin{cases}(1)x+(1)y+(1)z=9\\(1)x+(-1)y+(0)z=-1\\(0)x+(1)y+(-1)z=-1\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

After elimination gives y=3 and z=4, the original first equation is x+y+z=9. What remains to finish the solution?

  1. Find x=2, then check all three original equations.
  2. Stop; two values are enough.
  3. Choose any x without checking.
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+(1)z=7(1)x+(−1)y+(0)z=2(0)x+(1)y+(−1)z=1\begin{cases}(1)x+(1)y+(1)z=7\\(1)x+(-1)y+(0)z=2\\(0)x+(1)y+(-1)z=1\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.

{(1)x+(1)y+(1)z=3(2)x+(−1)y+(0)z=0(0)x+(1)y+(−1)z=−12\begin{cases}(1)x+(1)y+(1)z=3\\(2)x+(-1)y+(0)z=0\\(0)x+(1)y+(-1)z=\frac{-1}{2}\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.

{(2)x+(1)y+(−1)z=1(1)x+(−1)y+(1)z=2(1)x+(1)y+(1)z=6\begin{cases}(2)x+(1)y+(-1)z=1\\(1)x+(-1)y+(1)z=2\\(1)x+(1)y+(1)z=6\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+(−2)y+(0)z=0(0)x+(1)y+(−2)z=0(1)x+(1)y+(1)z=21\begin{cases}(1)x+(-2)y+(0)z=0\\(0)x+(1)y+(-2)z=0\\(1)x+(1)y+(1)z=21\end{cases}
Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Extension15

Three ticket types cost 2,3,5 dollars. Their counts are x,y,z. Nine tickets cost 33 dollars, and the first two types total 5 tickets. Form and solve a three-variable system; check integer counts.

Check your work
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Extension16

Do three written equations always determine three unique values? Describe the real solutions of this system.

{(1)x+(1)y+(1)z=6(2)x+(2)y+(2)z=12(1)x+(−1)y+(0)z=0\begin{cases}(1)x+(1)y+(1)z=6\\(2)x+(2)y+(2)z=12\\(1)x+(-1)y+(0)z=0\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

Use substitution/elimination with two unknowns and check a solution in every original equation.

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. 67–70 · textbook lessons 34–35

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

Three-variable systems: eliminate, solve and check

Year 7, lower volume · text pp. 67–70 · textbook lessons 34–35 · 1/2

Name: __________ Class: ______ Date: __________

Worked example1

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

{(1)x+(1)y+(1)z=9(1)x+(−1)y+(0)z=−1(0)x+(1)y+(−1)z=−1\begin{cases}(1)x+(1)y+(1)z=9\\(1)x+(-1)y+(0)z=-1\\(0)x+(1)y+(-1)z=-1\end{cases}
Guided attempt2

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

{(1)x+(1)y+(1)z=6(1)x+(−1)y+(0)z=0(0)x+(1)y+(−1)z=0\begin{cases}(1)x+(1)y+(1)z=6\\(1)x+(-1)y+(0)z=0\\(0)x+(1)y+(-1)z=0\end{cases}
  1. From x−y=0x-y=0, write y=□y=\square.
  2. From y−z=0y-z=0, write z=y=□z=y=\square.
  3. Substitute in the first equation: x+x+x=6x+x+x=6, so x=□x=\square; recover y,z and check all three.
Prerequisite3

Does x=1,y=2,z=3x=1, y=2, z=3 solve the displayed system? Check every equation.

{(1)x+(1)y+(1)z=6(1)x+(−1)y+(0)z=−1(0)x+(1)y+(−1)z=−1\begin{cases}(1)x+(1)y+(1)z=6\\(1)x+(-1)y+(0)z=-1\\(0)x+(1)y+(-1)z=-1\end{cases}
  1. Yes, every original equation holds.
  2. No, checking one equation is enough to reject it.
Prerequisite4

To reduce three equations in x,y,z to a two-variable system, what should the two derived equations have in common?

  1. They both eliminate the same unknown and retain the same other two.
  2. One eliminates x and the other only y, leaving three unknowns together.
  3. Their right sides must both be zero.
g7-three-variable-systems · 2026-10-15-lower-core-v1 · Conditions and reasons matter; measurement is not proof.

Three-variable systems: eliminate, solve and check

Year 7, lower volume · text pp. 67–70 · textbook lessons 34–35 · 2/2

Name: __________ Class: ______ Date: __________

Independent practice5

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

{(1)x+(1)y+(1)z=12(1)x+(−1)y+(0)z=2(0)x+(1)y+(−1)z=−1\begin{cases}(1)x+(1)y+(1)z=12\\(1)x+(-1)y+(0)z=2\\(0)x+(1)y+(-1)z=-1\end{cases}
Independent practice6

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

{(1)x+(1)y+(1)z=10(2)x+(−1)y+(0)z=−1(0)x+(1)y+(−1)z=2\begin{cases}(1)x+(1)y+(1)z=10\\(2)x+(-1)y+(0)z=-1\\(0)x+(1)y+(-1)z=2\end{cases}
Independent practice7

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

{(1)x+(1)y+(0)z=5(0)x+(1)y+(1)z=7(1)x+(0)y+(1)z=6\begin{cases}(1)x+(1)y+(0)z=5\\(0)x+(1)y+(1)z=7\\(1)x+(0)y+(1)z=6\end{cases}
Independent practice8

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

{(1)x+(1)y+(1)z=3(1)x+(−1)y+(0)z=1(0)x+(1)y+(−1)z=1\begin{cases}(1)x+(1)y+(1)z=3\\(1)x+(-1)y+(0)z=1\\(0)x+(1)y+(-1)z=1\end{cases}
Exit question9

Does x=2,y=3,z=4x=2, y=3, z=4 solve the displayed system? Check every equation.

{(1)x+(1)y+(1)z=9(1)x+(−1)y+(0)z=−1(0)x+(1)y+(−1)z=−1\begin{cases}(1)x+(1)y+(1)z=9\\(1)x+(-1)y+(0)z=-1\\(0)x+(1)y+(-1)z=-1\end{cases}
  1. Yes, every original equation holds.
  2. No, checking one equation is enough to reject it.
Exit question10

After elimination gives y=3 and z=4, the original first equation is x+y+z=9. What remains to finish the solution?

  1. Find x=2, then check all three original equations.
  2. Stop; two values are enough.
  3. Choose any x without checking.
g7-three-variable-systems · 2026-10-15-lower-core-v1 · Conditions and reasons matter; measurement is not proof.

Three-variable systems: eliminate, solve and check — Further practice

Year 7, lower volume · text pp. 67–70 · textbook lessons 34–35

Further practice11

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

{(1)x+(1)y+(1)z=7(1)x+(−1)y+(0)z=2(0)x+(1)y+(−1)z=1\begin{cases}(1)x+(1)y+(1)z=7\\(1)x+(-1)y+(0)z=2\\(0)x+(1)y+(-1)z=1\end{cases}
Further practice12

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

{(1)x+(1)y+(1)z=3(2)x+(−1)y+(0)z=0(0)x+(1)y+(−1)z=−12\begin{cases}(1)x+(1)y+(1)z=3\\(2)x+(-1)y+(0)z=0\\(0)x+(1)y+(-1)z=\frac{-1}{2}\end{cases}
Further practice13

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

{(2)x+(1)y+(−1)z=1(1)x+(−1)y+(1)z=2(1)x+(1)y+(1)z=6\begin{cases}(2)x+(1)y+(-1)z=1\\(1)x+(-1)y+(1)z=2\\(1)x+(1)y+(1)z=6\end{cases}
Further practice14

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

{(1)x+(−2)y+(0)z=0(0)x+(1)y+(−2)z=0(1)x+(1)y+(1)z=21\begin{cases}(1)x+(-2)y+(0)z=0\\(0)x+(1)y+(-2)z=0\\(1)x+(1)y+(1)z=21\end{cases}
Extension15

Three ticket types cost 2,3,5 dollars. Their counts are x,y,z. Nine tickets cost 33 dollars, and the first two types total 5 tickets. Form and solve a three-variable system; check integer counts.

Extension16

Do three written equations always determine three unique values? Describe the real solutions of this system.

{(1)x+(1)y+(1)z=6(2)x+(2)y+(2)z=12(1)x+(−1)y+(0)z=0\begin{cases}(1)x+(1)y+(1)z=6\\(2)x+(2)y+(2)z=12\\(1)x+(-1)y+(0)z=0\end{cases}

Three-variable systems: eliminate, solve and check — Teacher answers

Year 7, lower volume · text pp. 67–70 · textbook lessons 34–35

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: Use substitution/elimination with two unknowns and check a solution in every original equation.

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

Inquiry and discussion: 18 minutes on paper: start with x+y+z=9, x−y=−1, y−z=−1. Predict the order of the three values (3 min). Two groups eliminate a different chosen unknown consistently, each producing a two-variable system and then the complete triple (7 min). Compare results and check all three original equations (4 min). Change only the first total from 9 to 12; predict and verify how every value changes (3 min). State why obtaining only two values does not complete a three-variable solution (1 min). No three-variable interactive engine is claimed.

Worked example1

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

{(1)x+(1)y+(1)z=9(1)x+(−1)y+(0)z=−1(0)x+(1)y+(−1)z=−1\begin{cases}(1)x+(1)y+(1)z=9\\(1)x+(-1)y+(0)z=-1\\(0)x+(1)y+(-1)z=-1\end{cases}
Reference answer

x=2,y=3,z=4x=2,\quad y=3,\quad z=4.

x=2,y=3,z=4x=2,\quad y=3,\quad z=4
  1. Equation 2 minus 11 times equation 1 gives (0)x+(−2)y+(−1)z=−10(0)x+(-2)y+(-1)z=-10.
  2. Equation 3 minus −12\frac{-1}{2} times equation 2 gives (0)x+(0)y+(−32)z=−6(0)x+(0)y+(\frac{-3}{2})z=-6.
  3. Back-substitute to obtain z=4z=4.
  4. Back-substitute to obtain y=3y=3.
  5. Back-substitute to obtain x=2x=2.
  6. 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.
Guided attempt2

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

{(1)x+(1)y+(1)z=6(1)x+(−1)y+(0)z=0(0)x+(1)y+(−1)z=0\begin{cases}(1)x+(1)y+(1)z=6\\(1)x+(-1)y+(0)z=0\\(0)x+(1)y+(-1)z=0\end{cases}
  1. From x−y=0x-y=0, write y=□y=\square.
  2. From y−z=0y-z=0, write z=y=□z=y=\square.
  3. Substitute in the first equation: x+x+x=6x+x+x=6, so x=□x=\square; recover y,z and check all three.
Reference answer

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

x=2,y=2,z=2x=2,\quad y=2,\quad z=2
  1. Equation 2 minus 11 times equation 1 gives (0)x+(−2)y+(−1)z=−6(0)x+(-2)y+(-1)z=-6.
  2. Equation 3 minus −12\frac{-1}{2} times equation 2 gives (0)x+(0)y+(−32)z=−3(0)x+(0)y+(\frac{-3}{2})z=-3.
  3. Back-substitute to obtain z=2z=2.
  4. Back-substitute to obtain y=2y=2.
  5. Back-substitute to obtain x=2x=2.
  6. 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=1,y=2,z=3x=1, y=2, z=3 solve the displayed system? Check every equation.

{(1)x+(1)y+(1)z=6(1)x+(−1)y+(0)z=−1(0)x+(1)y+(−1)z=−1\begin{cases}(1)x+(1)y+(1)z=6\\(1)x+(-1)y+(0)z=-1\\(0)x+(1)y+(-1)z=-1\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 6 and right side 6.
  2. Equation 2 gives left side -1 and right side -1.
  3. Equation 3 gives left side -1 and right side -1.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Prerequisite4

To reduce three equations in x,y,z to a two-variable system, what should the two derived equations have in common?

  1. They both eliminate the same unknown and retain the same other two.
  2. One eliminates x and the other only y, leaving three unknowns together.
  3. Their right sides must both be zero.
Reference answer

They both eliminate the same unknown and retain the same other two.

  1. The reduced pair must involve just two common unknowns.
  2. Otherwise it is not yet a two-variable system.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Independent practice5

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

{(1)x+(1)y+(1)z=12(1)x+(−1)y+(0)z=2(0)x+(1)y+(−1)z=−1\begin{cases}(1)x+(1)y+(1)z=12\\(1)x+(-1)y+(0)z=2\\(0)x+(1)y+(-1)z=-1\end{cases}
Reference answer

x=5,y=3,z=4x=5,\quad y=3,\quad z=4.

x=5,y=3,z=4x=5,\quad y=3,\quad z=4
  1. Equation 2 minus 11 times equation 1 gives (0)x+(−2)y+(−1)z=−10(0)x+(-2)y+(-1)z=-10.
  2. Equation 3 minus −12\frac{-1}{2} times equation 2 gives (0)x+(0)y+(−32)z=−6(0)x+(0)y+(\frac{-3}{2})z=-6.
  3. Back-substitute to obtain z=4z=4.
  4. Back-substitute to obtain y=3y=3.
  5. Back-substitute to obtain x=5x=5.
  6. 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 practice6

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

{(1)x+(1)y+(1)z=10(2)x+(−1)y+(0)z=−1(0)x+(1)y+(−1)z=2\begin{cases}(1)x+(1)y+(1)z=10\\(2)x+(-1)y+(0)z=-1\\(0)x+(1)y+(-1)z=2\end{cases}
Reference answer

x=2,y=5,z=3x=2,\quad y=5,\quad z=3.

x=2,y=5,z=3x=2,\quad y=5,\quad z=3
  1. Equation 2 minus 22 times equation 1 gives (0)x+(−3)y+(−2)z=−21(0)x+(-3)y+(-2)z=-21.
  2. Equation 3 minus −13\frac{-1}{3} times equation 2 gives (0)x+(0)y+(−53)z=−5(0)x+(0)y+(\frac{-5}{3})z=-5.
  3. Back-substitute to obtain z=3z=3.
  4. Back-substitute to obtain y=5y=5.
  5. Back-substitute to obtain x=2x=2.
  6. 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.

{(1)x+(1)y+(0)z=5(0)x+(1)y+(1)z=7(1)x+(0)y+(1)z=6\begin{cases}(1)x+(1)y+(0)z=5\\(0)x+(1)y+(1)z=7\\(1)x+(0)y+(1)z=6\end{cases}
Reference answer

x=2,y=3,z=4x=2,\quad y=3,\quad z=4.

x=2,y=3,z=4x=2,\quad y=3,\quad z=4
  1. Equation 3 minus 11 times equation 1 gives (0)x+(−1)y+(1)z=1(0)x+(-1)y+(1)z=1.
  2. Equation 3 minus −1-1 times equation 2 gives (0)x+(0)y+(2)z=8(0)x+(0)y+(2)z=8.
  3. Back-substitute to obtain z=4z=4.
  4. Back-substitute to obtain y=3y=3.
  5. Back-substitute to obtain x=2x=2.
  6. 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+(1)z=3(1)x+(−1)y+(0)z=1(0)x+(1)y+(−1)z=1\begin{cases}(1)x+(1)y+(1)z=3\\(1)x+(-1)y+(0)z=1\\(0)x+(1)y+(-1)z=1\end{cases}
Reference answer

x=2,y=1,z=0x=2,\quad y=1,\quad z=0.

x=2,y=1,z=0x=2,\quad y=1,\quad z=0
  1. Equation 2 minus 11 times equation 1 gives (0)x+(−2)y+(−1)z=−2(0)x+(-2)y+(-1)z=-2.
  2. Equation 3 minus −12\frac{-1}{2} times equation 2 gives (0)x+(0)y+(−32)z=0(0)x+(0)y+(\frac{-3}{2})z=0.
  3. Back-substitute to obtain z=0z=0.
  4. Back-substitute to obtain y=1y=1.
  5. Back-substitute to obtain x=2x=2.
  6. 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=3,z=4x=2, y=3, z=4 solve the displayed system? Check every equation.

{(1)x+(1)y+(1)z=9(1)x+(−1)y+(0)z=−1(0)x+(1)y+(−1)z=−1\begin{cases}(1)x+(1)y+(1)z=9\\(1)x+(-1)y+(0)z=-1\\(0)x+(1)y+(-1)z=-1\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 9 and right side 9.
  2. Equation 2 gives left side -1 and right side -1.
  3. Equation 3 gives left side -1 and right side -1.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.
Exit question10

After elimination gives y=3 and z=4, the original first equation is x+y+z=9. What remains to finish the solution?

  1. Find x=2, then check all three original equations.
  2. Stop; two values are enough.
  3. Choose any x without checking.
Reference answer

Find x=2, then check all three original equations.

  1. Substitute 3 and 4 into x+y+z=9.
  2. x=9−3−4=2; verify the complete triple.
  • 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+(1)z=7(1)x+(−1)y+(0)z=2(0)x+(1)y+(−1)z=1\begin{cases}(1)x+(1)y+(1)z=7\\(1)x+(-1)y+(0)z=2\\(0)x+(1)y+(-1)z=1\end{cases}
Reference answer

x=4,y=2,z=1x=4,\quad y=2,\quad z=1.

x=4,y=2,z=1x=4,\quad y=2,\quad z=1
  1. Equation 2 minus 11 times equation 1 gives (0)x+(−2)y+(−1)z=−5(0)x+(-2)y+(-1)z=-5.
  2. Equation 3 minus −12\frac{-1}{2} times equation 2 gives (0)x+(0)y+(−32)z=−32(0)x+(0)y+(\frac{-3}{2})z=\frac{-3}{2}.
  3. Back-substitute to obtain z=1z=1.
  4. Back-substitute to obtain y=2y=2.
  5. Back-substitute to obtain x=4x=4.
  6. 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.

{(1)x+(1)y+(1)z=3(2)x+(−1)y+(0)z=0(0)x+(1)y+(−1)z=−12\begin{cases}(1)x+(1)y+(1)z=3\\(2)x+(-1)y+(0)z=0\\(0)x+(1)y+(-1)z=\frac{-1}{2}\end{cases}
Reference answer

x=12,y=1,z=32x=\frac{1}{2},\quad y=1,\quad z=\frac{3}{2}.

x=12,y=1,z=32x=\frac{1}{2},\quad y=1,\quad z=\frac{3}{2}
  1. Equation 2 minus 22 times equation 1 gives (0)x+(−3)y+(−2)z=−6(0)x+(-3)y+(-2)z=-6.
  2. Equation 3 minus −13\frac{-1}{3} times equation 2 gives (0)x+(0)y+(−53)z=−52(0)x+(0)y+(\frac{-5}{3})z=\frac{-5}{2}.
  3. Back-substitute to obtain z=32z=\frac{3}{2}.
  4. Back-substitute to obtain y=1y=1.
  5. Back-substitute to obtain x=12x=\frac{1}{2}.
  6. 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.

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

x=1,y=2,z=3x=1,\quad y=2,\quad z=3.

x=1,y=2,z=3x=1,\quad y=2,\quad z=3
  1. Equation 2 minus 12\frac{1}{2} times equation 1 gives (0)x+(−32)y+(32)z=32(0)x+(\frac{-3}{2})y+(\frac{3}{2})z=\frac{3}{2}.
  2. Equation 3 minus 12\frac{1}{2} times equation 1 gives (0)x+(12)y+(32)z=112(0)x+(\frac{1}{2})y+(\frac{3}{2})z=\frac{11}{2}.
  3. Equation 3 minus −13\frac{-1}{3} times equation 2 gives (0)x+(0)y+(2)z=6(0)x+(0)y+(2)z=6.
  4. Back-substitute to obtain z=3z=3.
  5. Back-substitute to obtain y=2y=2.
  6. Back-substitute to obtain x=1x=1.
  7. 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+(−2)y+(0)z=0(0)x+(1)y+(−2)z=0(1)x+(1)y+(1)z=21\begin{cases}(1)x+(-2)y+(0)z=0\\(0)x+(1)y+(-2)z=0\\(1)x+(1)y+(1)z=21\end{cases}
Reference answer

x=12,y=6,z=3x=12,\quad y=6,\quad z=3.

x=12,y=6,z=3x=12,\quad y=6,\quad z=3
  1. Equation 3 minus 11 times equation 1 gives (0)x+(3)y+(1)z=21(0)x+(3)y+(1)z=21.
  2. Equation 3 minus 33 times equation 2 gives (0)x+(0)y+(7)z=21(0)x+(0)y+(7)z=21.
  3. Back-substitute to obtain z=3z=3.
  4. Back-substitute to obtain y=6y=6.
  5. Back-substitute to obtain x=12x=12.
  6. 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

Three ticket types cost 2,3,5 dollars. Their counts are x,y,z. Nine tickets cost 33 dollars, and the first two types total 5 tickets. Form and solve a three-variable system; check integer counts.

Reference answer

x=2,y=3,z=4x=2,\quad y=3,\quad z=4.

x=2,y=3,z=4x=2,\quad y=3,\quad z=4
  1. Counts give x+y+z=9x+y+z=9, revenue gives 2x+3y+5z=332x+3y+5z=33, and x+y=5x+y=5.
  2. Equation 2 minus 22 times equation 1 gives (0)x+(1)y+(3)z=15(0)x+(1)y+(3)z=15.
  3. Equation 3 minus 11 times equation 1 gives (0)x+(0)y+(−1)z=−4(0)x+(0)y+(-1)z=-4.
  4. Back-substitute to obtain z=4z=4.
  5. Back-substitute to obtain y=3y=3.
  6. Back-substitute to obtain x=2x=2.
  7. 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.
Extension16

Do three written equations always determine three unique values? Describe the real solutions of this system.

{(1)x+(1)y+(1)z=6(2)x+(2)y+(2)z=12(1)x+(−1)y+(0)z=0\begin{cases}(1)x+(1)y+(1)z=6\\(2)x+(2)y+(2)z=12\\(1)x+(-1)y+(0)z=0\end{cases}
Reference answer

x=t, y=t, z=6−2t, t∈Rx=t,\ y=t,\ z=6-2t,\ t\in\mathbb{R}; infinitely many.

x=t, y=t, z=6−2t, t∈Rx=t,\ y=t,\ z=6-2t,\ t\in\mathbb{R}
  1. The second equation is twice the first. The third says x=y.
  2. Put x=y=t; then z=6−2t and every original equation holds.
  • Use the given conditions and show a justified intermediate step.
  • State and check the requested result accurately.