Module 8 of 47Lesson 1 of 7 · Equations 1 · Two equations, two unknowns
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Solving equations with technology · Essential understanding 1

Two equations, two unknowns

IB AI SL 1.8with Prof. Kitso
💻 Using the TI-Nspire CX Student Software on a laptop
  • The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
  • menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
  • enter = Enter  ·  tab = Tab (next box)  ·  esc = Esc  ·  del = Backspace.
  • Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
  • Fraction: type /  ·  √: type sqrt(  ·  π: type pi  ·  eˣ: type exp(  ·  store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
  • New page: Ctrl+I (or the Insert menu)  ·  change page: Ctrl+← / Ctrl+→  ·  graph entry line: Ctrl+G  ·  var: Ctrl+L.
  • Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
  • IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
  • If you type nDeriv(, the software shows it as centralDiff( — it is the same command.
The big idea: A system of equations is a set of equations that must all be true at the same time. To find two unknowns you need two equations. The solution is the pair of numbers that makes both true, and on a graph it is the point where the two lines cross.
By the end of this lesson you can:
  • check whether a pair of numbers is a solution of a system
  • write equations in standard form ax + by = c
  • solve two equations in two unknowns with the GDC (and by elimination)
  • write a system from a story, solve it and answer in context
  • link the solution to the point where two straight lines meet
🧑🏾 Imagine this…
Neo says: “I bought a loaf and a bottle of milk and paid P20.” That is one equation with two unknowns, so many answers fit (P5 and P15, P8 and P12, …). Now Neo adds: “The milk cost P4 more than the loaf.” That is a second equation, and now only one answer fits: loaf P8, milk P12. Two unknowns need two facts.
Unknown
a letter (like x or y) standing for a number we do not know yet.
Equation
a statement that two things are equal, for example 2x + y = 10.
System of equations
two or more equations that must all be true together. Also called simultaneous equations.
Solution
the values of the unknowns that make every equation in the system true.
Standard form
all the unknowns on the left, only a number on the right: ax + by = c.
Coefficient
the number in front of an unknown. In 5x − 2y = 7 the coefficients are 5 and −2.
★ The method in four steps
  • 1. Define the unknowns (what does x mean? what does y mean?).
  • 2. Write one equation for each fact. Two unknowns need two equations.
  • 3. Tidy each equation into standard form ax + by = c.
  • 4. Solve with the GDC, then check in both equations and answer the question that was asked.
★ What the solution means on a graph
Each equation is a straight line. The solution is the point where the lines cross. Two lines with different gradients cross exactly once, so there is one solution.
Prof. Kitso says
In the exam you are allowed to go straight to the GDC. But writing the equations in standard form first is where the marks for method are, and it stops you typing a number in the wrong place.
⚠️ Watch out
  • Move terms carefully: 3x = 4y − 7 becomes 3x − 4y = −7. The sign of 4y flips when it crosses the equals sign.
  • Keep the same order of unknowns in every equation (always x first, then y). Mixed order is the most common typing mistake.
  • Do not stop at x and y. Read the question: it may ask for a total, a cost or a time.
🧪 Try it yourselfWhere do two lines meet?
Two lines from Example 1: 2x + y = 10 (the line y = 10 − 2x) and x + 3y = c. With c = 20 they cross at (2, 6). Drag the slider for c: the second line moves, and so does the solution. Where is the crossing for c = 20? Does it stay on the first line?

Worked examples

Example 1.1Is it a solution?
Two equations are 2x + y = 10 and x + 3y = 20. Which of these pairs is a solution of the system? (i) (x, y) = (3, 4) (ii) (x, y) = (2, 6)
🧠 Think it through
  1. A solution must make both equations true. Substitute the pair into each one.
  2. If even one equation fails, the pair is not a solution of the system.
✍️ By hand
(i) ✓ but ✗   → not a solution.
(ii) ✓ and ✓   → a solution.
🔢 On your TI-Nspire CX II
  1. You can check on a Calculator page: type 2*2+6 enter ↵💻 Enter and 2+3*6 enter ↵💻 Enter.
1.1*DocRAD2·2+6102+3·6203+3·415
Both equations hold for (2, 6)
✅ Answer (i) No: it satisfies only the first equation. (ii) Yes: (2, 6) satisfies both, so x = 2, y = 6.
(3, 4) is on the first line but not on the second, so it is not where the two lines cross. Only (2, 6) is on both lines.
Example 1.2Solving a system with the GDC
Solve the system   5x + 2y = 19   and   3x − 4y = 1.
🧠 Think it through
  1. Both equations are already in standard form, so read off the coefficients: (5, 2 | 19) and (3, −4 | 1).
  2. Use the GDC system solver with 2 equations and the variables x, y.
  3. Check your answer in both equations.
✍️ By hand
Elimination: multiply the first equation by 2: .
Add to the second equation: .
Substitute into the first: .
🔢 On your TI-Nspire CX II
  1. On a Calculator page press menu💻 click menu on keypad and choose Algebra ▸ Solve System of Linear Equations (menu 3, then 2).
    1.1*DocRAD⚡1Actions½2Numberx=3Algebra∫4Calculus⚄5Probabilityx̄6Statistics▦7Matrix & Vector$€8Finance019Functions & Programs1Numerical Solve2Solve System of Linear Equations...3Polynomial Tools
  2. Set Number of equations = 2 and Variables = x, y. Press enter ↵💻 Enter.
    1.1*DocRADSolve a System of Linear EquationsNumber of equations:2Variables:x, yOKCancel
  3. A template appears. Type the left side and right side of each equation in its boxes, with the coefficients in the same order each time: 5 2 19 and 3 -4 1. Press enter ↵💻 Enter.
  4. The GDC shows the values in the order of your variable list {x, y}.
    1.1*DocRADlinSolve5·x+2·y=193·x−4·y=1,{x,y}{3,2}
    {3, 2} means x = 3 and y = 2
✅ Answer x = 3, y = 2
The GDC command behind the menu is called linSolve. You can type it directly if you like: linSolve({equation 1, equation 2},{x,y}). Another route is simult( ), which takes the coefficient matrix and the constants.
Example 1.3Equations not in standard form
Solve   y = 2x + 1   and   3x = 4y − 7.
🧠 Think it through
  1. Write each equation as ax + by = c first.
  2. First: y = 2x + 1 → subtract 2x from both sides → −2x + y = 1 (or 2x − y = −1).
  3. Second: 3x = 4y − 7 → subtract 4y from both sides → 3x − 4y = −7.
  4. The answers will not be whole numbers, which is fine.
✍️ By hand
Substitute into the second: .
So , , and .
🔢 On your TI-Nspire CX II
  1. Menu: Algebra ▸ Solve System of Linear Equations, 2 equations, variables x, y.
  2. Enter the standard form: 2 -1 -1 for the first and 3 -4 -7 for the second.
  3. Read the answer.
1.1*DocRADlinSolve2·x−y=-13·x−4·y=-7,{x,y}{0.6,2.2}
✅ Answer x = 0.6, y = 2.2
Check: 3(0.6) = 1.8 and 4(2.2) − 7 = 1.8 ✓.
Example 1.4A shopping problem
Thato buys 3 loaves of bread and 2 bottles of milk for P53. Neo buys 2 loaves and 5 bottles of milk for P80.25.
  1. (a) Write two equations for the cost of a loaf (a pula) and a bottle of milk (m pula).
  2. (b) Find the cost of one loaf and of one bottle of milk.
  3. (c) Find the cost of 4 loaves and 3 bottles of milk.
🧠 Think it through
  1. Define: a = cost of one loaf (P), m = cost of one bottle of milk (P).
  2. Each shopper gives one equation: (number of loaves) × a + (number of bottles) × m = total paid.
  3. Solve, then use the values in the last part. Keep the full values in the GDC rather than rounding early.
✍️ By hand
(a)   and   .
(c) .
🔢 On your TI-Nspire CX II
  1. Menu: Algebra ▸ Solve System of Linear Equations, 2 equations, variables a, m.
  2. Enter 3 2 53 and 2 5 80.25.
  3. Then type 4*9.5+3*12.25 for the last part.
1.1*DocRADlinSolve3·a+2·m=532·a+5·m=80.25,{a,m}{9.5,12.25}4·9.5+3·12.2574.75
✅ Answer (a) 3a + 2m = 53 and 2a + 5m = 80.25. (b) A loaf costs P9.50 and a bottle of milk P12.25. (c) P74.75.
Always write what your letters stand for. Without “a = cost of one loaf” the examiner cannot give you the method marks.
Example 1.5Break-even: two phone plans
Plan A costs P40 per month plus P0.80 per minute of calls. Plan B costs P15 per month plus P1.30 per minute. Let t be the number of minutes and C the monthly cost in pula.
  1. (a) Write an equation for C for each plan.
  2. (b) Find the number of minutes for which the two plans cost the same, and that cost.
  3. (c) Kagiso makes about 80 minutes of calls a month. Which plan should she choose?
🧠 Think it through
  1. Fixed charge + rate × minutes: Plan A: C = 40 + 0.8t. Plan B: C = 15 + 1.3t.
  2. “Cost the same” means both equations are true together, so this is a system. Put it in standard form: −0.8t + C = 40 and −1.3t + C = 15.
  3. The unknowns here are t and C (in that order).
  4. For (c) compare the costs at 80 minutes, or use the picture: after the crossing point the plan with the smaller rate is cheaper.
✍️ By hand
At the crossing: , and .
(c) Plan A at 80 min: . Plan B: .
🔢 On your TI-Nspire CX II
  1. Menu: Algebra ▸ Solve System of Linear Equations, 2 equations, variables t, c.
  2. Enter -0.8 1 40 and -1.3 1 15.
  3. Or graph both plans (f1(x) = 40 + 0.8x and f2(x) = 15 + 1.3x), then use menu💻 click menu on keypad ▸ 6: Analyze Graph ▸ 4: Intersection.
1.1*DocRADlinSolve-0.8·t+c=40-1.3·t+c=15,{t,c}{50,80}
1.11.2*DocRAD01001600xy2020f1(x)=40+0.8·xf2(x)=15+1.3·x(50, 80)
The plans meet at 50 min
✅ Answer (a) Plan A: C = 40 + 0.8t; Plan B: C = 15 + 1.3t. (b) 50 minutes, costing P80. (c) Plan A: it costs P104 against P119 for Plan B.
Interpret the crossing point: below 50 minutes Plan B is cheaper (low fixed charge), above 50 minutes Plan A is cheaper (low rate). A common exam follow-up.

Challenge questions

⭐ Challenge 1.6Taxi faresPaper 1 style
A taxi in Gaborone charges a fixed starting fee of Pf plus Pk for each kilometre. A 6 km journey costs P38 and a 15 km journey costs P83.
  1. (a) Write down two equations in f and k.
  2. (b) Find f and k.
  3. (c) Find the cost of a 20 km journey.
✏️ Try it on paper and on your calculator first. Then check.
Show the full solution
🧠 Think it through
  1. Fare = f + k × distance, so each journey gives one equation.
  2. Standard form: f + 6k = 38 and f + 15k = 83.
  3. Solve with the GDC, then substitute into the fare rule.
🔢 On your TI-Nspire CX II
  1. Menu: Algebra ▸ Solve System of Linear Equations, variables f, k. Enter 1 6 38 and 1 15 83.
  2. Then type 8+5*20.
1.1*DocRADlinSolvef+6·k=38f+15·k=83,{f,k}{8,5}8+5·20108
✅ Answer (a) f + 6k = 38 and f + 15k = 83. (b) f = 8, k = 5. (c) P108.
⭐ Challenge 1.7River cruise ticketsPaper 2 style
A river cruise at Kasane sells adult tickets at P180 and child tickets at P95. On one trip, 38 tickets were sold.
  1. (a) Let a be the number of adult tickets and c the number of child tickets. Write down an equation for the total number of tickets.
  2. (b) The money collected was P5480. Write down a second equation.
  3. (c) Find a and c.
  4. (d) On another trip 38 tickets brought in P5000. Show that this is impossible.
  5. (e) Explain how your GDC answer in part (d) tells you this.
✏️ Try it on paper and on your calculator first. Then check.
Show the full solution
🧠 Think it through
  1. Number of tickets: a + c = 38.
  2. Money: 180a + 95c = 5480.
  3. Solve. For (d) solve the new system and look at the answer: tickets must be whole numbers.
🔢 On your TI-Nspire CX II
  1. Menu: Solve System of Linear Equations, variables a, c. Enter 1 1 38 and 180 95 5480.
  2. Repeat with the last constant 5000.
1.1*DocRADlinSolvea+c=38180·a+95·c=5480,{a,c}{22,16}linSolvea+c=38180·a+95·c=5000,{a,c}{16.3529,21.6471}
✅ Answer (a) a + c = 38. (b) 180a + 95c = 5480. (c) 22 adult tickets and 16 child tickets. (d)–(e) The solution is a ≈ 16.35 and c ≈ 21.65. You cannot sell part of a ticket, so P5000 is impossible.
A correct solution of the equations is not always a sensible answer in context. Always ask: can this quantity be fractional or negative?
Next: Lesson 2 · Three equations, three unknowns
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