Tasks 06 - Systems of Linear Equations

Mathematics for Master Students

These problems accompany Session 06 of the tutorial, the last session set. On the website, every problem has a fully worked solution you can expand to check yourself. The PDF version contains only the problems, so you can print it and use it as a worksheet.

The problems are staged: the first two are warm-ups, the middle ones drill the core techniques (substitution, elimination, Gaussian elimination on the augmented matrix, and reading off how many solutions a system has), and the last two practice the mock-exam skill of translating a business story into a system first. See the Mock Exam. Since this rounds off the tutorial, treat the closing problems as a mock-exam warm-up, and keep the Cheatsheet beside you.

Problem 1: Linear or Not, and Is It a Solution?

  1. Decide for each equation whether it is linear in its unknowns:
    1. \(3x - 2y = 7\)
    2. \(x^2 + y = 5\)
    3. \(4a + 7b - c = 0\)
    4. \(xy = 12\)
    5. \(\sqrt{x} + y = 1\)
  2. Consider the system \(\;2x + y = 8\;\) and \(\;x - y = 1\). Check whether each pair is a solution:
    1. \((x, y) = (3, 2)\)
    2. \((x, y) = (4, 0)\)

An equation is linear when every unknown appears on its own, to the first power: no squares, no products of two variables, no variables under a root or in an exponent.

    1. Linear: a weighted sum of \(x\) and \(y\).
    2. Not linear: \(x^2\) is a square.
    3. Linear: three variables, each to the first power.
    4. Not linear: \(xy\) is a product of two variables.
    5. Not linear: \(\sqrt{x}\) puts a variable under a root.
  2. A pair is a solution only if it satisfies every equation at once.
    1. \(2(3) + 2 = 8\) ✓ and \(3 - 2 = 1\) ✓: both hold, so \((3, 2)\) is a solution.
    2. \(2(4) + 0 = 8\) ✓, but \(4 - 0 = 4 \neq 1\) ✗: the second equation fails, so \((4, 0)\) is not a solution.

Problem 2: A Quick Substitution

A depot ships goods in small crates (\(x\)) and large crates (\(y\)). Two conditions fix the plan: \[y = 2x - 1, \qquad 3x + y = 14.\] One equation is already solved for \(y\), so use substitution.

Substitution: the first equation already gives \(y\) in terms of \(x\), so drop it into the second.

Step 1 - Substitute \(y = 2x - 1\) into \(3x + y = 14\): \[3x + (2x - 1) = 14 \;\Rightarrow\; 5x - 1 = 14 \;\Rightarrow\; 5x = 15 \;\Rightarrow\; x = 3.\]

Step 2 - Back-substitute into \(y = 2x - 1\): \[y = 2(3) - 1 = 5.\]

Solution: \((x, y) = (3, 5)\): 3 small crates and 5 large ones. Check: \(y = 2(3) - 1 = 5\) ✓ and \(3(3) + 5 = 14\) ✓.

Problem 3: One System, Two Methods

Solve the system \[3x + 2y = 14, \qquad x - y = 3\] twice (once by substitution and once by elimination) and confirm the two roads reach the same point.

Method 1 - Substitution. Solve the simpler equation for a variable, then substitute.

Step 1 From \(x - y = 3\) we get \(x = y + 3\). Step 2 Substitute into \(3x + 2y = 14\): \[3(y + 3) + 2y = 14 \;\Rightarrow\; 3y + 9 + 2y = 14 \;\Rightarrow\; 5y = 5 \;\Rightarrow\; y = 1.\] Step 3 Then \(x = y + 3 = 4\).

Method 2 - Elimination. Scale a row so one variable cancels on adding.

Step 1 Multiply \(x - y = 3\) by \(2\): \(\;2x - 2y = 6\). Step 2 Add it to \(3x + 2y = 14\), so the \(y\)-terms cancel: \[(3x + 2y) + (2x - 2y) = 14 + 6 \;\Rightarrow\; 5x = 20 \;\Rightarrow\; x = 4.\] Step 3 Back-substitute: \(\;4 - y = 3 \;\Rightarrow\; y = 1\).

Both methods give \((x, y) = (4, 1)\). Check: \(3(4) + 2(1) = 14\) ✓ and \(4 - 1 = 3\) ✓.

Problem 4: Gaussian Elimination in Three Unknowns

A workshop makes three product types in quantities \(x\), \(y\), \(z\) and wants to use every resource fully:

  • Packing: one box per unit, \(6\) boxes available \(\;\Rightarrow\; x + y + z = 6\)
  • Labour: \(1, 2, 3\) hours per product, \(11\) hours available \(\;\Rightarrow\; x + 2y + 3z = 11\)
  • Machine: \(2, 1, 1\) hours per product, \(9\) hours available \(\;\Rightarrow\; 2x + y + z = 9\)

Write the augmented matrix, reduce it to triangular form by forward elimination, solve by back-substitution, and verify.

Gaussian elimination: forward-eliminate to a triangular staircase, then read it from the bottom up.

Step 1 - Augmented matrix. Coefficients on the left, right-hand sides after the bar: \[\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 1 & 2 & 3 & 11 \\ 2 & 1 & 1 & 9 \end{array}\right]\]

Step 2 - Clear the first column using the top row as pivot: \[\xrightarrow[\;R_3 \,\to\, R_3 - 2R_1\;]{\;R_2 \,\to\, R_2 - R_1\;}\; \left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 5 \\ 0 & -1 & -1 & -3 \end{array}\right]\]

  • \(R_2 - R_1\): \(\;(1, 2, 3 \,|\, 11) - (1, 1, 1 \,|\, 6) = (0, 1, 2 \,|\, 5)\)
  • \(R_3 - 2R_1\): \(\;(2, 1, 1 \,|\, 9) - 2(1, 1, 1 \,|\, 6) = (0, -1, -1 \,|\, -3)\)

Step 3 - Clear the second column using the new second row: \[\xrightarrow[\;\;]{\;R_3 \,\to\, R_3 + R_2\;}\; \left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 1 & 2 \end{array}\right]\]

  • \(R_3 + R_2\): \(\;(0, -1, -1 \,|\, -3) + (0, 1, 2 \,|\, 5) = (0, 0, 1 \,|\, 2)\)

Step 4 - Back-substitute. The matrix is triangular, so read the equations bottom-up: \[x + y + z = 6, \qquad y + 2z = 5, \qquad z = 2.\] - Bottom row: \(\;z = 2\). - Middle row: \(\;y + 2(2) = 5 \;\Rightarrow\; y = 1\). - Top row: \(\;x + 1 + 2 = 6 \;\Rightarrow\; x = 3\).

Solution: \((x, y, z) = (3, 1, 2)\): make 3 of the first product, 1 of the second, 2 of the third.

Step 5 - Verify in the original equations: \[x + y + z = 3 + 1 + 2 = 6 \ \checkmark, \quad x + 2y + 3z = 3 + 2 + 6 = 11 \ \checkmark, \quad 2x + y + z = 6 + 1 + 2 = 9 \ \checkmark.\]

WarningCommon Mistake

Every row operation acts on the whole row, the right-hand side included. When you form \(R_3 - 2R_1\), you must also compute \(9 - 2 \cdot 6 = -3\) for the last entry. Updating the coefficients but forgetting the right-hand side silently corrupts the entire solve.

Problem 5: How Many Solutions?

For each \(2 \times 2\) system, decide whether it has a unique solution, no solution, or infinitely many. Where there are infinitely many, describe all solutions with a parameter \(t\).

  1. \(\;x + y = 5, \qquad x - y = 1\)
  2. \(\;x + 2y = 4, \qquad 2x + 4y = 10\)
  3. \(\;x - 3y = 2, \qquad 2x - 6y = 4\)

Eliminate one variable and read the leftover row: a clean value means unique; \(0 = (\text{non-zero})\) means no solution; \(0 = 0\) means infinitely many.

  1. Add the two equations: \(\;2x = 6 \;\Rightarrow\; x = 3\), then \(y = 3 - 1 = 2\). Unique solution \((x, y) = (3, 2)\).

  2. The second equation is nearly twice the first. Compute \(R_2 - 2R_1\): \[(2x + 4y) - 2(x + 2y) = 10 - 2(4) \;\Rightarrow\; 0 = 2.\] A false statement: the equations contradict each other. No solution (the lines are parallel).

  3. Here \(R_2 = 2R_1\) exactly. Compute \(R_2 - 2R_1\): \[(2x - 6y) - 2(x - 3y) = 4 - 2(2) \;\Rightarrow\; 0 = 0.\] The second equation was redundant, so infinitely many solutions. Parameterize: let \(y = t\) (any real number); then from \(x - 3y = 2\), \[x = 2 + 3t.\] Solution set: \((x, y) = (2 + 3t,\; t)\). For example \(t = 0 \to (2, 0)\) and \(t = 1 \to (5, 1)\), both check out.

WarningCommon Mistake

A vanished row \(\;0 = 0\;\) means one equation was redundant: the system has infinitely many solutions, not “no solution”. “No solution” appears only as \(\;0 = (\text{non-zero})\), a genuine contradiction. Always read the right-hand side before you decide.

Problem 6: Market Equilibrium

For a product at price \(p\) (in €), monthly demand and supply (in units) are \[\text{demand:}\ q = 90 - 3p, \qquad \text{supply:}\ q = 2p - 10.\] Find the equilibrium price and quantity (the point where the market clears) by solving the system.

The market clears where supply equals demand: one price and quantity satisfying both curves at once.

Step 1 - Write it as a system. Both equations give \(q\), so bring \(p\) across: \[q + 3p = 90, \qquad q - 2p = -10.\]

Step 2 - Eliminate \(q\) by subtracting the second from the first: \[(q + 3p) - (q - 2p) = 90 - (-10) \;\Rightarrow\; 5p = 100 \;\Rightarrow\; p = 20.\]

Step 3 - Back-substitute into the demand equation: \[q = 90 - 3(20) = 90 - 60 = 30.\]

Equilibrium: price €20, quantity \(30\) units.

Check: the supply side gives \(\;2(20) - 10 = 30\) ✓: supply and demand agree.

Problem 7: A Blending Plan

A factory produces two blends, \(x\) and \(y\) (in batches), from two raw materials. Each batch of blend \(x\) uses \(3\) units of material A and \(5\) units of material B; each batch of blend \(y\) uses \(4\) units of A and \(2\) units of B. This week \(44\) units of A and \(50\) units of B are available and must be used in full: \[3x + 4y = 44, \qquad 5x + 2y = 50.\] Find the number of batches of each blend.

Elimination: scale one equation so a variable cancels.

Step 1 - Line up the \(y\)-terms. Multiply the second equation by \(2\) so both carry \(4y\): \[5x + 2y = 50 \;\xrightarrow{\times 2}\; 10x + 4y = 100.\]

Step 2 - Subtract the first equation to cancel \(y\): \[(10x + 4y) - (3x + 4y) = 100 - 44 \;\Rightarrow\; 7x = 56 \;\Rightarrow\; x = 8.\]

Step 3 - Back-substitute into \(3x + 4y = 44\): \[3(8) + 4y = 44 \;\Rightarrow\; 24 + 4y = 44 \;\Rightarrow\; 4y = 20 \;\Rightarrow\; y = 5.\]

Solution: \(8\) batches of blend \(x\) and \(5\) batches of blend \(y\).

Check: material A: \(3(8) + 4(5) = 24 + 20 = 44\) ✓; material B: \(5(8) + 2(5) = 40 + 10 = 50\) ✓: both materials used to the last unit.