Pre-Calculus - Week 6

Systems of Equations and Matrices

Systems of equations appear whenever two or more constraints must be satisfied simultaneously — pricing models, mixture problems, network flows, and equilibrium in physics all reduce to systems. This week you solve 2x2 and 3x3 systems by substitution, elimination, and matrix row reduction, building the algebraic fluency that calculus and linear algebra will extend.

Learning Objectives

- Solve 2x2 linear systems by substitution. - Solve 2x2 linear systems by elimination (addition method). - Identify inconsistent systems (no solution) and dependent systems (infinite solutions). - Set up and solve applied problems using linear systems. - Represent a linear system as an augmented matrix. - Use elementary row operations to put a matrix in row echelon form. - Solve 3x3 linear systems using Gaussian elimination. - Solve 2x2 nonlinear systems algebraically.

1. What Is a System of Equations?

A system of equations is a set of two or more equations that share the same variables. A solution is a set of values that satisfies ALL equations simultaneously. For two linear equations in x and y, the solution is the intersection point of their lines: - One intersection: one solution (consistent independent). - Parallel lines: no solution (inconsistent). - Same line: infinite solutions (consistent dependent).

2. Solving by Substitution

Solve one equation for one variable, then substitute into the other. Example: Solve the system: 2x + y = 7 and x - y = 2. From equation 2: x = y + 2. Substitute into equation 1: 2(y + 2) + y = 7. 2y + 4 + y = 7. 3y = 3. y = 1. Back-substitute: x = 1 + 2 = 3. Solution: (3, 1). Check: 2(3)+1=7 ✓ and 3-1=2 ✓.

3. Solving by Elimination

Add (or subtract) equations to eliminate one variable. Example 1: Solve 3x + 2y = 8 and x - 2y = 0. Add the equations: 4x = 8. x = 2. Substitute: 2 - 2y = 0. y = 1. Solution: (2, 1). Example 2: Solve 2x + 3y = 5 and 4x - y = 3. Multiply first eq by 2: 4x + 6y = 10. Subtract second: (4x+6y) - (4x-y) = 10-3. 7y = 7. y = 1. Then: 2x + 3 = 5. x = 1. Solution: (1, 1).

4. Inconsistent and Dependent Systems

Inconsistent (no solution): you arrive at a FALSE statement like 0 = 5. Example: Solve x + y = 3 and x + y = 7. Subtract: 0 = -4. No solution. The lines are parallel. Dependent (infinite solutions): you arrive at a TRUE statement like 0 = 0. Example: Solve 2x - y = 4 and 6x - 3y = 12. Divide second by 3: 2x - y = 4 (identical). 0 = 0. Infinite solutions. General solution: y = 2x - 4, for any x.

5. Applied Systems: Mixture and Rate Problems

Example 1 (Mixture): A chemist mixes a 20% acid solution with a 50% acid solution to get 90 mL of 30% acid. How many mL of each? Let x = mL of 20%, y = mL of 50%. x + y = 90 and 0.20x + 0.50y = 0.30(90) = 27. From eq 1: y = 90 - x. Substitute: 0.20x + 0.50(90-x) = 27. 0.20x + 45 - 0.50x = 27. -0.30x = -18. x = 60. y = 30. 60 mL of 20% and 30 mL of 50%. Example 2 (Rate): A boat travels 24 miles upstream in 4 hours and 24 miles downstream in 2 hours. Find the boat speed and current speed. Upstream: b - c = 6. Downstream: b + c = 12. Add: 2b = 18. b = 9. c = 3.

6. Augmented Matrices

A system of equations can be written as an augmented matrix [A|b]. For the system: 2x + y = 5 and x - 3y = 7: [ 2 1 | 5 ] [ 1 -3 | 7 ] Each row represents one equation. Each column (before the bar) represents one variable's coefficients. The column after the bar is the right-hand side. Row operations that preserve solutions: 1. Swap two rows. 2. Multiply a row by a nonzero constant. 3. Add a multiple of one row to another.

7. Gaussian Elimination

Use row operations to get the matrix to row echelon form (upper triangular with leading 1s), then back-substitute. Example: Solve x + y + z = 6, 2x - y + z = 3, x + 2y - z = 2. Augmented matrix: [ 1 1 1 | 6 ] [ 2 -1 1 | 3 ] [ 1 2 -1 | 2 ] R2 = R2 - 2*R1: [0, -3, -1 | -9] R3 = R3 - R1: [0, 1, -2 | -4] Swap R2 and R3: [0, 1, -2 | -4] [0, -3, -1 | -9] R3 = R3 + 3*R2: [0, 0, -7 | -21]. So z = 3. Back-sub: y - 6 = -4, y = 2. x + 2 + 3 = 6, x = 1. Solution: (1, 2, 3).

8. Nonlinear Systems

A system where at least one equation is nonlinear. Solve by substitution. Example 1: Solve x^2 + y^2 = 25 and y = x + 1. Substitute: x^2 + (x+1)^2 = 25. x^2 + x^2 + 2x + 1 = 25. 2x^2 + 2x - 24 = 0. x^2 + x - 12 = 0. (x+4)(x-3) = 0. x = -4, y = -3 and x = 3, y = 4. Example 2: Solve y = x^2 and y = x + 2. x^2 = x + 2. x^2 - x - 2 = 0. (x-2)(x+1) = 0. x = 2, y = 4 and x = -1, y = 1.

9. Interpreting Solutions Graphically

Each solution to a nonlinear system corresponds to an intersection point of the graphs. For y = x^2 and y = x + 2: - One solution at (2, 4): the parabola and line intersect at (2, 4). - Another at (-1, 1). For a circle and a line, you can have 0, 1, or 2 solutions (no intersection, tangent, or two crossing points). This geometric intuition matters in calculus when finding where two curves intersect (computing areas between curves).

10. Common Mistakes

- Substituting incorrectly: if you solved for x, substitute into the other equation (not the same one). - Losing negative signs during elimination when subtracting rows. - Declaring 'no solution' when you get 0 = 0 (that's infinite solutions, not no solution). - In nonlinear systems, discarding a valid negative solution because it 'seems wrong.' - Not checking solutions in all original equations.