College Algebra - Week 6

Systems of Equations

A single equation in two variables has infinitely many solutions. A system of two equations in two variables usually has exactly one — the point where the lines cross. This week you master substitution, elimination, and graphical interpretation of systems, and apply them to real problems involving mixtures, investments, speed, and cost analysis.

Learning Objectives

- Solve a 2x2 linear system using the substitution method. - Solve a 2x2 linear system using the elimination (addition) method. - Identify systems with no solution (parallel lines) or infinitely many solutions (same line). - Solve applied problems using systems of equations. - Solve 3x3 linear systems using back-substitution or elimination. - Interpret the solution of a system graphically.

1. What is a System of Equations?

A system of equations is two or more equations sharing the same variables. The solution is a set of values that satisfies ALL equations simultaneously. For two linear equations in two variables: - One solution: the lines intersect at one point. (Consistent and independent) - No solution: the lines are parallel. (Inconsistent) - Infinitely many solutions: the equations describe the same line. (Consistent and dependent) Graphically, the solution is the intersection point.

2. Substitution Method

Step 1: Solve one equation for one variable. Step 2: Substitute the expression into the other equation. Step 3: Solve the resulting one-variable equation. Step 4: Back-substitute to find the other variable. Step 5: Check in BOTH original equations. Example: Solve { 2x + y = 7; x - y = 2 } From equation 2: x = y + 2. Substitute into equation 1: 2(y + 2) + y = 7 → 3y + 4 = 7 → y = 1. Back-substitute: x = 1 + 2 = 3. Solution: (3, 1). Check: 2(3)+1 = 7. 3-1 = 2. Both true.

3. Elimination Method

Step 1: Multiply equations (if needed) so that one variable has equal and opposite coefficients. Step 2: Add the equations to eliminate that variable. Step 3: Solve for the remaining variable. Step 4: Substitute back to find the other variable. Example: Solve { 3x + 2y = 16; 5x - 2y = 8 } The y-coefficients are already +2 and -2. Add the equations: 8x = 24 → x = 3. Substitute: 3(3) + 2y = 16 → 9 + 2y = 16 → y = 7/2. Solution: (3, 7/2).

4. Elimination with Multiplication

Example: Solve { 2x + 3y = 12; 5x - y = 7 } Multiply equation 2 by 3: 15x - 3y = 21. Now y-coefficients are 3 and -3. Add: 17x = 33 → x = 33/17. Substitute back: 2(33/17) + 3y = 12 → 66/17 + 3y = 12 → 3y = 138/17 → y = 46/17. Answer: (33/17, 46/17). Check by substituting both values into each equation.

5. No Solution and Infinitely Many Solutions

No solution (parallel lines): Solve { 2x - y = 4; 4x - 2y = 3 } Multiply equation 1 by 2: 4x - 2y = 8. Subtract equation 2: 0 = 5. FALSE. No solution — parallel lines. Infinitely many solutions (same line): Solve { x + y = 5; 2x + 2y = 10 } Multiply equation 1 by 2: 2x + 2y = 10. Same as equation 2. 0 = 0. TRUE always. Infinitely many solutions — same line. Solution: { (x, y) : x + y = 5 }

6. Mixture Problems

Mixture problems set up systems based on total amount and total value (or concentration). Example: How many liters of 20% acid and 50% acid must be mixed to make 30 L of 30% acid? Let x = liters of 20%, y = liters of 50%. Total amount: x + y = 30 Total acid: 0.20x + 0.50y = 0.30(30) = 9 From equation 1: x = 30 - y. Substitute: 0.20(30-y) + 0.50y = 9 6 - 0.20y + 0.50y = 9 0.30y = 3 → y = 10, x = 20.

7. Investment and Interest Problems

Example: $10,000 is split between two accounts. One earns 4% annually, the other 6%. Together they earn $520 in interest. How much is in each? Let x = amount at 4%, y = amount at 6%. x + y = 10000 0.04x + 0.06y = 520 From equation 1: x = 10000 - y. 0.04(10000-y) + 0.06y = 520 400 - 0.04y + 0.06y = 520 0.02y = 120 → y = 6000, x = 4000. $4,000 at 4%, $6,000 at 6%.

8. Speed and Distance Problems

Recall: distance = rate * time. Example: A boat travels 60 miles downstream in 2 hours and 60 miles upstream in 3 hours. Find the boat's speed in still water and the current's speed. Downstream: (b + c)(2) = 60 → b + c = 30 Upstream: (b - c)(3) = 60 → b - c = 20 Add: 2b = 50 → b = 25. Subtract: 2c = 10 → c = 5. Boat speed: 25 mph, current: 5 mph.

9. 3x3 Systems: Introduction

A 3x3 system has three equations and three unknowns. The solution is a point in 3D space. Strategy: eliminate one variable from two pairs of equations, creating a 2x2 system. Solve the 2x2 system. Back-substitute. Example: Solve { x + y + z = 6; 2x - y + z = 3; x + 2y - z = 4 } Add equations 1 and 2: 3x + 2z = 9 ... (A) Add equations 1 and 3: 2x + 3y = 10 ... (B) Subtract eq 2 from eq 3: -x + 3y - 2z = 1 ... (C) Add (A) and 2*(C): x + 8y = 11 ... (D) [Continue with equations B and D...]

10. Graphical Interpretation

Each linear equation in two variables graphs as a straight line. For a 2x2 system: - One solution → lines cross at one point. - No solution → lines are parallel (same slope, different y-intercept). - Infinitely many → lines are the same (same slope, same y-intercept). For a 3x3 system, each equation is a plane in 3D space. The solution is a single point where all three planes meet (if consistent and independent).

11. Choosing the Best Method

Substitution works best when one variable has coefficient 1 or -1 in at least one equation. Elimination works best when the coefficients are larger and harder to isolate by hand. Both methods produce the same answer. Choose whichever involves less arithmetic for a given problem.

12. Common Mistakes

- Back-substituting into the wrong equation (always use an original equation, not a derived step). - Stopping after finding x without finding y. - Forgetting to distribute negatives when subtracting equations. - Declaring 'no solution' when variables cancel and a FALSE statement results vs. 'infinitely many' when a TRUE statement results. - Not checking the answer in BOTH original equations.