Solving systems, graphical and algebraic methods
High School Essentials • 9-12
What happens when two conditions must be satisfied at the same time? Systems of linear equations model exactly this: two or more equations that share the same variables, and we need to find values that make all equations true simultaneously. Applications range from comparing phone plans to balancing chemical reactions.
When you graph two linear equations on the same coordinate plane, the solution to the system is the point where the lines intersect. This method gives a visual understanding but can be imprecise when the intersection does not fall on grid points.
| Graph Result | Number of Solutions | System Type |
|---|---|---|
| Lines intersect at one point | Exactly one solution | Independent (consistent) |
| Lines are parallel (never meet) | No solution | Inconsistent |
| Lines are identical (overlap completely) | Infinitely many solutions | Dependent (consistent) |
Substitution works best when one equation is already solved for a variable (or can be easily solved for one).
Solve the system: y = 2x + 1 and 3x + y = 11.
The solution is (2, 5). Check in both equations: 5 = 2(2) + 1 = 5, and 3(2) + 5 = 11. Both true.
Elimination (also called the addition method) works by adding or subtracting equations to eliminate one variable. It is especially efficient when both equations are in standard form.
Solve: 2x + 3y = 12 and 4x - 3y = 6.
The solution is (3, 2).
Solve: x + 2y = 4 and 2x + 4y = 5.
Graphically, these lines are parallel (same slope, different intercepts).
When multiplying an equation by a constant to set up elimination, you must multiply every term -- including the constant on the right side. For example, multiplying x + 2y = 4 by -2 gives -2x - 4y = -8, not -2x - 4y = 4.
Use graphing for a visual estimate or to understand the geometry. Use substitution when one variable is already isolated. Use elimination when both equations are in standard form and coefficients align conveniently. All three methods yield the same answer for a given system.
1. Solve by substitution: y = x - 3 and 2x + y = 9.
Substitute y = x - 3 into the second equation: 2x + (x - 3) = 9, so 3x - 3 = 9, giving 3x = 12, x = 4. Then y = 4 - 3 = 1. Solution: (4, 1).
2. Solve by elimination: 3x + 2y = 16 and 3x - 2y = 8.
Add the equations: 6x = 24, so x = 4. Substitute into the first: 3(4) + 2y = 16, so 12 + 2y = 16, 2y = 4, y = 2. Solution: (4, 2).
3. Solve: 2x + y = 7 and 4x + 2y = 14.
Notice the second equation is exactly 2 times the first. These are the same line. The system has infinitely many solutions. Any point on y = -2x + 7 is a solution.
4. Solve by substitution: x = 3y + 2 and 5x - 15y = 4.
Substitute x = 3y + 2 into the second: 5(3y + 2) - 15y = 4, so 15y + 10 - 15y = 4, giving 10 = 4. This is a contradiction -- no solution.
5. Two pizzas and three salads cost $27. One pizza and two salads cost $16. Find the cost of each item.
Let p = pizza cost, s = salad cost. System: 2p + 3s = 27 and p + 2s = 16.
Multiply the second equation by -2: -2p - 4s = -32. Add to the first: -s = -5, so s = 5. Then p + 2(5) = 16, giving p = 6.
A pizza costs $6 and a salad costs $5.
A system of linear equations can have one solution (intersecting lines), no solution (parallel lines), or infinitely many solutions (identical lines). The three solving methods -- graphing, substitution, and elimination -- each have strategic advantages depending on the form of the equations. Always verify your solution by substituting back into both original equations.