2.3 Systems of Equations & Inequalities

What a System Means

A system of equations is two or more equations considered together. Its solution is the ordered pair (x, y) that makes every equation true at the same time. Graphically, that is the point of intersection of the lines.

The Regents expects fluency with three methods, summarized here:

All three give the same answer; pick the one that matches how the equations are written.

Solving by Substitution

Substitution works well when one equation is already solved for a variable.

Worked example: Solve the system y = 2x - 1 and 3x + y = 14. Substitute 2x - 1 for y in the second equation: 3x + (2x - 1) = 14. Combine: 5x - 1 = 14, so 5x = 15 and x = 3. Back-substitute into y = 2x - 1: y = 2(3) - 1 = 5. The solution is (3, 5).

Check in the second equation: 3(3) + 5 = 14. It works. Always substitute the found value back to get the partner coordinate and to confirm.

Substitution also handles equations not yet solved for a variable, as long as one is easy to isolate. From x + y = 7 you can write y = 7 - x and substitute that into the other equation. Choose the variable with a coefficient of 1 to avoid fractions, which keeps the algebra clean.

Solving by Elimination

Elimination adds or subtracts equations to cancel one variable. Scale an equation first if needed so a variable's coefficients are opposites.

Worked example: Solve 2x + 3y = 12 and 4x - 3y = 6. The y-coefficients are +3 and -3, so add the equations directly: (2x + 4x) + (3y - 3y) = 12 + 6, which gives 6x = 18, so x = 3. Substitute into 2x + 3y = 12: 6 + 3y = 12, so 3y = 6 and y = 2. The solution is (3, 2).

When coefficients do not match, multiply one or both equations first. For 3x + 2y = 16 and 2x + 5y = 18, multiply the first by 5 and the second by 2 to make the y-terms 10y and 10y, then subtract. The decision to add or subtract depends on the signs: add when the matching coefficients are opposites, subtract when they are identical, so the chosen variable cancels cleanly.

Number of Solutions

The relationship between the two lines tells you how many solutions exist:

• One solution: the lines have different slopes and cross at a single point.
• No solution: the lines are parallel (same slope, different y-intercept); solving produces a false statement like 0 = 5.
• Infinitely many solutions: the equations are the same line; solving produces a true statement like 0 = 0.

Worked example: In 2x + y = 4 and 4x + 2y = 9, the second is almost double the first but the constant 9 is not double 4. Doubling the first gives 4x + 2y = 8, which contradicts 4x + 2y = 9. The lines are parallel, so there is no solution.

You can also judge solution count by rewriting both equations in slope-intercept form y = mx + b. Different slopes give one solution; equal slopes with different b-values give none; equal slopes and equal b-values give infinitely many. On a graphing calculator, two lines that never cross on screen and stay the same distance apart confirm a parallel, no-solution system.

Systems of Inequalities and Modeling

A system of inequalities is graphed by shading each inequality's half-plane; the feasible region is where the shadings overlap. Any point in that overlap satisfies all conditions. Boundary lines are solid for >= or <= and dashed for > or <.

Modeling example: A student has at most 8 hours to spend, where math practice (x) takes 1 hour each and reading (y) takes 2 hours each, and they want at least 5 sessions total. The system is x + 2y <= 8 and x + y >= 5, with x >= 0 and y >= 0. A point like (4, 2) gives 4 + 4 = 8 hours (allowed) and 6 sessions (at least 5), so it lies in the feasible region.

The Regents reward interpreting the solution: it is the combination of real quantities where every constraint holds at once.

Word problems that pair two relationships almost always become a two-equation system. "Adult tickets cost 12 dollars, child tickets 8 dollars, 90 tickets sold for 920 dollars" gives a + c = 90 and 12a + 8c = 920. Solving (here a = 50 adults, c = 40 children) and then stating the answer in the problem's units is the full Regents expectation, not just finding x and y.