7.4 Simultaneous Equations & Inequalities

7.4 Simultaneous Equations & Inequalities

Quick Answer: Solve 2x2 linear systems by substitution (solve one equation for one variable, plug into the other) or elimination (align coefficients, then add). Parallel lines (same slope, different intercept) give no solution; coincident lines (one equation is a nonzero multiple of the other) give infinitely many solutions. Systems of inequalities shade the overlap of half-planes.

Why Simultaneous Systems Matter on the GRE

ETS routinely embeds systems into word problems, Quantitative Comparison (QC) items, and Numeric Entry questions. Recognizing when a question is really asking about a system — and spotting the "no solution" or "infinitely many solutions" trap — is one of the most reliable separators between a 160 and a 165 on Quant.

Linear Systems in Two Variables

A system of two linear equations in two variables has the form:

• a1 x + b1 y = c1
• a2 x + b2 y = c2

Geometrically, each equation is a line; the solution is their intersection point. The Graphical data and coordinate systems are drawn to scale convention (per ETS Math Conventions) means that when a system is presented on a coordinate grid, you can estimate the intersection visually — but exact algebra is safer.

Method 1: Substitution

Solve one equation for one variable, then substitute into the other.

Worked Example 1: Solve x + y = 10 and 2x - y = 8.

• Step 1: From the first equation, y = 10 - x.
• Step 2: Substitute into the second: 2x - (10 - x) = 8, so 3x - 10 = 8, 3x = 18, x = 6.
• Step 3: Back-substitute: y = 10 - 6 = 4.
• Solution: (6, 4).

Method 2: Elimination

Multiply one or both equations so a variable's coefficients align, then add to eliminate it.

Worked Example 2: Solve 3x + 2y = 16 and 5x - 2y = 8.

• The y-coefficients already oppose. Add the equations: 8x = 24, so x = 3.
• Substitute back: 3(3) + 2y = 16, 9 + 2y = 16, 2y = 7, y = 3.5.
• Solution: (3, 3.5).

Choose substitution when one equation already isolates a variable; choose elimination when coefficients align or can be made to align with a small multiplier.

The Trap: No Solution and Infinitely Many Solutions

When both equations describe parallel lines (same slope, different y-intercept), the system has no solution — it is inconsistent. When both equations are secretly the same line (one is a nonzero multiple of the other), the system has infinitely many solutions — it is dependent.

The GRE favorite trap: giving two equations that look different but reduce to the same line, then asking for "the unique solution." There is none.

Worked Example 3: How many solutions does 2x + 3y = 12 and 4x + 6y = 30 have?

• Multiply the first equation by 2: 4x + 6y = 24.
• Compare to 4x + 6y = 30. Same left side, different right side -> parallel lines -> no solution.

Worked Example 4: How many solutions does 2x + 3y = 12 and 4x + 6y = 24 have?

• Multiply the first by 2: 4x + 6y = 24, identical to the second -> coincident -> infinitely many solutions.

Quick check: compute the ratio of coefficients a1/a2, b1/b2, c1/c2. If the first two ratios are equal but the third differs, the system is inconsistent. If all three ratios are equal, the system is dependent.

Nonlinear Systems

The GRE occasionally pairs a linear equation with a quadratic (parabola) or, more rarely, a circle. Solve by substitution.

Worked Example 5: Solve y = x^2 and y = x + 2.

• Substitute: x^2 = x + 2, so x^2 - x - 2 = 0, (x - 2)(x + 1) = 0.
• x = 2 or x = -1.
• Points: (2, 4) and (-1, 1).

A line can intersect a parabola at 0, 1 (tangent), or 2 points — useful for QC items asking "how many intersection points?" Count solutions to the substituted quadratic by checking its discriminant.

Systems of Inequalities: Shading the Region

A linear inequality like y < 2x + 1 shades a half-plane. To graph:

1. Draw the boundary line: solid for <= or >= (boundary included), dashed for < or > (boundary excluded).
2. Pick a test point not on the line (the origin is easiest when the line doesn't pass through it).
3. If the test point satisfies the inequality, shade its side; otherwise, shade the opposite side.

For a system of inequalities, the solution is the intersection of all shaded half-planes — the region satisfying every inequality simultaneously.

Worked Example 6: Shade y >= x - 2 and y <= -x + 4.

• Boundary 1: y = x - 2 (solid). Test (0, 0): 0 >= -2, true -> shade above.
• Boundary 2: y = -x + 4 (solid). Test (0, 0): 0 <= 4, true -> shade below.
• Solution region: the wedge above the first line and below the second, including their intersection point.
• Intersection corner: x - 2 = -x + 4, so 2x = 6, x = 3, y = 1. The corner of the shaded region is (3, 1).

A QC variant: "Which point lies in the solution region of both inequalities?" Substitute each candidate point into both inequalities; the one that satisfies both is correct.

GRE Strategic Tips

• Line up coefficients before eliminating to avoid arithmetic slips; the on-screen calculator computes operations in the order you enter them, so handle each multiplication and addition as separate keystrokes.
• Check for parallel/coincident before solving by comparing slope and intercept — this can save you from computing a non-existent intersection.
• For QC questions comparing two systems, solve both; never assume "looks like" means "equals."
• For inequality shading questions, test the origin first (when it isn't on a boundary); the origin is the fastest test point.
• For nonlinear systems, substitute the simpler equation into the more complex one, then count the quadratic's real roots via the discriminant.
• Watch the inequality flip: when you eliminate a variable by multiplying an inequality by a negative number, flip the inequality sign.