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GRE · Algebra · TOPIC LESSON

Simultaneous equations and inequalities

Find values that satisfy every condition, and recognize when a system fixes only part of the information.

GRE General Test40 min study guide9 indexed sections

BY THE END, YOU CAN

  • Solve simultaneous equations with substitution and elimination.
  • Classify intersections and dependent or inconsistent equations.
  • Combine inequalities and model constrained quantities.

BEFORE YOU START

  • Linear equations
  • Slope and basic inequality rules

Original educational lesson. Work through the examples before revealing the practice answers. This original lesson supports current GRE General Test study; it does not reproduce official questions, certify a writing score or predict test content.

LEARN THROUGH ORIGINAL PRACTICE

Original GRE-style worked examples and practice

All teaching passages, writing prompts, examples and exercises in this lesson are original. No official ETS questions, prompts or explanations are reproduced. Current structure and observed public sample metadata are separate from this original teaching; no official score or question probability is implied.

01 / UNDERSTAND & APPLY

A solution must satisfy all equations together

A system is a collection of conditions that apply simultaneously. A pair (x,y) solving one equation is not enough; it must satisfy every equation and all domain restrictions. Two linear equations in two variables can have one solution, no solutions, or infinitely many.

Choose variables with units, and keep their meanings unchanged. An equation for total objects and another for total cost are independent conditions; writing the same relationship twice in different forms supplies no additional information.

02 / UNDERSTAND & APPLY

Replace a variable by an equal expression

Substitution is efficient when one equation already isolates a variable or has coefficient 1 or −1. Solve that equation for a variable, then replace every occurrence of that variable in the other equation. Parenthesize the substituted expression so all signs and factors distribute correctly.

Once one value is found, substitute back for the other, and verify the ordered pair in both originals. The order in (x,y) matters.

ORIGINAL WORKED EXAMPLE

A substitution system

Solve y=2x+1 and 3x+y=16.

  1. Replace y in the second equation: 3x+(2x+1)=16.
  2. 5x=15 gives x=3, then y=7.
  3. Check 7=2·3+1 and 3·3+7=16.

Answer (x,y)=(3,7)

03 / UNDERSTAND & APPLY

Choose combinations that remove a variable

Adding equations preserves common solutions. Multiply one or both equations by nonzero constants so coefficients of a chosen variable are opposites, then add. Multiply every term, including the right side. Subtraction is equally valid but requires careful handling of the entire second equation.

If the question asks for a sum or difference, an equation combination may produce it directly. Solving for each individual variable can be unnecessary.

ORIGINAL WORKED EXAMPLE

Eliminate with scaling

Solve 2x+3y=17 and 5x−2y=14.

  1. Multiply the first equation by 2 and the second by 3: 4x+6y=34 and 15x−6y=42.
  2. Add to get 19x=76, so x=4.
  3. Then 8+3y=17 gives y=3; the second original checks as 20−6=14.

Answer (4,3)

04 / UNDERSTAND & APPLY

Parallel, intersecting or identical lines

For nonvertical lines written y=mx+b, different slopes give one intersection. Equal slopes with different intercepts give distinct parallel lines and no solution. Equal slopes and equal intercepts give the same line and infinitely many solutions. Algebraically these cases yield a unique value, a contradiction, or an identity.

A zero coefficient requires care; vertical lines cannot be written with a finite slope. More generally, compare whole equations, including constants. Proportional left-hand coefficients do not guarantee dependent equations if their right-hand constants are not proportional.

ORIGINAL WORKED EXAMPLE

A contradiction after elimination

Classify 2x−y=5 and 4x−2y=13.

  1. Double the first equation: 4x−2y=10.
  2. The second claims the identical left side equals 13.
  3. Subtracting yields 0=3, so no ordered pair satisfies both.

Answer No solution

05 / UNDERSTAND & APPLY

The requested expression may be fixed even when variables are not

If every solution satisfies x+y=8, the sum is known even though x and y may vary. A dependent system can therefore determine a requested expression without determining the pair. Conversely, knowing a sum does not fix the product or difference.

For QC, distinguish lack of a unique variable from lack of a unique comparison. Use equation combinations to see what remains fixed, then use two valid solutions to test what varies.

ORIGINAL WORKED EXAMPLE

A fixed sum from a dependent system

Suppose 2x+2y=14 and x+y=7. Find 3x+3y.

  1. The first equation is twice the second, so they do not uniquely determine x and y.
  2. The requested expression is 3(x+y).
  3. Since x+y=7 for every solution, the expression is always 21.

Answer 21

06 / UNDERSTAND & APPLY

Intersect the feasible regions

Simultaneous inequalities require their intersection. In one variable, solve each and keep only their common interval. In two variables, each linear inequality describes a half-plane; include its boundary for ≤ or ≥ and exclude it for < or >.

To identify the side of a line, test a point not on the boundary, such as the origin if allowed. A dashed boundary signals exclusion, a solid one inclusion. The overlap of shaded regions contains feasible points; a boundary intersection alone does not establish that a point satisfies all restrictions.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Intersection of two linear inequality regionsEqual-scale coordinate axes show the solid lower line y=x−1 and dashed upper line y=−x+3. Their common region is shaded between the lines to the left of (2,1), which is excluded. The point (0,0) is inside and (0,4) outside.-2-101234-3-2-112345(2, 1) excluded(0, 0) feasible(0, 4) not feasibleSolid: y = x − 1Dashed: y = −x + 3xyOverlap: x − 1 ≤ y < −x + 3Equal unit scales; the shaded wedge continues to the left
Both inequalities must hold: y≥x−1 and y<−x+3. The lower boundary is included only left of the crossing; the upper boundary and crossing are excluded. Shading continues beyond the displayed left edge.

ORIGINAL WORKED EXAMPLE

Intersect one-variable constraints

Solve 2x−1>5 and 3x+2≤20 simultaneously.

  1. The first gives x>3.
  2. The second gives x≤6.
  3. Their intersection is 3<x≤6; neither entire original range alone is sufficient.

Answer (3,6]

ORIGINAL WORKED EXAMPLE

Intersect two half-planes

Describe the solution region of y≥x−1 and y<−x+3, and test (0,0) and (0,4).

  1. The first inequality includes the solid boundary y=x−1 and points above it. The second excludes the dashed boundary y=−x+3 and includes points below it.
  2. The lines meet where x−1=−x+3, so (x,y)=(2,1). A shared vertical interval exists only when x−1<−x+3, giving x<2. Thus the region is the wedge to the left of the crossing, between the lines.
  3. At (0,0), both 0≥−1 and 0<3 hold, so the point is feasible. At (0,4), 4≥−1 holds but 4<3 fails, so the point is not feasible.
  4. The crossing (2,1) is excluded because it lies on the strict upper boundary. The lower boundary is included only where it also lies strictly below the upper boundary, namely for x<2.

Answer x<2 and x−1≤y<−x+3; (0,0) is feasible, (0,4) is not.

07 / UNDERSTAND & APPLY

Two kinds of totals often supply two equations

A standard model uses x+y=N for the count and px+qy=C for a cost, mass or revenue total. Another uses distance=rate×time for each traveler plus a shared distance or time relationship. Units must match across each equation.

Check feasibility after algebra. A ticket count of −2 or a fractional number of indivisible items cannot describe the stated situation. If a system is inconsistent, do not choose the “closest” pair unless the question explicitly permits approximation.

08 / UNDERSTAND & APPLY

Choose a method and answer the requested question

All examples and practice in this lesson are original. For quantitative comparison (QC), use A if Quantity A is greater, B if Quantity B is greater, C if they are equal, and D if the relationship cannot be determined from the information given. A, B or C must hold for every permitted case. To establish D, give two allowed cases producing different relationships; an inability to calculate is not itself evidence for D.

A single-answer question requires one choice. For multiple-answer questions, obey the stated selection count or select every correct option if instructed. Test each option separately; do not stop after finding one. Multiple-answer questions receive credit only when the complete correct set is selected, with no incorrect choices. Numeric entry requires the requested value, with units handled in the calculation; round only when the question directs you to. A single numeric-entry box accepts an integer or decimal; a pair of fraction boxes accepts an integer numerator and a nonzero integer denominator. Fractions need not be reduced, but reducing makes checks easier. Do not enter a percent sign when the requested number is a percentage. Preserve exact fractions until the last step.

A calculator can check arithmetic but cannot decide the domain, interpretation or denominator for you. Estimate sign and size first, enter parentheses carefully, and compare the result with the original conditions. Cancel factors or exploit structure when doing so is faster and more reliable than decimal entry.

09 / UNDERSTAND & APPLY

Original practice: simultaneous means every condition

Check answers in all original equations.

ORIGINAL PRACTICE · 1

Numeric entry: If x+y=11 and x−y=3, find x.

Reveal answer & reasoning

7

Add equations to obtain 2x=14. The other value is y=4, and both equations check. Reporting 4 would answer for the wrong variable.

ORIGINAL PRACTICE · 2

Single answer: Which pair solves 2x+y=9 and x−y=3? A (3,3); B (4,1); C (1,4); D (5,−1); E (2,5).

Reveal answer & reasoning

B

Adding gives 3x=12, hence x=4 and y=1. Other choices may satisfy one equation but fail the other.

ORIGINAL PRACTICE · 3

Select all that apply: Which pairs satisfy x+y≤5 and y≥2? A (1,2); B (4,2); C (0,5); D (3,1); E (−2,6).

Reveal answer & reasoning

A, C and E

A has sum 3 and y=2; C has sum 5 and y=5; E has sum 4 and y=6. B has sum 6, too large; D has y=1, too small. Equality on either boundary is permitted.

ORIGINAL PRACTICE · 4

QC: x+y=6, with x,y real. Quantity A: x. Quantity B: y.

Reveal answer & reasoning

D

Pairs (5,1) and (1,5) satisfy the condition and reverse the comparison. A fixed sum does not order its two parts.

ORIGINAL PRACTICE · 5

Numeric entry: If 3x+2y=19 and x+2y=9, find x.

Reveal answer & reasoning

5

Subtract the second whole equation from the first to get 2x=10. Substitution gives y=2; every constant must be subtracted as well.

ORIGINAL PRACTICE · 6

Single answer: The system y=3x+1 and y=3x−4 has A no solutions; B one solution; C two solutions; D infinitely many solutions; E insufficient information to count.

Reveal answer & reasoning

A

The slopes match but intercepts differ, so the lines are distinct and parallel. Equal slopes alone do not mean identical lines.

ORIGINAL PRACTICE · 7

Numeric entry: A shop sells 20 notebooks: small at $3 and large at $5, for total $76. How many large notebooks?

Reveal answer & reasoning

8

Let L be large and S small. S+L=20 and 3S+5L=76. Substitute S=20−L: 60+2L=76, so L=8 and S=12.

ORIGINAL PRACTICE · 8

QC: 4x−2y=10. Quantity A: 2x−y. Quantity B: 5.

Reveal answer & reasoning

C

Dividing the equation by 2 fixes the requested expression at 5, even though x and y individually are not unique. Indeterminate variables do not automatically imply QC choice D.

KEEP THE BIG IDEAS

Quick revision

  • Satisfy every condition and preserve variable meanings.
  • Choose substitution or elimination to exploit coefficients.
  • An identity may leave variables free but still fix a requested expression.
  • Systems of inequalities use the intersection of allowed regions.
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