GRE Systems of Equations Questions Practice Questions with Answers

Two or more algebraic equations sharing the same set of variables form a system of equations, requiring a simultaneous solution for all unknowns. On the GRE Quantitative Reasoning section, these problems often appear as word problems or direct algebraic expressions. Success depends on your ability to manipulate variables through substitution or elimination to find where two lines intersect on a coordinate plane. These concepts are foundational for more advanced topics you might encounter in GRE Prep, where efficiency and accuracy are paramount.

Concept Explanation

A system of equations is a set of two or more equations that you deal with at the same time to find the common values for the variables involved. In the context of the GRE, you will primarily encounter systems of two linear equations with two variables, typically $x$ and $y$. There are three primary methods to solve these systems:

• Substitution: Isolate one variable in one equation and plug its equivalent expression into the other equation. This is most effective when one variable has a coefficient of 1 or -1.
• Elimination (Addition/Subtraction): Add or subtract the equations to cancel out one variable. You may need to multiply one or both equations by a constant to align the coefficients.
• Graphic Interpretation: The solution to a system is the point $(x, y)$ where the two lines intersect. If the lines are parallel, there is no solution; if they are the same line, there are infinitely many solutions.

For more rigorous practice with mathematical logic, you might find the AI Question Generator useful for creating custom drills. According to Khan Academy, mastering these linear relationships is a prerequisite for understanding complex data sets and functions.

Solved Examples

Example 1: Substitution Method
Solve for $x$ and $y$:
$$x + y = 10$$
$$2x - y = 8$$

1. Isolate $x$ in the first equation: $x = 10 - y$.
2. Substitute this into the second equation: $2(10 - y) - y = 8$.
3. Distribute: $20 - 2y - y = 8$, which simplifies to $20 - 3y = 8$.
4. Solve for $y$: $-3y = -12$, so $y = 4$.
5. Find $x$: $x = 10 - 4 = 6$. The solution is $(6, 4)$.

Example 2: Elimination Method
Solve for $a$ and $b$:
$$3a + 2b = 16$$
$$2a - 2b = 4$$

1. Add the two equations together to eliminate $b$: $(3a + 2a) + (2b - 2b) = 16 + 4$.
2. Simplify: $5a = 20$.
3. Solve for $a$: $a = 4$.
4. Substitute $a = 4$ into the first equation: $3(4) + 2b = 16$.
5. Simplify and solve: $12 + 2b = 16 \rightarrow 2b = 4 \rightarrow b = 2$. The solution is $(4, 2)$.

Example 3: Word Problem Application
A coffee shop sells small coffees for $2 and large coffees for $3. If they sold 50 coffees in total for $120, how many small coffees were sold?

1. Define variables: Let $s$ be small and $l$ be large.
2. Set up the system: $s + l = 50$ and $2s + 3l = 120$.
3. Multiply the first equation by 2: $2s + 2l = 100$.
4. Subtract this from the second equation: $(2s + 3l) - (2s + 2l) = 120 - 100$.
5. Solve for $l$: $l = 20$.
6. Find $s$: $s + 20 = 50$, so $s = 30$. The shop sold 30 small coffees.

Practice Questions

1. Solve the system for $x$:
$$4x + 3y = 25$$
$$x - 3y = -5$$

2. If $2x + y = 12$ and $x + 2y = 9$, what is the value of $x + y$?

3. A person buys 10 stamps, some costing $0.50 and others costing $0.20. If the total cost is $3.20, how many $0.50 stamps were purchased?

4. Solve for $y$:
$$5x - 2y = 1$$
$$3x + 4y = 11$$

5. If $3a + 7b = 26$ and $6a - 2b = 4$, find the value of $b$.

6. At a movie theater, adult tickets cost $10 and child tickets cost $6. If 100 tickets were sold for a total of $840, how many child tickets were sold?

7. Solve the system for $x$:
$$\frac{1}{2}x + \frac{1}{3}y = 5$$
$$x - y = 0$$

8. If $4x - y = 10$ and $2x + 3y = 26$, what is the value of $xy$?

9. A rectangle has a perimeter of 40 cm. If the length is 4 cm more than the width, what is the length? (Hint: $2L + 2W = 40$)

10. Determine the number of solutions for the system:
$$2x + 4y = 10$$
$$x + 2y = 5$$

Answers & Explanations

1. Answer: 4. Add the equations together: $(4x + x) + (3y - 3y) = 25 - 5$. This gives $5x = 20$, so $x = 4$.

2. Answer: 7. Instead of solving for $x$ and $y$ individually, add the two equations: $(2x + x) + (y + 2y) = 12 + 9$, which is $3x + 3y = 21$. Divide by 3 to get $x + y = 7$.

3. Answer: 4. Let $x$ be $0.50 stamps and $y$ be $0.20 stamps. Equations: $x + y = 10$ and $0.5x + 0.2y = 3.2$. Multiply the second by 10: $5x + 2y = 32$. From the first, $y = 10 - x$. Substitute: $5x + 2(10 - x) = 32 \rightarrow 5x + 20 - 2x = 32 \rightarrow 3x = 12 \rightarrow x = 4$.

4. Answer: 2. Multiply the first equation by 2: $10x - 4y = 2$. Add the second equation: $(10x + 3x) + (-4y + 4y) = 2 + 11$. So $13x = 13$, meaning $x = 1$. Substitute $x = 1$ into $3(1) + 4y = 11$, giving $4y = 8$, so $y = 2$.

5. Answer: 3. Multiply the first equation by 2: $6a + 14b = 52$. Subtract the second equation: $(6a - 6a) + (14b - (-2b)) = 52 - 4$. This gives $16b = 48$, so $b = 3$.

6. Answer: 40. Let $a + c = 100$ and $10a + 6c = 840$. From the first, $a = 100 - c$. Substitute: $10(100 - c) + 6c = 840 \rightarrow 1000 - 10c + 6c = 840 \rightarrow -4c = -160 \rightarrow c = 40$.

7. Answer: 6. From the second equation, $x = y$. Substitute into the first: $\frac{1}{2}x + \frac{1}{3}x = 5$. Find a common denominator: $\frac{3}{6}x + \frac{2}{6}x = 5 \rightarrow \frac{5}{6}x = 5$. Multiply by $\frac{6}{5}$ to get $x = 6$.

8. Answer: 24. Multiply the first equation by 3: $12x - 3y = 30$. Add the second equation: $14x = 56$, so $x = 4$. Substitute $x = 4$ into $4(4) - y = 10 \rightarrow 16 - y = 10 \rightarrow y = 6$. Then $xy = 4 \times 6 = 24$.

9. Answer: 12. Equations: $2L + 2W = 40$ (or $L + W = 20$) and $L = W + 4$. Substitute: $(W + 4) + W = 20 \rightarrow 2W = 16 \rightarrow W = 8$. Then $L = 8 + 4 = 12$.

10. Answer: Infinitely many. Notice that if you divide the first equation by 2, you get $x + 2y = 5$, which is identical to the second equation. Since they are the same line, they overlap at every point.

1. If a system of two linear equations consists of two lines that are parallel, how many solutions does the system have?

• AZero
• BOne
• CTwo
• DInfinitely many

Frequently Asked Questions

What is the most common way systems of equations appear on the GRE?

Most GRE systems of equations appear as word problems involving rates, mixtures, or costs, requiring you to translate text into algebraic expressions. You will also see them in Quantitative Comparison questions where you must determine which variable is larger.

How do I know whether to use substitution or elimination?

Use substitution when one variable is already isolated or has a coefficient of 1. Use elimination when the variables are already lined up in $Ax + By = C$ form and can be easily cancelled by addition or subtraction.

Can a system of two linear equations have exactly two solutions?

No, a system of two linear equations can only have zero solutions (parallel lines), one solution (intersecting lines), or infinitely many solutions (identical lines). Only non-linear systems, such as a line and a parabola, can have exactly two solutions.

What should I do if the equations have fractions?

The easiest strategy is to multiply the entire equation by the least common multiple of the denominators to clear the fractions. This turns the system into a standard integer-based system that is much easier to solve using elimination.

Are there systems with three variables on the GRE?

While rare, three-variable systems can appear. They are usually designed to be solved quickly by adding all three equations together or by noticing a pattern that allows you to find a specific expression like $x + y + z$ without solving for each variable individually.

Can I use a calculator for these on the GRE?

Yes, the GRE provides an on-screen calculator, but relying on it for systems of equations can sometimes be slower than doing the mental math or algebraic manipulation. Use it primarily for final arithmetic steps like large multiplications or divisions.