Hard GRE Linear Equations Questions Practice Questions

Concept Explanation

Linear equations are algebraic statements where the highest power of the variable is one, typically representing a straight line when graphed on a coordinate plane. In the context of the GRE, these problems often evolve beyond simple one-step calculations into complex systems of equations, word problems involving rates or mixtures, and quantitative comparison tasks. Success on Hard GRE Linear Equations Questions requires the ability to translate dense text into mathematical symbols and solve for multiple variables simultaneously using substitution or elimination methods.

A standard linear equation in one variable takes the form $$ax + b = c$$ while a system of two linear equations in two variables is typically expressed as:

$$a_1x + b_1y = c_1$$

$$a_2x + b_2y = c_2$$

On the GRE Prep journey, you will encounter "hard" variations that include fractional coefficients, variables in the denominator that can be linearized, or constraints that require integer solutions. Understanding the properties of lines, such as slope-intercept form $y = mx + b$ and the conditions for parallel or perpendicular lines, is also essential for coordinate geometry-based linear questions. For more comprehensive practice, you can explore GRE Practice Questions with Answers to refine your foundational skills.

Solved Examples

1. Example 1: The Mixture Problem
   A chemist has two solutions of sulfuric acid. Solution A is 20% acid and Solution B is 45% acid. How many liters of Solution B must be added to 15 liters of Solution A to create a mixture that is 30% acid?
   1. Define the variable: Let $x$ be the liters of Solution B.
   2. Set up the equation based on the total amount of pure acid: $0.20(15) + 0.45(x) = 0.30(15 + x)$.
   3. Multiply by 100 to clear decimals: $20(15) + 45x = 30(15 + x)$.
   4. Simplify: $300 + 45x = 450 + 30x$.
   5. Isolate $x$: $15x = 150$, so $x = 10$.
   6. Answer: 10 liters.
2. Example 2: Systems with Three Variables
   Solve for $z$ given the system:
   $$x + y + z = 12$$
   $$2x - y + z = 7$$
   $$x + 2y - z = 9$$
   1. Add the first two equations to eliminate $y$: $(x + y + z) + (2x - y + z) = 12 + 7 \Rightarrow 3x + 2z = 19$.
   2. Multiply the first equation by 2 and subtract the third: $2(x + y + z) - (x + 2y - z) = 2(12) - 9 \Rightarrow 2x + 2y + 2z - x - 2y + z = 15 \Rightarrow x + 3z = 15$.
   3. Now solve the new system: (1) $3x + 2z = 19$ and (2) $x + 3z = 15$.
   4. From (2), $x = 15 - 3z$. Substitute into (1): $3(15 - 3z) + 2z = 19$.
   5. Simplify: $45 - 9z + 2z = 19 \Rightarrow 45 - 7z = 19 \Rightarrow 26 = 7z$.
   6. Answer: $z = \frac{26}{7}$.
3. Example 3: Quantitative Comparison
   Quantity A: The value of $k$ if the lines $3x + 4y = 12$ and $kx - 6y = 10$ are perpendicular.
   Quantity B: 8
   1. Find the slope of the first line: $4y = -3x + 12 \Rightarrow y = -\frac{3}{4}x + 3$. Slope $m_1 = -\frac{3}{4}$.
   2. Perpendicular lines have slopes that are negative reciprocals. So, the slope of the second line $m_2$ must be $\frac{4}{3}$.
   3. Find the slope of the second line: $-6y = -kx + 10 \Rightarrow y = \frac{k}{6}x - \frac{10}{6}$. Slope $m_2 = \frac{k}{6}$.
   4. Set the slopes equal: $\frac{k}{6} = \frac{4}{3}$.
   5. Solve for $k$: $3k = 24 \Rightarrow k = 8$.
   6. Answer: The two quantities are equal.

Practice Questions

1. A rental car company charges a flat daily fee plus a fixed cost per mile driven. If a customer pays $55 for a 100-mile trip and $85 for a 250-mile trip, what is the flat daily fee?
2. Solve for $x$ in the equation: $$\frac{2x - 3}{4} - \frac{x + 1}{3} = \frac{x - 2}{6}$$
3. In a certain piggy bank, there are only quarters ($0.25) and dimes ($0.10). If there are 45 coins in total and the total value is $8.25, how many quarters are in the bank?

4. If $5a + 3b = 21$ and $2a + 7b = 20$, what is the value of $a - b$?
5. A line passes through the points $(2, -3)$ and $(5, k)$. If the slope of the line is 4, what is the value of $k$?
6. Working alone, Machine A can produce 500 widgets in 4 hours. Machine B can produce 500 widgets in 6 hours. If both machines work together, how many hours will it take to produce 1,000 widgets?
7. Find the value of $y$ such that the system of equations has no solution:
   $$4x - 6y = 12$$
   $$6x + ay = 15$$
8. A merchant mixes coffee that costs $12 per pound with coffee that costs $18 per pound. If the merchant wants 30 pounds of a blend that costs $14 per pound, how many pounds of the $18 coffee should be used?
9. If $\frac{1}{x} + \frac{1}{y} = \frac{1}{4}$ and $x = 3y$, what is the value of $x$?
10. Quantity A: The x-intercept of the line $2x - 5y = 20$.
    Quantity B: The y-intercept of the line $3x + 2y = 16$.

Answers & Explanations

1. $35: Let $f$ be the flat fee and $m$ be the cost per mile. Equation 1: $f + 100m = 55$. Equation 2: $f + 250m = 85$. Subtract Eq 1 from Eq 2: $150m = 30$, so $m = 0.20$. Substitute into Eq 1: $f + 100(0.20) = 55 \Rightarrow f + 20 = 55 \Rightarrow f = 35$.
2. $x = 9$: Multiply the entire equation by the least common multiple (12): $3(2x - 3) - 4(x + 1) = 2(x - 2)$. Expand: $6x - 9 - 4x - 4 = 2x - 4$. Simplify: $2x - 13 = 2x - 4$. Wait, if we simplify further we get $-13 = -4$, which is impossible. *Correction*: Let's re-evaluate. $3(2x-3) - 4(x+1) = 6x - 9 - 4x - 4 = 2x - 13$. The right side is $2(x-2) = 2x - 4$. This equation has no solution. (Note: On the GRE, ensure you don't misread the LCM).
3. 25 quarters: Let $q$ be quarters and $d$ be dimes. $q + d = 45$ and $0.25q + 0.10d = 8.25$. From the first, $d = 45 - q$. Substitute: $0.25q + 0.10(45 - q) = 8.25 \Rightarrow 0.25q + 4.5 - 0.10q = 8.25 \Rightarrow 0.15q = 3.75 \Rightarrow q = 25$.
4. 1: Subtract the two equations: $(5a + 3b) - (2a + 7b) = 21 - 20 \Rightarrow 3a - 4b = 1$. This doesn't give $a-b$ directly. Let's solve for $a$ and $b$. Multiply Eq 1 by 7 and Eq 2 by 3: $35a + 21b = 147$ and $6a + 21b = 60$. Subtract: $29a = 87 \Rightarrow a = 3$. Substitute back: $2(3) + 7b = 20 \Rightarrow 7b = 14 \Rightarrow b = 2$. Thus, $a - b = 3 - 2 = 1$.
5. 9: Slope $m = \frac{y_2 - y_1}{x_2 - x_1}$. So, $4 = \frac{k - (-3)}{5 - 2} \Rightarrow 4 = \frac{k + 3}{3}$. Multiply by 3: $12 = k + 3 \Rightarrow k = 9$.
6. 4.8 hours: Rate A = 500/4 = 125 widgets/hr. Rate B = 500/6 = 83.33 widgets/hr. Combined rate = $125 + \frac{500}{6} = \frac{750+500}{6} = \frac{1250}{6}$ widgets/hr. Time = Distance/Rate = $1000 / (\frac{1250}{6}) = \frac{6000}{1250} = 4.8$ hours.
7. -9: For no solution, the slopes must be equal but the y-intercepts different. Slope 1: $-6y = -4x + 12 \Rightarrow y = \frac{2}{3}x - 2$. Slope 2: $ay = -6x + 15 \Rightarrow y = -\frac{6}{a}x + \frac{15}{a}$. Set slopes equal: $\frac{2}{3} = -\frac{6}{a} \Rightarrow 2a = -18 \Rightarrow a = -9$.
8. 10 pounds: Let $x$ be the pounds of $18 coffee. Then $30 - x$ is the pounds of $12 coffee. $18x + 12(30 - x) = 14(30)$. $18x + 360 - 12x = 420 \Rightarrow 6x = 60 \Rightarrow x = 10$.
9. 16: Substitute $x = 3y$ into the first equation: $\frac{1}{3y} + \frac{1}{y} = \frac{1}{4}$. Find common denominator: $\frac{1}{3y} + \frac{3}{3y} = \frac{4}{3y} = \frac{1}{4}$. Cross multiply: $3y = 16$. Since $x = 3y$, then $x = 16$.
10. Quantity A is larger: For Quantity A, set $y = 0$ in $2x - 5y = 20 \Rightarrow 2x = 20 \Rightarrow x = 10$. For Quantity B, set $x = 0$ in $3x + 2y = 16 \Rightarrow 2y = 16 \Rightarrow y = 8$. 10 is greater than 8.

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

• AZero
• BOne
• CTwo
• DInfinitely many

Frequently Asked Questions

What makes a linear equation question "hard" on the GRE?

Hard questions typically involve multiple variables, require setting up equations from complex word problems, or incorporate other concepts like coordinate geometry and probability. They often use non-integer coefficients or require you to recognize when a system has no solution or infinite solutions.

How do I solve systems of equations quickly?

The elimination method is usually faster for GRE-style questions where coefficients can be easily matched. If one variable is already isolated, substitution is a better choice to save time during the exam.

Can linear equations have more than one solution?

A single linear equation in two variables has infinitely many solutions, but a system of two distinct linear equations can have exactly one solution, no solution (parallel lines), or infinitely many solutions (coincident lines). On the GRE, you must identify which case applies.

How do I identify a linear equation in a word problem?

Look for keywords implying a constant rate of change, such as "per hour," "fixed fee," or "additional cost." These indicate a relationship where the total value increases or decreases by a steady amount for every unit of change.

Is the slope-intercept form always the best way to graph?

While the $y = mx + b$ form is great for identifying slope and intercept quickly, the standard form $Ax + By = C$ is often more efficient for finding both intercepts by setting one variable to zero at a time.