GRE Algebra & Functions Practice & Questions

1. Solve for $x$: $$3x - 7 = 2x + 8$$ $$3x - 2x = 8 + 7 \implies x = 15$$ 2. Calculate $x^2 - 10$: $$15^2 - 10 = 225 - 10 = 215$$ Thus, the correct answer is $215$.

Always read the prompt carefully: ETS often asks for an expression like $x^2 - 10$ rather than $x$ itself.

1. Find the slope of line $L_1$: $$m_1 = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2$$ 2. Since $L_2 \perp L_1$, its slope $m_2$ is the negative reciprocal: $$m_2 = -\frac{1}{m_1} = -\frac{1}{2}$$ 3. Use point-slope form with $(4, -1)$: $$y - (-1) = -\frac{1}{2}(x - 4)$$ $$y + 1 = -\frac{1}{2}x + 2 \implies y = -\frac{1}{2}x + 1$$ 4. The $y$-intercept ($b$) is $1$.

Perpendicular lines have negative reciprocal slopes: $m_1 \cdot m_2 = -1$.

1. Expand both quantities: - Quantity A: $x^2 + 2x + 1$ - Quantity B: $x^2 - 2x + 1$ 2. Subtract $x^2 + 1$ from both quantities: - Quantity A becomes $2x$ - Quantity B becomes $-2x$ 3. Test cases based on sign: - If $x = 2$ (positive): Quantity A $= 4$, Quantity B $= -4 \implies \text{Quantity A} > \text{Quantity B}$. - If $x = -2$ (negative): Quantity A $= -4$, Quantity B $= 4 \implies \text{Quantity B} > \text{Quantity A}$. Because the relative order flips when $x$ switches signs, the relationship cannot be determined.

Standard ETS trap: examinees unconsciously test only positive integers ($x = 1, 2$), where Quantity A is larger, and fail to test negative values where the inequality inverts.

Always test numbers from the "ZONE F" framework: Zero, One, Negatives, Extremes, and Fractions.

1. Determine the individual rates of work (fraction of reservoir filled per hour): - Rate of Pump $A$: $R_A = \frac{1}{6}$ reservoir/hour. - Rate of Pump $B$: $R_B = \frac{1}{4}$ reservoir/hour. 2. Add the rates to determine the combined simultaneous rate $R_{\text{combined}}$: $$R_{\text{combined}} = \frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12} \text{ reservoir/hour}$$ 3. The total time $T$ required to fill $1$ reservoir is the reciprocal of the combined rate: $$T = \frac{1}{R_{\text{combined}}} = \frac{12}{5} = 2.4 \text{ hours}$$ Thus, both pumps working together will fill the reservoir in $2.4$ hours.

Examinees often intuitively average the times: $(6 + 4)/2 = 5$ hours. Two pumps working together must take less time than the fastest pump working alone (4 hours). Always sum rates, never average times.

For combined work rate problems: $\frac{1}{T} = \frac{1}{t_1} + \frac{1}{t_2} \implies T = \frac{t_1 \times t_2}{t_1 + t_2}$.

1. Expand the absolute value inequality $|2x - 5| \le 7$: $$-7 \le 2x - 5 \le 7$$ 2. Add $5$ to all three parts: $$-2 \le 2x \le 12$$ 3. Divide by $2$: $$-1 \le x \le 6$$ 4. Count the integer values of $x$ in the inclusive range $[-1, 6]$: $$\text{Count} = \text{Last} - \text{First} + 1 = 6 - (-1) + 1 = 6 + 1 + 1 = 8$$ The integer solutions are $\{-1, 0, 1, 2, 3, 4, 5, 6\}$, exactly $8$ integers. Therefore, Quantity A ($8$) is equal to Quantity B ($8$).

Test takers often calculate $6 - (-1) = 7$, forgetting to add $1$ when counting integers inclusively in an interval.

The number of consecutive integers from $a$ to $b$ inclusive is always $(b - a + 1)$. Never forget to add $1$.