What algebra should you prepare?
GMAC’s Quantitative Reasoning page lists linear and quadratic equations, systems, powers, roots, functions, inequalities, and absolute value, among other topics. This workshop focuses on validity conditions that apply across those topics; it is not a complete study program on its own. The unknowns in the three exercises below are real numbers.
First distinguish the question from a routine classroom instruction. Finding a value, identifying an interval, or deciding whether information is sufficient may require different work. Substituting answer choices may be faster than solving generally, but it must respect every restriction.
Leave one line in your scratch work for conditions. This prevents a result at the end of your calculation from making you forget that an earlier division was not allowed. For your final check, return to the original problem, not just the last transformed equation.
What remains after simplifying?
Simplification can make a denominator disappear from view, but its restriction still applies.
For x ≠ 2, (x² − 4)/(x − 2) = 7. What is x?
Choose an answer.
The solution, step by step
5.
The method in action
- Factor x² − 4 as (x − 2)(x + 2). For x ≠ 2, the fraction equals x + 2. The equation becomes x + 2 = 7, so x = 5.
- In the original expression, (25 − 4)/(5 − 2) = 21/3 = 7. The condition x ≠ 2 is satisfied.
- 2 is excluded; 3 and 4 give 5 and 6; 7 confuses the value of the fraction with the value of the unknown.
When should you reverse an inequality?
Adding the same quantity preserves the inequality. Multiplying or dividing both sides by a negative quantity reverses it. If the sign is unknown, separate cases before that operation.
Which statement is equivalent to 5 − 2x < 11?
Choose an answer.
The solution, step by step
x > −3.
The method in action
- Subtracting 5 gives −2x < 6. Dividing by −2 reverses the inequality: x > −3.
- Check with x = 0: 5 < 11 is true, and 0 > −3 is also true. At the boundary x = −3, you get 11 < 11, which is false; equality is excluded.
- x < −3 gives the opposite solution set; the choices around 3 lose the negative sign; equality alone does not satisfy the strict inequality.
The pitfall
Testing a few values does not prove a universal rule
A trial can disprove an equality claimed to hold for all numbers if it yields a counterexample. Three successful trials do not prove the equality. Distinguish a numerical check from a proof.
For example, substituting positive numbers for unknowns in an inequality may hide the negative case. If the prompt does not require positivity, your method must also cover zero, negative numbers, or fractions when allowed.
How can you avoid missing a root?
An equation with a squared unknown may have two solutions. A sign restriction can eliminate one, but it must come from the prompt.
If (x − 1)² = 16 and x < 0, what is x?
Choose an answer.
The solution, step by step
−3.
The method in action
- x − 1 equals 4 or −4. The two solutions to the equation alone are 5 and −3.
- The condition x < 0 leaves only −3. Check: (−3 − 1)² = 16.
- 5 satisfies the equation but not the condition; −4 misses the shift by 1; −5 and 3 do not satisfy the equation.
Key takeaways
- Keep restrictions even after simplifying.
- Check the sign when dividing.
- A counterexample disproves a universal rule; a few successful trials do not prove it.