Plugging in values and testing answer choices

2.1 Distinguish two different substitution strategies

Choosing your own values replaces a general variable situation with a convenient numerical instance. This is useful when the question asks which expression represents a relationship or is equivalent to another expression.

Testing answer choices treats the supplied options as candidate solutions. This is useful when inserting a number into the original conditions is easier than deriving that number from scratch.

The SAT uses both four-option multiple-choice questions and student-produced responses. A student-produced response may allow more than one correct value, but you submit only one.^[5] Backsolving from choices is unavailable when no choices are supplied; choosing a useful illustrative value also cannot replace finding a specific unknown that is fixed by the problem.

Strategy	Good trigger	What must stay true
Choose numbers	Variables appear in a general relationship and in the answer choices.	The numbers satisfy every stated restriction.
Test candidates	The options are possible inputs, dimensions, or counts.	A surviving option satisfies the original problem and the requested qualifier.
Find a counterexample	A choice makes a claim that must hold for all allowed values.	One allowed failure disproves the claim.
Use strategic inputs in an identity	A polynomial or function equality holds for all allowed inputs.	Substitution produces information rather than an automatic $0 = 0.$

A numerical test has a logical direction

If two expressions are supposed to be equal for every allowed input, one allowed input giving different outputs proves they are not equivalent. But matching at one input does not prove equivalence.

In a single-correct-answer multiple-choice item, if a valid test eliminates three choices, the surviving choice is identified. If two or more survive, test a different allowed input or use algebra. Do not stop merely because the first option you tested happened to match.

Keep the contract of the question

A value that violates “positive,” “integer,” “distinct,” or “nonzero” is not a valid test. A candidate that satisfies an equation but violates the original domain is not a valid answer.

Use these strategies as alternatives, not rituals. If an equation is already one step from the target, direct algebra is usually shorter than evaluating four options.

2.2 Choose useful values, not merely easy ones

Good test values make arithmetic manageable and make the choices behave differently. The easiest-looking values, especially 0 and 1, often make unrelated expressions coincide.

Preserve the constraints

If $a = 3b,$ you may choose $b = 2$ and $a = 6,$ but not independently choose $a = 4$ and $b = 5.$ If x and y are distinct positive integers, (2, 5) is more useful than (1, 1), which is not allowed. If a denominator contains $x - 2,$ do not choose $x = 2.$

Use the base of 100 deliberately

For a percentage question with an unspecified original amount, choosing 100 often makes each percent an easy number. The choice is valid only when the relationship is independent of the scale. If an actual original amount is supplied, replacing it with 100 changes the problem unless you preserve the required proportional relationship.

Separate the roles of different letters

Choose unequal values for quantities with different roles. If a price and a discount rate are both represented by letters, assigning both the value 10 may conceal a reversed numerator or denominator. A price of 80 and a rate of 25 makes the roles easier to track.

A test value causes this problem	A better next move
Several options give the same number	Use a second input chosen to separate those survivors.
A denominator becomes zero	Choose another input in the original domain.
The result depends on an arbitrary choice	Recheck whether the target was supposed to be fixed or general.
The arithmetic becomes cumbersome	Choose a smaller allowed integer or a convenient percent base.
A symmetry makes two quantities identical	Choose unequal values that still satisfy the conditions.

A compact testing ledger

Record the chosen values once, compute the target value from the problem, and evaluate every surviving option using those same values. Keep a row for each test so that evidence from different inputs is not mixed.

You do not need many random trials. One purposeful test may eliminate most options; one carefully selected second test may finish the job. If the surviving expressions remain hard to separate, their algebraic difference may reveal the issue more directly.

2.3 Backsolve with a reasoned order

Backsolving starts with a candidate and asks whether it produces the stated output. Keep the original units and all stages of the model. If a candidate is an original price, apply the specified discount to it; do not substitute it as the discounted price.

When a middle choice can save work

Suppose numerical choices are ordered, and the modeled output increases as the candidate increases throughout the allowed domain. If a middle candidate produces too small an output, all smaller candidates can be eliminated at once. This is a monotonicity argument, not a universal testing rule. For example, with positive radius r, the expression $r 2 (r + 4)$ increases as r increases because both positive factors increase. By contrast, $x 2 - 10x + 16$ decreases and then increases. A “too high” or “too low” result on that quadratic does not, by itself, eliminate an entire side of the option list.

Order is useful only with a valid direction

Before rejecting several choices at once, explain why the relevant output must move in one direction over the allowed inputs. If you cannot justify that, test candidates individually or use another method.

Filter before substituting

A length must be positive. A probability must be between 0 and 1. A solution to $\sqrt{x + 5}$ must have $x \geq 1$ because the square root is nonnegative. Such filters can remove candidates before any detailed arithmetic.

Test in the original conditions

A transformed equation may introduce extraneous candidates. Squaring an equation removes sign information. Multiplying by an expression that can be zero can hide a domain restriction. A correct candidate must pass the original statement, not just the last line of your work.

Read the qualifier after finding a match

A question asking for the greater solution may present both roots as choices. Finding the smaller root verifies that it is a solution, not that it is the answer. Similarly, a minimum-count problem requires checking that the preceding integer fails.

A stopping condition: all requirements are met, and either the options’ single-answer structure or a mathematical argument establishes that your selected response is the requested one.

2.4 Use testing to discover structure, then justify it

Substitution and algebra are partners. A numerical test can reveal a relationship, locate a plausible candidate, or disprove a claim. Algebra can then explain why the result holds generally.

Existence: one valid example proves that something can happen.

Counterexample: one allowed failure disproves a claim that something must happen.

Identity: an equality intended to hold for all inputs needs a general justification, unless the multiple-choice elimination logic already identifies the unique answer.

Use strategic evaluation in polynomial identities

In an identity, choose an input that removes unwanted terms while preserving the target. Substituting $x = 0$ often exposes a constant term. Substituting a known root can expose another factor. If the desired expression is $a + b,$ substituting $x = 1$ may reveal it directly.

The useful input is not always a root. If both sides automatically vanish at that root, the result $0 = 0$ supplies no new information. Choose an input that isolates what you actually need.

When numerical evidence is not enough

Matching graphs in a small window does not establish equality for all real inputs. Two formulas that agree for one row of a table may disagree elsewhere. Two algebraic expressions can produce the same outputs wherever both are defined but still have different original domains.

What you have shown	What you may conclude
An option fails at one allowed input	That option is not a universal identity.
An option works at one input	It remains a candidate; it is not thereby proved universal.
Three of four options are eliminated validly	The remaining option answers a single-correct-answer item.
A proposed root satisfies the original equation	That root is valid, but another root may also exist.
A factorization is algebraically verified	It holds throughout the domain where the expressions are defined.

Section 2 exit standard

You can explain why your chosen inputs are allowed, why they are informative, and what the results actually establish. You can distinguish finding a candidate from proving that it answers the question.

15 worked examples

2.01. Use a percent base to compare expressions

For positive numbers p and q, which expression represents p% of q?
A) $\frac{pq}{100}$ B) $\frac{100q}{p}$
C) $\frac{q}{100p}$ D) 100pq

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Recognize. This is a general expression question. Choose different, convenient values that satisfy positivity: $p = 20$ and $q = 50.$

Work. Twenty percent of 50 is 10. The choices give, respectively, 10, 250, 0.025, and 100,000. Only A matches.

Answer: A, $pq/100.$

Check. Algebra confirms the result: $p% = p/100,$ so p% of q is $(p/100)q.$

Avoid the trap. The percent sign already divides the rate by 100. Multiplying by 100 instead reverses the conversion and gives an implausibly large result.

2.02. Choose 100 without changing the scale relationship

A positive quantity x is decreased by 20% and then increased by 25%. Which expression equals the final quantity? A) 0.95x B) x C) 1.05x D) 1.25x

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Recognize. Both operations are proportional, so a convenient original value can distinguish the proposed multipliers.

Work. Choose $x = 100.$ The first change gives 80; the second gives $80(1.25) = 100.$ The options then give 95, 100, 105, and 125, so only B survives.

Answer: B, x.

Check. For arbitrary x, the final amount is $x(0.80)(1.25) = x.$

Avoid the trap. Adding −20% and 25% would suggest a 5% increase, but the increase uses the reduced base. Here the two multipliers happen to be reciprocals.

2.03. Pick a scale while respecting similarity

Two similar triangles have corresponding side lengths in the ratio 2 : 5, smaller to larger. What is the ratio of the smaller triangle’s area to the larger triangle’s area? A) 2/5 B) 4/25 C) 5/2 D) 8/125

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Recognize. Similarity permits a convenient representative pair of dimensions. Both base and height scale by the same factor.

Work. Take a smaller triangle with base 2 and height 2. Its area is 2. The similar larger triangle has base 5 and height 5, with area $25/2.$ The area ratio is $2/(25/2) = 4/25.$

Answer: B, $4/25.$

Check. In general, multiplying each length by $5/2$ multiplies area by $(5/2) 2 .$ The requested ratio is the reciprocal, $(2/5) 2 .$

Avoid the trap. You may choose the scale of similar figures, but not independently change their shapes. The area ratio is not the unsquared side ratio.

2.04. Test an original price through the actual discount

After a 30% discount, a bag costs $84. What was its original price?
A) $105 B) $112 C) $120 D) $126

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Recognize. Each option is a candidate original price. A 30% discount leaves 70% of that price.

Work. Test C: $0.70(120) = 84.$ It matches exactly. Because $0.70p$ is strictly increasing in p, no different original price can give the same discounted amount.

Answer: C, $120.

Alternative. Direct algebra is equally efficient: $0.70p = 84$ gives $p = 84/0.70 = 120.$

Check. The discount is $36, and $120 - 36 = 84.$

Avoid the trap. Increasing $84 by 30% uses the wrong base. It does not reverse a 30% discount from the original price.

2.05. A matching root may not be the requested root

What is the greater solution of x² − 10x + 16 = 0?
A) 2 B) 4 C) 6 D) 8

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Recognize. The qualifier is “greater.” The quadratic is not monotonic over all the listed choices, so a one-direction backsolving rule is unsafe.

Work. Substitution shows $x = 2$ gives $4 - 20 + 16 = 0.$ That is a root, but testing $x = 8$ gives $64 - 80 + 16 = 0$ as well. Factoring confirms $(x - 2)(x - 8) = 0.$

Answer: D, 8.

Check. A nonzero quadratic has at most two distinct real roots, and the factorization identifies both. Eight is the greater one.

Avoid the trap. Stopping at the first option that satisfies the equation answers “a solution,” not “the greater solution.” Read the final qualifier again before selecting.

2.06. Use distinct values to test an equivalent expression

For real numbers a ≠ b, which expression is equivalent to $\frac{a^{2} − b^{2}}{a − b}$ ?
A) a − b B) a + b C) a² + b² D) (a + b)²

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Recognize. Choose values with a nonzero denominator and outputs that separate the choices. Let $a = 5$ and $b = 2.$

Work. The original expression becomes $(25 - 4)/(5 - 2) = 7.$ The options become 3, 7, 29, and 49. Only B matches.

Answer: B, $a + b,$ on the stated domain $a \neq b.$

Check. The factorization $a 2 - b 2 = (a - b)(a + b)$ proves the equivalence wherever $a - b \neq 0.$

Avoid the trap. Choosing $a = b$ is not merely unhelpful; it violates the original domain. Cancellation does not make that forbidden input valid.

2.07. Recognize when an easy input is inconclusive

Which expression is equivalent to x(x + 1) for all real x? A) 2x B) x² + x C) x² − x D) x³ + x

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Recognize. Testing $x = 1$ is allowed but not very informative: the target and choices A, B, and D all equal 2.

Work. Use a second input, $x = 3.$ The target equals 12. Choices A, B, C, and D give 6, 12, 6, and 30. Only B survives both tests.

Answer: B, $x 2 + x.$

Check. Distribution proves $x(x + 1) = x 2 + x$ for every real x.

Avoid the trap. A match at $x = 1$ does not establish an identity. Special values can collapse powers and hide structural differences. Record all survivors, then choose a more discriminating input.

2.08. Choose numbers that satisfy a relationship

Positive numbers a and b satisfy a = 3b. What is the value of $\frac{2a + b}{a − b}$ ?
A) 5/2 B) 3 C) 7/2 D) 7

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Recognize. The choices are constants, and the relation fixes the ratio. Choose $b = 2,$ which forces $a = 6.$

Work. Substitute into the target: $\frac{2(6) + 2}{6 − 2}$

Answer: C, $7/2.$

Check. More generally, replacing a by $3b$ gives $(6b + b)/(3b - b) = 7b/(2b) = 7/2,$ since $b > 0.$

Avoid the trap. Choosing a and b independently would create a different problem. A valid numerical instance must preserve every relationship, not just positivity.

2.09. Use more than one table row to distinguish models

A function has the following values. Which proposed model matches all three rows?
t
0
2
4
F(t)
80
100
125
A) 80(1.25)^t/2 B) 80(1.25)^t C) 80 + 10t D) 80(1.125)^t/2

t	0	2	4
F(t)	80	100	125

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Recognize. At $t = 0,$ all options give 80, so that row alone cannot distinguish them.

Work. At $t = 2,$ A and C give 100; B gives 125 and D gives 90. At $t = 4,$ A gives 125 while C gives 120. Only A matches the entire table.

Answer: A, $80(1.25) t/2 .$

Check. The outputs multiply by 1.25 for each two-unit increase in t, matching the exponent $t/2.$

Avoid the trap. Two matching rows do not establish which model fits a third. Use all supplied data before deciding that a pattern is linear.

2.10. Separate two variables in a difference quotient

For h ≠ 0, which expression is equivalent to $\frac{(x + {h)}^{2} − x^{2}}{h}$ ?
A) 2x + h B) 2x + h² C) x + h D) 2x

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Recognize. Choose different values for the two roles and avoid $h = 0.$ Let $x = 3$ and $h = 2.$

Work. The original value is $(5 2 - 3 2)/2 = 8.$ The choices give 8, 10, 5, and 6, leaving A.

Answer: A, $2x + h.$

Check. Expanding the numerator gives $2xh + h 2 = h(2x + h);$ dividing by nonzero h gives the result.

Avoid the trap. With $h = 1,$ choices A and B would coincide. A first test may be valid yet insufficient; the problem is lack of distinction, not lack of arithmetic skill.

2.11. Backsolve a geometric model instead of solving a cubic

A cylinder has radius r centimeters and height (r +4) centimeters. Its volume is 128π cubic centimeters. Which value is r? A) 3 B) 4 C) 6 D) 8

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Recognize. The model is $r 2 (r + 4) = 128$ with $r > 0.$ Testing the few radius choices is simpler than expanding and solving a cubic.

Work. For $r = 4,$ the height is 8 and $r 2 (r + 4) = 16(8) = 128.$ Thus B fits. Over positive r, both $r 2$ and $r + 4$ increase, so their positive product increases; no other positive radius gives the same volume.

Answer: B, 4 centimeters.

Check. Directly, $V = π(4) 2 (8) = 128π$ cubic centimeters.

Avoid the trap. The quantity $r + 4$ is the height, not the diameter. Backsolving is only reliable when the candidate is put into the correctly interpreted model.

2.12. Use a counterexample and then prove the survivor

For every real number x > 1, which expression must be greater than x? A) x² B) 2 − x C) 1/x D) $\sqrt{x}$

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Recognize. One allowed failure eliminates a universal claim. The input $x = 4$ sharply separates the options.

Work. At $x = 4,$ the four expressions are 16, −2, $1/4,$ and 2. Choices B, C, and D fail to exceed 4. For A, $x > 1$ implies $x(x - 1) > 0,$ so $x 2 - x > 0$ for every allowed x.

Answer: A, $x 2 .$

Check. The proof uses the condition $x > 1.$ Without it, A would not always exceed x; for instance, $x = 1/2$ fails.

Avoid the trap. Numerical success at $x = 4$ alone would not prove A universal. The counterexamples eliminate the others, and the inequality establishes the general result.

2.13. Reverse a percent comparison with a useful example

Amount A is r% greater than positive amount B, where r > 0. By what percent is B less than A? A) r B) $\frac{r}{100 + r}$
C) $\frac{100r}{100 + r}$ D) $\frac{100r}{100 − r}$

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Recognize. The reversed comparison uses A as its base. Choose $B = 100$ and $r = 25,$ so $A = 125.$

Work. The decrease from 125 to 100 is $25/125 = 20%.$ The choices give 25, 0.2, 20, and $100/3,$ respectively; only C is the requested percent.

Answer: C, $100r/(100 + r)$ percent.

Check. Generally, $A = (1 + r/100)B.$ Then $100(A - B)/A = 100r/(100 + r).$

Avoid the trap. Choice B is a fractional rate, not the numerical percent requested. A correct ratio still needs the correct response scale.

2.14. Reject a candidate before squaring introduces it

Which of the following is the solution set of $\sqrt{x + 5}$ = x − 1? A) {−1, 4} B) {4} C) {−1} D) {1}

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Recognize. The left side is nonnegative, so a solution must have $x \geq 1.$ That immediately excludes −1 as a valid root.

Work. Squaring yields $x + 5 = (x - 1) 2,$ or $(x - 4)(x + 1) = 0.$ Check the candidates in the original equation: at $x = 4,$ both sides equal 3; at $x = -1,$ the sides are 2 and −2.

Answer: B, {4}.

Check. At $x = 1,$ the left side is $\sqrt{6}$ and the right side is 0, so D fails as well.

Avoid the trap. A solution of the squared equation is only a candidate. Squaring has erased the sign distinction that rules out −1.

2.15. Choose the input that exposes the requested combination

For constants a and b, the identity
(x − 2)(ax + b) = 3x² − 5x − 2
holds for every real x. What is a + b?

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Recognize. The target is the combination $a + b,$ not the two constants separately. Setting $x = 1$ makes $ax + b$ equal the target while keeping the other factor nonzero.

Work. Substitute $x = 1$ into the identity:

$(1 - 2)(a + b) = 3 - 5 - 2 = -4.$

Thus $-(a + b) = -4,$ and $a + b = 4.$

Answer: 4.

Check. Coefficient comparison gives $a = 3$ from the quadratic term. The constant term gives $-2b = -2,$ so $b = 1.$ The linear coefficient is then $b - 2a = 1 - 6 = -5,$ confirming $a = 3,$ $b = 1,$ and $a + b = 4.$

Avoid the trap. Setting $x = 2$ produces $0 = 0,$ which is true but supplies no information about $a + b.$ Choosing a root is useful only when it removes unwanted terms without also removing the target.

Alternative. Expanding the left side gives $ax 2 + (b - 2a)x - 2b.$ Matching coefficients is a valid full solution, but it solves for more than the question requires.

Transfer. In an identity, look for an input that turns an expression into the requested combination: $x = 1$ often reveals a sum, $x = -1$ a signed difference, and $x = 0$ a constant. These are structural choices, not arbitrary numerical trials.

Testing versus proving

A carefully selected input can extract exact information from a known identity. That is different from observing that two unrelated formulas happen to agree at one input. The wording “for every real x” licenses the substitution here.

2.6 Section exit check

For each question, use substitution or backsolving when useful, and explain why the test is valid.

Topic practice

Review prompt

Were your values allowed? Did they separate the surviving choices? Did you prove a general claim, disprove it, or only find a candidate?