Shortcut or Trap? SAT shortcuts and when they fail

#1 · Graph it, then zoom out

The question. How many solutions does this system of equations have?

The shortcut. A solution to a system is a point where the two graphs meet: an intersection point. (A point where they touch without crossing still counts as one.) So instead of solving, graph both equations in the calculator built into Bluebook and count the intersection points. It's often the fastest way to answer a "how many solutions" question.

The fix

Don't count only the intersections in the first window. Zoom out and check farther along both graphs. If the view is still ambiguous, set the equations equal:

That confirms intersections at (0, 0) and (10, 20): two solutions.

Two related cases

Two question types from our own bank that look like this one, and how the shortcut changes for each:

• An unknown constant. "For what value of k does the system have exactly one solution?" There's nothing concrete to graph until you know k. You can add a slider and drag it until the graphs just touch (that's shortcut #11), or set the discriminant equal to zero.
• No solution or infinitely many solutions for two lines. When every number is known, graphing works here too: lines that look parallel mean no solution, and if you entered two equations but see only one line, they're the same line (infinitely many solutions; hide one to check). But "looks parallel" is this episode's trap again: lines with slopes 2 and 2.01 look parallel up close and still meet far off the screen, so zoom out or compare the slopes before you answer. When the question has an unknown constant, compare slopes and intercepts, or, in ax + by = c form, check whether the coefficients are in the same ratio.

The takeaway: if you graph a system, check beyond the first window before counting intersections.

#2 · Match the subject to the description

The question. Which choice completes the text so that it conforms to the conventions of Standard English?

The shortcut. The second sentence opens with a description: "Weighing up to 500 pounds." Whatever that phrase describes has to be the subject that comes right after the comma. So you don't need to read every word of every choice. Look at how each choice begins and ask: what weighs up to 500 pounds? Zookeepers don't, and plants don't, so A and D are out.

The fix: read as far as the subject

Read each choice as far as its subject, the noun the verb belongs to, and check that noun against the opening description. Here that leaves B: "Weighing up to 500 pounds, the gorilla eats leaves, stems and fruit throughout the day."

The takeaway: when a sentence opens with a description, match the subject to the description.

#3 · And, so, or but? Decide first

The question. Which choice completes the text with the most logical transition?

The shortcut. Don't read the choices yet. First decide how the two sentences connect, using three plain words:

• and: the second sentence adds more of the same.
• so: the second sentence is a result of the first.
• but: the second sentence is a surprise after the first.

These three cover most transition questions. When none fits, the second sentence is usually an example of the first ("for example").

Then peek at the choices. "Furthermore" and "Moreover" both mean "and." A question has one right answer, so two choices that mean the same thing can't both be right, and neither is. That crosses off A and B before you've weighed anything.

Now test the two that are left by saying the sentence out loud with each plain word:

• "It's the largest fish, so it eats tiny food?" No. Eating tiny food isn't a result of being big.
• "It's the largest fish, but it eats tiny food?" Yes. The second fact is a surprise after the first.

"However" is just a fancy "but," so the answer is D.

The fix: if you're not sure, keep both

Cross off two choices only when they mean exactly the same thing. If you're not sure, keep both and try each one in the sentence.

The takeaway: and, so, or but? Decide first.

#4 · Which way does it shift?

The question. What is the vertex of the graph of y = (x − 3)² + 5?

1. (−3, 5)
2. (3, 5)
3. (3, −5)
4. (5, 3)

The shortcut. Before you pick, ask which way it shifts. The original graph, y = x², has its vertex where x is zero: (0, 0). The new graph's vertex is where x − 3 is zero, and that happens when x = 3. The + 5 lifts the whole graph 5 up. So the vertex is (3, 5). The answer is B.

Check it on the graph

If you're not sure, graph y = x² and y = (x − 3)² + 5 in the calculator built into Bluebook and look where the new vertex sits. It's at (3, 5): 3 to the right and 5 up.

The same move elsewhere

• Any function: g(x) = f(x − 4) is the graph of f moved 4 to the right.
• Circles: (x − 2)² + (y + 1)² = 9 has its center at (2, −1).
• Factors: if (x − 5) is a factor, then 5 is a zero.

The takeaway: ask which way it shifts.

#5 · A root isn't always a solution

The question. What is the sum of the solutions to (3x² + 2)/(x − 2) = 7x/(x − 2)?

1. 1/3
2. 2/3
3. 2
4. 7/3

The algebra. Both sides have the same denominator, so you only need the numerators: 3x² + 2 = 7x, which is 3x² − 7x + 2 = 0. There are three ways to finish:

• Factor it: (3x − 1)(x − 2) = 0, so x = 1/3 or x = 2.
• Use the quadratic formula: x = (7 ± 5)/6, which gives the same two roots.
• Use the shortcut: the solutions of ax² + bx + c = 0 add up to −b/a, which here is 7/3.

All three give a sum of 7/3.

Or graph it from the start

Type the whole equation into the graphing calculator built into Bluebook. It draws a vertical line at each real solution. You'll see one line, just right of zero, and nothing at x = 2: the calculator never shows the root that doesn't work.

When the solution isn't a whole number

Here the line sits between gridlines, so you can't read it exactly. Graph the two sides separately, y = (3x² + 2)/(x − 2) and y = 7x/(x − 2), then click where they cross. The calculator shows the point (0.333333, −1.4), so x is 1/3.

One limit: the calculator rounds that point, so it can't turn 0.333333 back into 1/3 for you. Typing 0.333333 and switching it to a fraction gives 333333/1000000, so match the decimal to the answer choices yourself.

Where else this happens

Watch for it whenever x is in a denominator, like this question, or under a square root. In both cases the usual algebra can hand you a root that doesn't work in the original equation. So put each root back into the original equation, or graph it.

The takeaway: a root isn't always a solution. Check your roots, or graph the equation.

#6 · Find the subject first.

The question. Which choice completes the text so that it conforms to the conventions of Standard English?

The shortcut. Match the verb to the noun in front of the blank. It's the move your ear makes on its own, and when the subject sits right before the verb it's all you need. Here the noun in front of the blank is "fossils," so the ear says "are." (C and D are plural forms too, so they don't agree with the singular subject. The real choice is A or B.)

The fix: flip it

Put the subject first and read it straight into the verb: "A single complete skeleton ______ hidden among the thousands of fossils." Now your ear is right: a single complete skeleton is. (Khan Academy's advice, word for word: "try flipping the sentence.")

The same move with "there."

There ______ a dozen reasons to visit.

"There" isn't the subject; it's a placeholder that pushes the real subject after the verb. Ask what's there: a dozen reasons. Plural, so are, whatever the "a" in "a dozen" suggests.

When to reach for it. Flip when a sentence opens with a describing phrase ("Among…", "Beneath…"), starts with "there," or leads with a comparison ("less obvious is…"). If the choices differ only in number, find the subject before you trust the noun next to the blank.

The takeaway: find the subject first. When the verb comes before it, flip the sentence so the subject leads, then choose the verb.

#7 · Let the opener decide

The question. Which choice completes the text so that it conforms to the conventions of Standard English?

The quick move, and where it breaks. On a punctuation question, the quick move is to read the words on each side of the blank. Here that gives "blue blood, can change color," which sounds fine, so A looks right.

But this sentence has extra information dropped into the middle: an animal with three hearts and blue blood. Take it out and the sentence still works on its own: "The octopus can change its skin color in a moment." Extra information in the middle of a sentence gets a matching pair of marks, one to open it and one to close it. Here, a dash opened it, right after "octopus," so a dash has to close it. The answer is B.

The comma in A is the trap: it reads naturally next to the blank, but it would close with a comma what a dash opened. The semicolon in C would need a complete sentence on each side, and D leaves the dash with no partner.

The same rule, three ways

Change the opener, and the closing punctuation changes too:

• The octopus, an animal with three hearts and blue blood, can change its skin color in a moment.
• The octopus—an animal with three hearts and blue blood—can change its skin color in a moment.
• The octopus (an animal with three hearts and blue blood) can change its skin color in a moment.

All three are correct. What is never correct is a mismatch: a dash on one side and a comma on the other. So don't decide at the blank. Find the mark that opened the extra information and let it decide.

Two cases the video leaves out

• An aside at the end of a sentence has no closer. "The octopus has three hearts, a fact that surprises most people." The period does the closing. The rule is about extra information in the middle.
• A single dash can work like a colon. "The octopus has one hard part—its beak." One dash, no partner, correct. A student who learns "dashes come in pairs" as an absolute will second-guess this.

The takeaway: let the opener decide. When extra information sits in the middle of a sentence, the mark that closes it has to match the mark that opened it.

#8 · Test one value per unknown

The question. In the given equation, a, b, and c are constants. What is the value of a + b + c?

By hand. The two sides are the same expression written in different ways, so make the left side look like the right side.

Clear the fractions. The common denominator is (x − 1)(x + 3), which multiplies out to x² + 2x − 3: the right side's denominator. Multiply every term by it. Each fraction's own denominator cancels, which leaves two products on the left and only the top on the right.

Expand each product, then combine like terms.

Match it to the right side.

So a = 2, b = −7, and c = −10.

That makes a + b + c = 2 − 7 − 10 = −15.

The Desmos shortcut: a regression

The graphing calculator built into Bluebook is Desmos, and Desmos can find a, b, and c for you. Type the equation as a regression: a tilde (~) in place of the equals sign, and x₁ in place of x. (Plain x won't work: Desmos says x may not be used as a regression parameter.) On the next line, type a + b + c to see the sum.

Two more lists to avoid

• Avoid pairs that add to zero, like 2 and −2. x₁ = [−2, 0, 2] looks like three tests, but Desmos gives −9.75. Here a and b sit on x³ and x, so −2 gives the same numbers as 2 with the signs flipped, and Desmos learns nothing new from it. The safe habit: use values like 0, 2, and 3.
• Keep the values small. Huge values like a million push Desmos's rounding too far: [1000000, 2000000, 3000000] gives about −5. Small whole numbers are safest.

This question type is on the test: the College Board lists equivalent expressions under Advanced Math, and Desmos courses such as Test Prep Wizards teach the regression for it. The divide-by-zero check is our own addition.

The takeaway: make both sides look alike. In Desmos, test one value per unknown, and skip any value that makes a denominator zero.

#9 · Find the contrast first

The question. Which choice completes the text so that it conforms to the conventions of Standard English?

The usual pattern. With words like however, therefore, and instead between two independent clauses (each could stand on its own as a sentence), the usual pattern is a semicolon before the word and a comma after it: "It rained all day; however, the game went on." However signals a contrast, like but, and here the contrast is between the two clauses: it rained, but the game went on. So the semicolon goes before however.

The basic case: a comma alone isn't enough

Think but for the meaning, but don't copy its punctuation:

• "It rained all day, but the game went on." Correct. But is a joining word (so are and, or, so, and yet), so a comma is enough.
• "It rained all day, however, the game went on." Incorrect: two independent clauses joined by commas alone.
• "It rained all day; however, the game went on." Correct, and so is a period: "It rained all day. However, the game went on."

Inside one clause, and the other transition words

• In the middle of one clause, commas are right. "The reviews, however, were harsh." Here there is only one clause, so there is nothing for a semicolon to separate.
• The same check works for the other transition words. Therefore, instead, as a result, and in fact follow the same pattern. Before you place the semicolon, ask which clause the word belongs to.

This question type is on the test: the College Board's framework has students join clauses "using appropriate punctuation and, in some cases, a conjunction or conjunctive adverb," and its own question bank includes a hard question built exactly this way, with the semicolon after however. Prep guides such as UWorld teach the semicolon-before pattern.

The takeaway: find the contrast first. Decide which clause however belongs to, then place the semicolon between the two independent clauses.

#10 · Quadratic table? Look for twins.

The question. The table shows three values of the quadratic function f. What is the value of f(4)?

The slow way. Every quadratic can be written f(x) = ax² + bx + c. Plug in each row of the table and you get three equations:

Solve them, then plug in 4. It works, but there's a faster way.

Look for twins. Two inputs with the same output are twins. Here, 8 and 20 both give −12. A parabola is symmetric, so they sit equally far from its axis of symmetry, which is halfway between them: (8 + 20) ÷ 2 = 14.

Find the other twins. 4 is 10 to the left of 14, and 24 is 10 to the right, so 4 and 24 mirror each other too. That means f(4) = f(24), and the table already gives f(24) = 52.

So f(4) = 52. You never needed a, b, or c.

Or use Desmos

Enter the table (its columns are x₁ and y₁), then a regression on a new line: y₁ ~ ax₁² + bx₁ + c. Desmos finds a = 1, b = −28, c = 148. Then type a·4² + b·4 + c and Desmos shows 52. It works, but here spotting the twins is faster.

Go further: the same table, different questions

• What is the vertex's x-coordinate? Average the twins: (8 + 20) ÷ 2 = 14.
• What is the minimum value? Now you need a, which sets how stretched the parabola is. The twins give f(x) = a(x − 8)(x − 20) − 12 (slide the graph up 12 and the twins become zeros). The third row finds a: 64a − 12 = 52, so a = 1. Since a is positive the parabola opens upward, and f(14) = (6)(−6) − 12 = −48.

The halfway idea is a standard fact about parabolas: UWorld notes that the axis of symmetry is halfway between any two inputs with the same output, and Albert's quadratics review says the same.

The takeaway: quadratic table? Look for twins, two inputs with the same output. Average them to find the symmetry line, then look for the input the same distance away on the other side. The answer may already be in the table.

#11 · sin = cos? Angles add up to 90°.

The question. sin(x) = cos(3x − 10), where the angles are in degrees and both are acute. What is the value of x?

The fact. For an acute angle A, sin A = cos(90° − A): the sine of an angle equals the cosine of its complement. (In a right triangle, the two acute angles are complements: they add up to 90°.) So when the sine of one acute angle equals the cosine of another, the two angles add up to 90°.

Use it. The angles are x and 3x − 10, so x + (3x − 10) = 90. That gives 4x − 10 = 90, so x = 25.

Check it. The two angles are 25° and 3(25) − 10 = 65°, and 25 + 65 = 90. Sure enough, sin 25° = cos 65° ≈ 0.4226.

Desmos works too: switch to degrees

Switch Desmos to degrees first: it starts in radians, and in radians the curves don't cross at 25. Tap the wrench, then Degrees. Graph y = sin(x) and y = cos(3x − 10), then click where the curves cross: (25, 0.42262). Trig graphs cross again and again, so pick the crossing where both angles, x and 3x − 10, are between 0° and 90°. Here that's x = 25. It works, but if you know the fact, it's one line of math.

The takeaway: sin A = cos B, with A and B acute? Then A + B = 90°.