Chapter 1 — Linear Equations and Inequalities

Answer: A

Explanation

Two things were done to $x\mathclose{:}$ it was multiplied by 3, then 5 was added. Undo them in the opposite order.

Subtract 5 from both sides:

$$3x + 5 - 5 = 20 - 5 \quad\Longrightarrow\quad 3x = 15$$

Now divide both sides by 3:

$$\frac{3x}{3} = \frac{15}{3} \quad\Longrightarrow\quad x = 5$$

Check it, which takes three seconds: $3(5) + 5 = 15 + 5 = 20\mathclose{.}$ ✓

Why each wrong option is wrong:

• B, $\frac{20}{3}$ — divided by 3 first and ignored the $+5$ entirely. This is the "wrong order" mistake: you cannot divide by 3 until the $+5$ is gone, because the 3 is multiplying only the $x\mathclose{,}$ not the whole left-hand side.
• C, $\frac{25}{3}$ — added 5 to both sides instead of subtracting, giving $3x = 25\mathclose{.}$ Undoing means doing the opposite.
• D, $15$ — subtracted the 5 correctly to get $3x = 15\mathclose{,}$ then stopped and wrote down 15. It is the value of $3x\mathclose{,}$ not of $x\mathclose{.}$ Finishing the last step is worth more marks on this test than any clever trick.

Takeaway: Undo in reverse order — additions and subtractions first, multiplication and division last. Then check by substituting back; on a multiple-choice test a three-second check is nearly free.

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Answer: C

Explanation

The two charges are the same when $7(x - 2) = 3x + 6\mathclose{.}$

Clear the bracket first. The 7 multiplies everything inside:

$$7(x - 2) = 7x - 14$$

So the equation is $7x - 14 = 3x + 6\mathclose{.}$ Now gather the $x$ terms on one side and the numbers on the other. Subtract $3x$ from both sides:

$$4x - 14 = 6$$

Add 14 to both sides:

$$4x = 20$$

Divide by 4:

$$x = 5$$

Check: $7(5 - 2) = 7 \times 3 = 21\mathclose{,}$ and $3(5) + 6 = 21\mathclose{.}$ ✓

Why each wrong option is wrong:

• A, $-2$ — moved the $-14$ across as a $-14$ instead of a $+14\mathclose{.}$ When a term crosses the equals sign its sign flips.
• B, $2$ — distributed the 7 to the $x$ but not to the $-2\mathclose{,}$ giving $7x - 2 = 3x + 6\mathclose{.}$ This is the single most common algebra slip on the test. The 7 sits outside the bracket, so it multiplies both terms inside.
• D, $20$ — did everything right down to $4x = 20$ and then wrote 20. The same unfinished-last-step mistake that Q1 offers, and it is deliberately offered again because it is that common.

Takeaway: The number outside a bracket multiplies every term inside it. Write the expanded form on your paper rather than doing it in your head — this one slip accounts for more lost algebra marks than any other.

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Answer: A

Explanation

A third of the rod is $\frac{x}{3}$ and a quarter of it is $\frac{x}{4}\mathclose{.}$ Together they measure 7, so

$$\frac{x}{3} + \frac{x}{4} = 7$$

Add the two fractions. The lowest common denominator of 3 and 4 is 12:

$$\frac{x}{3} + \frac{x}{4} = \frac{4x}{12} + \frac{3x}{12} = \frac{7x}{12}$$

So $\dfrac{7x}{12} = 7\mathclose{.}$ Multiply both sides by 12:

$$7x = 84$$

Divide by 7:

$$x = 12$$

Check: $\frac{12}{3} + \frac{12}{4} = 4 + 3 = 7\mathclose{.}$ ✓

There is a faster route worth knowing: multiply every term by 12 at the very start, and the fractions vanish before you ever add them — $4x + 3x = 84\mathclose{,}$ so $7x = 84$ and $x = 12\mathclose{.}$

Why each wrong option is wrong:

• B, $\frac{49}{2}$ — added numerators and denominators, giving $\frac{2x}{7} = 7\mathclose{.}$ This is the same mistake as A wearing a different hat.
• C, $49$ — added the denominators, treating the sum as $\frac{x}{7} = 7\mathclose{.}$ Denominators are never added. $\frac12 + \frac12$ is 1, not $\frac24\mathclose{.}$
• D, $84$ — reached $7x = 84$ correctly, then stopped. The unfinished last step again.

Takeaway: Multiply every term by the common denominator immediately. It turns a fraction problem into an ordinary one, and it is the step a calculator cannot do for you.

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Answer: A

Explanation

Expand both sides first:

$$5(2x - 3) = 10x - 15 \qquad 3(x + 4) + 7 = 3x + 12 + 7 = 3x + 19$$

So $10x - 15 = 3x + 19\mathclose{.}$ Subtract $3x\mathclose{:}$

$$7x - 15 = 19$$

Add 15:

$$7x = 34$$

$$x = \frac{34}{7}$$

An ugly fraction is not a sign you went wrong. The SAT allows a calculator and often accepts fractions, so do not "fix" a correct answer into a rounder one.

Why each wrong option is wrong:

• B, $\frac{20}{7}$ — subtracted the 7 instead of adding it, treating $3(x+4) + 7$ as $3x + 12 - 7\mathclose{.}$ Read the sign in front of a term, not the sign you expect.
• C, $2$ — distributed each bracket to its first term only: $10x - 3 = 3x + 4 + 7\mathclose{,}$ which solves to $7x = 14\mathclose{.}$ Two distribution slips in one line, and they land on a tidy whole number — which is exactly why a tidy answer is not evidence of a correct one.
• D, $\frac{4}{7}$ — moved the $-15$ across as a $-15$ instead of a $+15\mathclose{,}$ giving $10x = 3x + 19 - 15$ and so $7x = 4\mathclose{.}$ When a term crosses the equals sign, its sign flips.

Takeaway: Expand every bracket before moving anything. And do not distrust an answer for being an awkward fraction — on this test, roughly a quarter of answers are typed in by hand, and fractions are accepted.

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Section 2 — No Solution, or Every Solution

Some SAT questions do not want a number. They give you an equation with an unknown coefficient and tell you it has no solution or infinitely many solutions, and ask what that coefficient must be. These look strange the first time and are formulaic once you see the idea.

Strip both sides down to the form $(\text{number})x + (\text{number})\mathclose{.}$ Then:

That table is the entire topic. The reasoning behind it: if the $x$ terms are identical they cancel, and you are left with a statement about numbers only — either a true one $(15 = 15\mathclose{,}$ always true) or a false one $(15 = 19\mathclose{,}$ never true).

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Answer: C

Explanation

Expand the left-hand side:

$$3(2x + 5) = 6x + 15$$

So the equation reads $6x + 15 = ax + 15\mathclose{.}$

For this to be true for every value of $x\mathclose{,}$ the two sides must be the same expression. The constants already match $(15$ on both sides), so the $x$ coefficients must match too:

$$a = 6$$

Sanity check with a number: put $a = 6$ and try $x = 100\mathclose{.}$ Left: $3(200 + 5) = 615\mathclose{.}$ Right: $600 + 15 = 615\mathclose{.}$ Equal — and it would be equal for any $x$ you tried, which is what "infinitely many solutions" means.

Why each wrong option is wrong:

• A, $2$ — read the coefficient of $x$ inside the bracket and stopped there. The bracket has not been expanded yet; the 3 still has to get in.
• B, $3$ — took the number outside the bracket instead of the product of the two.
• D, $15$ — matched the constant term rather than the $x$ coefficient. The constants do have to match, but they already do; $a$ multiplies $x\mathclose{.}$

Takeaway: "Infinitely many solutions" means the two sides are the same expression. Expand, then match the $x$ coefficients and the constants separately.

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Answer: 4

Explanation

Both sides are already in the form $(\text{number})x + (\text{number})\mathclose{:}$

$$4x + 9 \qquad\text{and}\qquad kx - 3$$

"No solution" means there is no value of $x$ that makes the two sides equal. That happens when the $x$ terms are identical — so they cancel — but the constants are not:

$$k = 4$$

Then the equation becomes $4x + 9 = 4x - 3\mathclose{.}$ Subtract $4x$ from both sides and you are left with

$$9 = -3$$

which is false, no matter what $x$ is. So there is no solution. ✓

Note the contrast with Q5. Same setup, opposite conclusion, and the only difference is whether the constants match. Here they cannot: $9 \neq -3$ regardless of $k\mathclose{,}$ so matching the coefficients guarantees no solution.

This is a student-produced response — one of the questions with no options to choose from, which make up about a quarter of the Math section. You work out the answer and type it in. Note what that removes: you cannot test the four choices, and a blank scores the same as a wrong guess, so always enter something.

Takeaway: No solution ⇒ same $x$ coefficient, different constants. Infinitely many ⇒ same coefficient and same constant. Set the coefficients equal and then check what the constants are doing.

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Section 3 — Turning Words into an Equation

Some questions never ask you to solve anything. They describe a situation and ask which equation represents it. The whole question is the translation.

Translate one phrase at a time, left to right, and be careful with two traps the test uses repeatedly:

• "7 more than 4 times a number" is $4x + 7\mathclose{,}$ not $4(x + 7)\mathclose{.}$ The "4 times" happens first; the 7 is added to the result.
• "5 less than $x$" is $x - 5\mathclose{,}$ not $5 - x\mathclose{.}$ The order reverses when you write it down.

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Answer: B

Explanation

Take the sentence in pieces:

So the answer is $2n - 9 = 35\mathclose{.}$ (If you want to check it, solving gives $2n = 44\mathclose{,}$ so $n = 22$ — and twice 22, less 9, really is 35.)

Why each wrong option is wrong:

• A, $2 \left(n - 9\right) = 35$ — this takes 9 off the number first and doubles afterwards. The sentence doubles first: the 9 comes off "the result". A bracket changes which operation happens first, so it is only right when the sentence says so.
• C, $9 - 2 n = 35$ — the subtraction is the wrong way round. "Reduced by 9" takes 9 away from $2n\mathclose{;}$ it does not take $2n$ away from 9.
• D, $\frac{n}{2} - 9 = 35$ — "doubled" means multiplied by 2. This halves the number instead.

Takeaway: Translate phrase by phrase and write each piece down before joining them. "More than" adds; "less than" subtracts and reverses the order; and a bracket belongs only where the sentence does the adding or subtracting first.

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Answer: D

Explanation

Sort the two numbers by what they do:

• $45 is fixed — it is charged once, whatever happens. It is not multiplied by anything.
• $28 is per hour — it is multiplied by the number of hours, $h\mathclose{.}$

Total charge = hourly part + fixed part:

$$28h + 45 = 185$$

(Solving is not required here, but it gives $28h = 140\mathclose{,}$ so $h = 5$ hours.)

Why each wrong option is wrong:

• B, $45 h + 28 = 185$ — swapped the roles. This charges $45 an hour and adds a $28 fee. The words "for each hour worked" attach to 28, not 45.
• C, $28 \left(h + 45\right) = 185$ — puts the fixed fee inside the bracket, so it gets multiplied by 28 as well. A fixed fee is added once, at the end.
• A, $28 h - 45 = 185$ — subtracts the call-out fee. The technician is charging it, not discounting it.

Takeaway: In any "fixed amount plus a rate" question, find which number is multiplied by the variable and which stands alone. The word "per" or "each" marks the one that gets multiplied.

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Answer: B

Explanation

In a linear model $b + mt\mathclose{,}$ the two numbers have fixed jobs:

• $b\mathclose{,}$ the constant — the value when $t = 0\mathclose{,}$ the starting amount.
• $m\mathclose{,}$ the number multiplied by $t$ — how much the quantity changes per unit of $t\mathclose{,}$ here per hour.

In $30 - 2.5t$ the number multiplied by $t$ is $-2.5\mathclose{.}$ The minus sign says the height is going down; the 2.5 says by how much each hour.

Test it. After 1 hour: $h(1) = 30 - 2.5 = 27.5\mathclose{.}$ After 2 hours: $h(2) = 30 - 5 = 25\mathclose{.}$ The candle is 2.5 centimetres shorter each hour. That is what the 2.5 means.

Why each wrong option is wrong:

• A — that is the 30, the value at $t = 0\mathclose{,}$ before anything has burned.
• C — the candle is gone when $30 - 2.5t = 0\mathclose{,}$ which is at $t = 12$ hours. The 2.5 is not a time at all.
• D — after 2.5 hours the height is $30 - 2.5(2.5) = 23.75$ centimetres. The 2.5 is a rate, in centimetres per hour, not a height. Watch the units: a rate and a height cannot both be right.

Takeaway: In $b + mt\mathclose{,}$ $b$ is where it starts and $m$ is how fast it changes — and a minus sign in front of $m$ means it is falling. When two options are close, check the units.

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Section 4 — Linear Inequalities

An inequality is solved exactly like an equation, with one extra rule:

When you multiply or divide both sides by a negative number, the inequality sign flips.

$\le$ becomes $\ge\mathclose{,}$ $<$ becomes $>\mathclose{.}$ Nothing else changes. Adding or subtracting a negative number does not flip anything — only multiplying or dividing by one does.

The reason is worth seeing once. Start with something true: $2 < 6\mathclose{.}$ Multiply both sides by $-1\mathclose{:}$ on the number line $-2$ sits to the right of $-6\mathclose{,}$ so $-2 > -6\mathclose{.}$ The order reversed.

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Answer: A

Explanation

Subtract 7 from both sides. Subtraction never flips the sign:

$$-4x \ge 12$$

Now divide both sides by $-4\mathclose{.}$ This is a division by a negative number, so the sign flips from $\ge$ to $\le\mathclose{:}$

$$x \le -3$$

Check with a number that should work: $x = -10$ satisfies $x \le -3\mathclose{.}$ Put it in the original: $-4(-10) + 7 = 40 + 7 = 47\mathclose{,}$ and $47 \ge 19\mathclose{.}$ ✓

Now check one that should fail: $x = 0\mathclose{.}$ Then $-4(0) + 7 = 7\mathclose{,}$ and $7 \ge 19$ is false. ✓ The boundary is in the right place and the inequality points the right way.

Why each wrong option is wrong:

• B, $x \geq -3$ — the arithmetic is right and the flip was forgotten. This is the mistake this topic tests, and it is why the option is offered.
• C, $x \leq 3$ — flipped the sign but dropped the minus on the 3. Both the sign of the inequality and the sign of the number come from that division.
• D, $x \geq 3$ — neither the flip nor the minus sign. Substituting $x = 4$ gives $-9\mathclose{,}$ which is not $\ge 19\mathclose{,}$ so this fails the check immediately.

Takeaway: Divide by a negative, flip the sign. Then test one number from your answer in the original inequality — it catches this mistake every time, and it takes five seconds.

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Answer: D

Explanation

Work out what is in the lift, then say what the limit does to it.

The load is the crates plus the operator, who rides up with them:

$$45n + 80$$

"A maximum load of 600" means the load must be 600 or less:

$$45n + 80 \le 600$$

That is the answer. (Solving it — not asked for here — gives $45n \le 520\mathclose{,}$ so $n \le 11.6\mathclose{,}$ meaning 11 crates. You cannot take 0.6 of a crate, so the answer to that question would round down, never up, no matter what the decimal is.)

Why each wrong option is wrong:

• A, $45 n \leq 600$ — forgot that the operator goes up with the crates. The sign is right and the crate weight is right, but the person standing in the lift counts towards the load. This option allows 13 crates, and 13 crates plus an operator is 665 kg.
• B, $45 n + 80 \geq 600$ — the correct left-hand side with the sign reversed. "Maximum" means at most, so the sign is $\le\mathclose{.}$
• C, $45 n - 80 \leq 600$ — subtracts the operator, as though the person reduces the load. They are standing in the lift; they add to it.

Takeaway: "Maximum", "at most", "no more than" all mean $\le\mathclose{.}$ "Minimum", "at least", "no fewer than" all mean $\ge\mathclose{.}$ Write down which one the sentence uses before you write the inequality — and when a count of physical objects comes out as a decimal, round down.

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Answer: 5

Explanation

Gather the $x$ terms on the side that keeps them positive — it avoids the sign flip altogether. Add $2x$ to both sides:

$$9 < 6x - 15$$

Add 15 to both sides:

$$24 < 6x$$

Divide by 6. Six is positive, so nothing flips:

$$4 < x$$

So $x$ is strictly greater than 4. The question asks for the least integer, and the inequality is strict — $x = 4$ does not satisfy it, because $4 < 4$ is false. The least integer greater than 4 is

$$x = 5$$

Check both sides of the boundary. At $x = 5\mathclose{:}$ left is $9 - 10 = -1\mathclose{,}$ right is $20 - 15 = 5\mathclose{,}$ and $-1 < 5\mathclose{.}$ ✓ At $x = 4\mathclose{:}$ left is $1\mathclose{,}$ right is $1\mathclose{,}$ and $1 < 1$ is false. ✓ So 5 is the first integer that works.

Two things this question is testing at once, and you need both:

1. Solving the inequality without a sign error.
2. Noticing that strict $(<)$ excludes the boundary, so the answer is the next integer up, not 4.

Takeaway: Move the variable to the side that keeps its coefficient positive and you never have to flip anything. Then read the sign again: $<$ excludes the boundary value, $\le$ includes it — and that difference is usually the whole question.

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Answer: A

Explanation

Expand the bracket. The $-3$ multiplies both terms, and $-3 \times -4 = +12\mathclose{:}$

$$-3x + 12 > 2x + 1$$

That plus sign is the whole question. Now move the $x$ terms to the side that keeps them positive — add $3x$ to both sides:

$$12 > 5x + 1$$

Subtract 1:

$$11 > 5x$$

Divide by 5, which is positive, so nothing flips:

$$\frac{11}{5} > x \qquad\text{that is}\qquad x < \frac{11}{5}$$

Check a value that should work, $x = 0\mathclose{:}$ left is $-3(-4) = 12\mathclose{,}$ right is $1\mathclose{,}$ and $12 > 1\mathclose{.}$ ✓ Check one that should not, $x = 3\mathclose{:}$ left is $-3(-1) = 3\mathclose{,}$ right is $7\mathclose{,}$ and $3 > 7$ is false. ✓

Why each wrong option is wrong:

• B, $x > \frac{11}{5}$ — right boundary, wrong direction. This comes from gathering the $x$ terms on the left instead, getting $-5x > -11\mathclose{,}$ and then dividing by $-5$ without flipping.
• C, $x < - \frac{13}{5}$ — distributed $-3(x-4)$ as $-3x - 12\mathclose{.}$ Multiplying two negatives gives a positive, so the $12$ is positive.
• D, $x > - \frac{13}{5}$ — both mistakes at once. Substituting $x = 0\mathclose{,}$ which satisfies this option, gives $12 > 1$ — true, so this option is not obviously wrong from one test. Test the boundary, not a convenient number.

Takeaway: $-3 \times -4 = +12\mathclose{.}$ When a negative multiplies a bracket, write the expansion down rather than doing it in your head, then move the variable to whichever side keeps it positive.

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Section 5 — Rearranging, and Reading a Model

Two things get asked constantly and neither requires you to find a number.

Rearranging — "solve for $x$ in terms of $y$" means get $x$ alone on one side. The other letters just come along for the ride; treat them exactly like numbers.

Reading a model — you are given a formula describing a situation and asked what one of its numbers means. The answer is always about the units.

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Answer: D

Explanation

Get the $x$ term on its own. The $-3y$ moves across and becomes $+3y\mathclose{:}$

$$4x = 3y + 12$$

Now divide everything on the right by 4:

$$x = \frac{3y + 12}{4}$$

Check it by picking a value. Let $y = 4\mathclose{.}$ The original says $4x - 12 = 12\mathclose{,}$ so $4x = 24$ and $x = 6\mathclose{.}$ The answer gives $\frac{12 + 12}{4} = 6\mathclose{.}$ ✓

That substitution trick is worth having: it turns a question in letters into a question in numbers, and it works on every option.

Why each wrong option is wrong:

• A, $\frac{3 y - 12}{4}$ — moved the $12$ across as if it needed a sign change. It was already on the right-hand side; nothing moved, so nothing flips.
• B, $\frac{12 - 3 y}{4}$ — moved the $-3y$ across but kept it negative. A term that crosses the equals sign changes sign.
• C, $3 y + 12$ — divided nothing by 4. Test it with $y = 4\mathclose{:}$ it gives 24, not 6.

Takeaway: Treat the other letters exactly as you would treat numbers. Then check by substituting one convenient value into both the original and your answer — it catches every sign slip.

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Answer: C

Explanation

"Consecutive" means each is one more than the last. Call the smallest $n\mathclose{;}$ then the three are $n\mathclose{,}$ $n+1$ and $n+2\mathclose{.}$ Their sum:

$$n + (n + 1) + (n + 2) = 3n + 3$$

Set that equal to 84:

$$3n + 3 = 84 \quad\Longrightarrow\quad 3n = 81 \quad\Longrightarrow\quad n = 27$$

So the three integers are $27, 28, 29\mathclose{,}$ and the largest is $29\mathclose{.}$

Check: $27 + 28 + 29 = 84\mathclose{.}$ ✓

Why each wrong option is wrong:

• A, $27$ — the smallest, not the largest. Everything was solved correctly and the last line of the question was not read.
• B, $28$ — the middle one. Worth noticing that the middle of three consecutive integers is always the sum divided by 3, which is a genuinely useful shortcut: $84 \div 3 = 28\mathclose{,}$ so the three are $27, 28, 29$ immediately.
• D, $42$ — halved 84. There are three numbers, not two.

Takeaway: With three consecutive integers, the middle one is the sum divided by 3 — that gets you there in one step. Then read the last line again and check which of the three the question wanted.

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Answer: B

Explanation

Sort the two numbers by the job they do:

• 400 is the amount at the start, before any time passes. It stands alone.
• 15 is a rate — litres per minute — so it is multiplied by $t\mathclose{.}$

The water is being removed, so the 15 per minute is subtracted:

$$V = 400 - 15t$$

Check at a couple of times. At $t = 0\mathclose{:}$ $V = 400\mathclose{,}$ the tank is full. ✓ At $t = 10\mathclose{:}$ $V = 400 - 150 = 250$ litres. ✓

Why each wrong option is wrong:

• A, $15 t + 400$ — this fills the tank rather than draining it. After ten minutes it claims 550 litres, in a tank that holds 400.
• C, $15 t - 400$ — the right two numbers, subtracted the wrong way round. At $t = 0$ it gives $-400$ litres, which is not a quantity of water.
• D, $15 - 400 t$ — swapped the roles, so it starts with 15 litres and drains 400 a minute. Checking $t = 0$ eliminates this in two seconds.

Takeaway: In any "starts at, changes by" model, the standalone number is the starting value and the number multiplied by the variable is the rate. Then substitute $t = 0\mathclose{:}$ it instantly kills every option with the two swapped.

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Answer: 260

Explanation

Let $m$ be the number of messages. The bill is the fixed part plus the message part:

$$18 + 0.06m = 33.60$$

Subtract the fixed charge:

$$0.06m = 15.60$$

Divide:

$$m = \frac{15.60}{0.06} = 260$$

Check: $260 \times 0.06 = 15.60\mathclose{,}$ and $15.60 + 18 = 33.60\mathclose{.}$ ✓

A word about the division, since a calculator is allowed and people still get this one wrong. $15.60 \div 0.06$ is not $2.6\mathclose{.}$ Multiply both numbers by 100 first — $1560 \div 6 = 260$ — and the decimal stops being frightening. If your answer to a "how many messages" question comes out as 2.6, the size of it is telling you something is wrong.

This is a grid-in. Enter 260. Not 260 messages, not $260.

Takeaway: Subtract the fixed charge before dividing, and sanity-check the size of your answer against the situation. A phone bill of $33.60 buying 2.6 text messages should stop you immediately.

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Section 6 — Absolute Value

$|x|$ means how far $x$ is from zero, ignoring direction. So $|5| = 5$ and $|-5| = 5\mathclose{.}$ Distance is never negative.

That one idea drives everything here. $|x - 3| = 7$ says "$x$ is 7 away from 3", and there are always two places 7 away from 3: one above and one below. This is why absolute-value equations usually have two answers, and why forgetting the second one is the mistake the test is looking for.

The method. To solve $|\text{something}| = k$ where $k$ is positive, write two equations and solve each:

$$\text{something} = k \qquad\text{and}\qquad \text{something} = -k$$

Two warnings worth having in advance:

• $|\text{something}| = k$ has no solution when $k$ is negative. Nothing is $-4$ units away from anywhere.
• Isolate the bars first. In $2|x - 1| + 3 = 11\mathclose{,}$ subtract the 3 and divide by the 2 before splitting into two cases.

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Answer: B

Explanation

$|x - 3| = 7$ says $x$ sits 7 away from 3. Split it into the two cases:

$$x - 3 = 7 \quad\Longrightarrow\quad x = 10$$ $$x - 3 = -7 \quad\Longrightarrow\quad x = -4$$

Check both. $|10 - 3| = |7| = 7$ ✓. $|-4 - 3| = |-7| = 7$ ✓. Both work, which is the point — there are genuinely two answers.

The question asks for their sum:

$$10 + (-4) = 6$$

Why each wrong option is wrong:

• A, $-4$ — solved only the second case. The mirror image of C.
• C, $10$ — solved only the first case and stopped. This is the mistake the whole question exists to catch: the bars always give you two cases.
• D, $14$ — treated the answers as $+7$ and $-7$ and added the sizes. The 7 is the distance, not either answer; you still have to undo the $-3\mathclose{.}$

There is a shortcut worth knowing. The two solutions of $|x - a| = k$ are always $a + k$ and $a - k\mathclose{,}$ so they are symmetric about $a$ — and their sum is always $2a\mathclose{.}$ Here $2 \times 3 = 6\mathclose{,}$ in one step and with no cases at all.

Takeaway: Absolute value gives two cases, always. And the two solutions of $|x - a| = k$ sum to $2a$ no matter what $k$ is, which answers a whole family of questions instantly.

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Answer: B

Explanation

Split into two cases. The bars come off, and the right-hand side takes each sign in turn:

$$2x + 1 = 9 \quad\Longrightarrow\quad 2x = 8 \quad\Longrightarrow\quad x = 4$$ $$2x + 1 = -9 \quad\Longrightarrow\quad 2x = -10 \quad\Longrightarrow\quad x = -5$$

Check the second one, since it is the one being asked for: $|2(-5) + 1| = |-10 + 1| = |-9| = 9$ ✓

The two solutions are $4$ and $-5\mathclose{,}$ and the least is $-5\mathclose{.}$

Why each wrong option is wrong:

• A, $-10$ — got to $2x = -10$ and wrote down $-10\mathclose{.}$ The unfinished last step.
• C, $-4$ — in the second case, moved the $+1$ across without changing its sign: $2x = -9 + 1$ instead of $-9 - 1\mathclose{.}$
• D, $4$ — the greatest solution. Both cases were solved correctly and the last line of the question was not read. On a test this well-behaved, that is the most expensive habit there is.

Takeaway: Solve both cases, then go back and read whether the question wants the greatest, the least, the sum, or the product. The maths is usually the easy half.

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Answer: D

Explanation

The bars are wrapped in other operations, so undo those first. Do not split into cases while there is still a $\times 2$ and a $+3$ attached.

Subtract 3:

$$2|x - 1| = 8$$

Divide by 2:

$$|x - 1| = 4$$

Now split:

$$x - 1 = 4 \quad\Longrightarrow\quad x = 5$$ $$x - 1 = -4 \quad\Longrightarrow\quad x = -3$$

The greatest is $x = 5\mathclose{.}$ Check: $2|5 - 1| + 3 = 2(4) + 3 = 11$ ✓

Why each wrong option is wrong:

• A, $-3$ — the least possible value. $|x - 1| = 4$ gives $x = 5$ or $x = -3\mathclose{,}$ and the question asks for the greatest.
• B, $\frac{7}{2}$ — divided by 2 before subtracting the 3, so the 3 got halved too, giving $|x - 1| + 3 = \frac{11}{2}\mathclose{.}$ Undo additions before multiplications, exactly as in Section 1.
• C, $4$ — reached $|x - 1| = 4$ and stopped. That is the distance from 1, not the value of $x\mathclose{;}$ the $-1$ still has to be undone.

Takeaway: Get the absolute value alone on one side before you split into cases. Everything outside the bars is undone first, in the usual reverse order.

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Answer: A

Explanation

Everything turns on one fact: an absolute value is never negative. $|4x - 7|$ is a distance, so whatever $x$ is, the left-hand side is zero or more.

So the equation has no solution exactly when the right-hand side is negative:

$$k - 2 < 0 \quad\Longrightarrow\quad k < 2$$

Of the options, only $k = 1$ satisfies that. With $k = 1$ the equation reads $|4x - 7| = -1\mathclose{,}$ and nothing is $-1$ units away from anything.

It is worth seeing the full picture, because the test asks about all three cases:

Why each wrong option is wrong:

• B, $2$ — gives $|4x - 7| = 0\mathclose{,}$ which has exactly one solution, $x = \frac{7}{4}\mathclose{.}$ One is not none, and this is the option that catches people who half-remember the rule.
• C, $5$ — gives $|4x - 7| = 3\mathclose{:}$ two solutions.
• D, $9$ — gives $|4x - 7| = 7\mathclose{:}$ two solutions.

Takeaway: An absolute value is never negative. Negative right-hand side means no solution; zero means exactly one; positive means two. The zero case is the one the test uses to separate people who know the rule from people who nearly do.

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Section 7 — Ranges and Compound Inequalities

Two inequalities at once describe a range. $2 < x < 8$ says $x$ is bigger than 2 and smaller than 8.

The method. Whatever you do, do it to all three parts:

$$1 < x + 4 < 9 \quad\Longrightarrow\quad -3 < x < 5$$

(subtract 4 from each part). And the flip rule still applies: multiplying or dividing all three parts by a negative reverses both signs.

Absolute value gives ranges too, and there are exactly two shapes:

The first is "less than, and it's between". The second is "greater than, and it's outside". Mixing those two up is most of what goes wrong here.

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Answer: B

Explanation

Read the picture first. The heavy line runs from $-3$ to $3\mathclose{.}$ The circle at $-3$ is open, so $-3$ is not a solution; the circle at $3$ is filled, so $3$ is. The number line shows

$$-3 < x \le 3$$

Now solve the options until one gives exactly that. A three-part inequality is solved by doing the same thing to all three parts at once. For B:

$$-5 < 3x + 4 \le 13$$

Subtract 4 from every part: $-9 < 3x \le 9\mathclose{.}$ Divide every part by 3, which is positive, so no sign flips: $-3 < x \le 3$ ✓

Each end keeps whatever sign it started with — the left stayed strict, the right stayed inclusive — and that is what the open and the filled circle record.

Why each wrong option is wrong:

• A, $-5 \le 3x + 4 < 13$ — the right numbers, with the two signs the wrong way round. It gives $-3 \le x < 3\mathclose{:}$ a filled circle at $-3$ and an open one at $3\mathclose{.}$
• C, $-5 < 3x - 4 \le 13$ — the 4 is subtracted, not added, so every part has 4 added back: $-1 < 3x \le 17\mathclose{,}$ and the line would run from $-\tfrac{1}{3}$ to $\tfrac{17}{3}\mathclose{.}$
• D, $-3 < 3x + 4 \le 3$ — copies $-3$ and $3$ off the number line into the inequality. Those are the ends for $x\mathclose{,}$ not for $3x + 4\mathclose{;}$ solving it gives $-\tfrac{7}{3} < x \le -\tfrac{1}{3}\mathclose{.}$

Takeaway: Turn the picture into an inequality for $x$ first — open circle strict, filled circle inclusive — and then solve the options until one matches. Do every step to all three parts.

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Answer: C

Explanation

Read the statement in words: $|x - 5| < 3$ says $x$ is less than 3 away from 5. Everything within 3 of 5 — so the answer is one interval, sitting around 5.

Go 3 below 5 and 3 above 5:

$$5 - 3 = 2 \qquad 5 + 3 = 8$$

The sign is strict $(<)\mathclose{,}$ so the endpoints themselves are excluded — a point exactly 3 away is not less than 3 away:

$$2 < x < 8$$

If you prefer the mechanical route, "less than" unwraps into a three-part inequality directly:

$$-3 < x - 5 < 3 \quad\Longrightarrow\quad 2 < x < 8$$

Check: $x = 5$ gives $|0| = 0 < 3$ ✓. $x = 2$ gives $|-3| = 3\mathclose{,}$ and $3 < 3$ is false ✓. $x = 9$ gives $|4| = 4\mathclose{,}$ not less than 3 ✓.

Why each wrong option is wrong:

• A, $x < 2$ or $x > 8$ — this is the answer to $|x - 5| > 3\mathclose{,}$ the "further than" version. It has the right boundaries pointing the wrong way. Less than gives you the inside; greater than gives you the outside.
• B, $2 \le x \le 8$ — right interval, but includes the endpoints. $x = 2$ is exactly 3 away, and the question asked for less than 3.
• D, $-3 < x < 3$ — used the 3 as the boundaries and ignored the 5. The 3 is the distance; the 5 is where you measure from.

Takeaway: $|x - a| < k$ is "within $k$ of $a$" — one interval, centred on $a\mathclose{,}$ reaching out $k$ each way. $|x - a| > k$ is "further than $k$ from $a$" — two pieces heading outwards. Say the statement in words before you write anything.

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Answer: D

Explanation

Work out what is being asked before reaching for notation. The bolt is acceptable when its diameter is within $0.04$ of $12.00\mathclose{:}$

$$|d - 12| \le 0.04$$

But the question asks when the bolt is rejected — the opposite. A bolt is rejected when it is further than $0.04$ from 12:

$$|d - 12| > 0.04$$

That is the whole question: spotting that "rejected" means you want the outside, not the inside.

Test it. A bolt at $12.05$ mm: $|12.05 - 12| = 0.05\mathclose{,}$ and $0.05 > 0.04\mathclose{,}$ so it is rejected ✓. A bolt at $12.03\mathclose{:}$ $|0.03| = 0.03\mathclose{,}$ not greater than $0.04\mathclose{,}$ so it is kept ✓. A bolt at $11.90\mathclose{:}$ $|-0.10| = 0.10 > 0.04\mathclose{,}$ rejected ✓ — and notice the absolute value handles the undersized bolt without any extra work. That is exactly why the notation is used here.

Why each wrong option is wrong:

• A, $|d - 12| < 0.04$ — describes the bolts that are accepted, not rejected. The arithmetic is right and the question was answered backwards.
• B, $|d - 0.04| > 12$ — swapped the target and the tolerance, asking whether $d$ is more than 12 away from $0.04\mathclose{.}$ Every bolt near 12 mm satisfies it, so this rejects the entire batch.
• C, $|d - 12| \le 0.04$ — the accepted set again, this time including the boundary. Two ways of answering the opposite question are offered because answering the opposite question is the mistake being tested.

Takeaway: Write the condition the words give you, then check whether the question wants that set or its opposite. "Within" is $\le$ and gives the inside; "further than", "outside tolerance" and "rejected" are $>$ and give the outside.

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Section 8 — When You Do Not Need to Find $x$

Some questions look like they want you to solve, and do not. They give you an equation and ask for the value of an expression — $2x + 6\mathclose{,}$ say, rather than $x\mathclose{.}$

You can always find $x$ and substitute. But often the expression is just a multiple of something you already have, and spotting that turns a two-minute question into a five-second one. The SAT sets these deliberately, and the wrong options are built from the values you pass through on the long route.

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Answer: D

Explanation

Look at the two expressions side by side before doing anything:

$$5x - 3 \qquad\text{and}\qquad 10x - 6$$

The second is exactly twice the first. So if $5x - 3 = 12\mathclose{,}$ then

$$10x - 6 = 2(5x - 3) = 2 \times 12 = 24$$

No solving required at all.

The long way works too and is worth checking against. $5x - 3 = 12$ gives $5x = 15\mathclose{,}$ so $x = 3\mathclose{.}$ Then $10(3) - 6 = 30 - 6 = 24$ ✓ — same answer, more steps.

Why each wrong option is wrong:

• A, $3$ — the value of $x\mathclose{.}$ Correct as far as it goes, and not what was asked. Every value you pass through on the long route is offered as an option here; that is not an accident.
• B, $15$ — the value of $5x\mathclose{.}$
• C, $18$ — doubled the 12 to get 24 and then subtracted the 6 again. The $-6$ is already inside the doubling: $10x - 6 = 2(5x - 3)\mathclose{.}$

Takeaway: Before solving, check whether the expression you want is a simple multiple of the one you were given. Look for the doubling — it is the most common version by far. And when the options include the intermediate values, that is a sign the question is testing whether you read the last line.

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Answer: D

Explanation

"True for all values of $x$" means the two sides are the same expression. Expand the left:

$$3(x + 2) = 3x + 6$$

So the equation reads

$$3x + 6 = 3x + k$$

The $3x$ matches on both sides already, so the constants must match too:

$$k = 6$$

There is a quicker route that works on every question of this type. If it is true for all $x\mathclose{,}$ it is true for the most convenient $x$ you can think of. Put $x = 0\mathclose{:}$

$$3(0 + 2) = 3(0) + k \quad\Longrightarrow\quad 6 = k$$

One line, no expanding.

Why each wrong option is wrong:

• A, $0$ — assumed everything has to cancel to $0 = 0\mathclose{.}$ Not so: an identity needs the two sides to be equal, not zero. With $k = 0$ the equation becomes $3x + 6 = 3x\mathclose{,}$ which is never true.
• B, $2$ — read the 2 from inside the bracket, before the 3 was distributed.
• C, $3$ — read the 3 from outside the bracket. $k$ is a constant term, and the 3 is a coefficient; they are not doing the same job.

Takeaway: "True for all $x$" means the two sides are identical, so match coefficients and constants separately — or just substitute $x = 0\mathclose{,}$ which is faster and harder to get wrong.

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