Chapter 3 — Algebra

Answer: D

Explanation

Expand both sides first:

$$10x - 15 = 3x + 12 + 7 = 3x + 19$$

Gather the $x$ terms on the left and the numbers on the right:

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

An awkward fraction is not a sign of an error. Check by substituting: the left gives $5\left(\frac{68}{7} - 3\right) = 5 \times \frac{47}{7} = \frac{235}{7}$, and the right gives $\frac{102}{7} + 19 = \frac{235}{7}$. Correct.

Why each wrong option is wrong:

• A, $\frac{4}{7}$ — multiplied the 5 by $2x$ but not by the $-3$.
• B, $2$ — dropped the $+7$ on the right.
• C, $\frac{20}{7}$ — expanded $3(x + 4)$ as $3x + 4$, multiplying only the first term.

Takeaway: the bracket multiplies everything inside it. Expand fully on both sides before you gather anything.

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Answer: F

Explanation

Subtract 7 from both sides — this does not affect the direction:

$$-4x \ge 12$$

Now divide by $-4$. Because $-4$ is negative, the sign turns round:

$$x \le -3$$

Check with a value. Take $x = -5$, which satisfies $x \le -3$: $-4(-5) + 7 = 27$, and $27 \ge 19$. Correct. Now take $x = 0$, which does not: $7 \ge 19$ is false. Correct again.

Why each wrong option is wrong:

• G, $x \ge -3$ — divided correctly but left the sign pointing the same way.
• H, $x \le 3$ — flipped the sign but lost the minus on the 3.
• J, $x \ge 3$ — did neither.

Takeaway: dividing an inequality by a negative reverses it. If you are ever unsure, test one number from your answer — it settles the question in five seconds.

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Answer: B

Explanation

Both sides are of the form $(\text{number})x + (\text{number})$.

"No solution" means the $x$ terms cancel — leaving no $x$ to solve for — while the constants disagree. The $x$ terms cancel exactly when their coefficients match:

$$k = 4$$

Then the equation becomes $4x + 9 = 4x - 3$. Subtract $4x$ from both sides:

$$9 = -3$$

which is false whatever $x$ is. So there is no solution.

Contrast with infinitely many solutions, which the ACT also asks: that needs the coefficients to match and the constants to match too. Same first condition, opposite conclusion — the constants decide which one you get.

Why each wrong option is wrong:

• A, $-4$ — right size, wrong sign; $k = -4$ gives $8x = -12$, which has one solution.
• C, $9$ and D, $-3$ — matched a constant rather than the coefficient.

Takeaway: no solution means the $x$ terms cancel and the numbers do not. Infinitely many means both cancel. Check the constants to tell them apart.

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Answer: J

Explanation

Treat it as one chain and do the same thing to all three parts.

Subtract 4 everywhere:

$$-9 < 3x \le 9$$

Divide by 3 everywhere — 3 is positive, so nothing flips:

$$-3 < x \le 3$$

The kind of each end matters as much as its value. The left was strict ($<$) and stays strict; the right allowed equality ($\le$) and still does.

Why each wrong option is wrong:

• F, $-3 \le x \le 3$, H, $-3 \le x < 3$ and G, $-3 < x < 3$ — all have the right numbers and the wrong ends. Copy each inequality sign down unchanged as you go; that is the whole difficulty.

Takeaway: in a three-part inequality, operate on all three parts together and carry each sign down exactly as it was.

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Section 2 — Systems of Equations

Two equations, two unknowns. Two methods, and the choice between them is worth making deliberately:

Substitution — best when one equation already gives a variable on its own. Put that expression into the other equation.

Elimination — best when neither does. Add or subtract the two equations so one variable disappears, scaling one or both first if needed.

The ACT also asks about systems with no solution (parallel lines: same slope, different intercept) and infinitely many (the same line twice).

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Answer: B

Explanation

Add the two equations. The $+2y$ and $-2y$ cancel:

$$(3x + 5x) + (2y - 2y) = 19 + 13$$

$$8x = 32 \qquad x = 4$$

Check by finding $y$ too: $3(4) + 2y = 19$ gives $2y = 7$, so $y = 3.5$. Put both into the second equation: $5(4) - 2(3.5) = 20 - 7 = 13$. Correct.

Why each wrong option is wrong:

• A, $\frac{7}{2}$ — solved correctly and gave $y$. Always check which letter the question asked for; this is the most common way to lose a mark you have earned.
• C, $32$ — added the two right-hand sides to get 32 and never divided by 8.
• D, $8$ — added the coefficients of $x$ and stopped.

Takeaway: when the coefficients of one variable are equal and opposite, add. When they are equal and the same, subtract. Then answer for the letter asked for.

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Answer: F

Explanation

To remove $x$, make its coefficients match. Multiply the first equation by 3 and the second by 2:

$$6x + 15y = 72$$ $$6x - 4y = -4$$

Now subtract the second from the first:

$$19y = 76 \qquad y = 4$$

Check: from the first equation $2x + 20 = 24$, so $x = 2$; and $3(2) - 2(4) = 6 - 8 = -2$. Correct.

Why each wrong option is wrong:

• H, $2$ — gave $x$, which is 2.
• G, $76$ — scaled the equations and then never combined them.
• J, $-2$ — copied the $-2$ from the right of the second equation.

Takeaway: to eliminate a variable, multiply each equation by the other's coefficient. Then subtract if the signs match and add if they do not.

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Answer: B

Explanation

Parallel lines have the same slope. Rearrange each into $y = mx + c$ form.

From the first: $-2y = -6x + 10$, so $y = 3x - 5$, slope 3.

From the second: $ky = -9x + 3$, so $y = -\frac{9}{k}x + \frac{3}{k}$, slope $-\frac{9}{k}$.

Set the slopes equal:

$$-\frac{9}{k} = 3 \qquad \Longrightarrow \qquad k = -3$$

Check that the lines are not identical, which would give infinitely many solutions instead: with $k = -3$ the second equation is $9x - 3y = 3$, or $y = 3x - 1$. Same slope, different intercept ($-1$ against $-5$), so they are parallel and distinct. No solution. Correct.

Why each wrong option is wrong:

• A, $3$ — got the size right and the sign wrong.
• C, $- \frac{4}{3}$ — divided the wrong pair of coefficients.
• D, $-2$ — took a coefficient from the first equation.

Takeaway: no solution means equal slopes and different intercepts. Always check the intercept too — equal slopes and equal intercepts is the opposite answer, infinitely many.

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Section 3 — Expanding and Factoring

Expanding and factoring are the same operation in opposite directions, and the ACT asks for both.

Three patterns are worth knowing on sight, because spotting them saves the time that the ACT is really testing:

$$(a + b)^2 = a^2 + 2ab + b^2$$ $$(a - b)^2 = a^2 - 2ab + b^2$$ $$(a + b)(a - b) = a^2 - b^2$$

The middle term in the first two is what people forget. $(a+b)^2$ is not $a^2 + b^2$.

To factor $x^2 + bx + c$, find two numbers that multiply to $c$ and add to $b$.

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Answer: J

Explanation

$(2x - 3)^2$ means $(2x - 3)(2x - 3)$. Multiply every term by every term:

$$2x \cdot 2x = 4x^2 \qquad 2x \cdot (-3) = -6x$$ $$-3 \cdot 2x = -6x \qquad -3 \cdot (-3) = +9$$

There are two middle terms, both $-6x$, so together they make $-12x$:

$$(2x - 3)^2 = 4x^2 - 12x + 9$$

Check at $x = 1$: the original is $(-1)^2 = 1$, and $4 - 12 + 9 = 1$. Correct.

Why each wrong option is wrong:

• F, $4 x^{2} + 9$ — squared each term separately. There is always a middle term.
• G, $4 x^{2} - 9$ — used the difference-of-squares pattern, which is for $(a+b)(a-b)$, not for a square.
• H, $4 x^{2} - 6 x + 9$ — found one middle term instead of two.

Takeaway: $(a \pm b)^2 = a^2 \pm 2ab + b^2$. Square the whole first term, coefficient included, and double the middle.

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Answer: C

Explanation

With a leading coefficient, the two numbers must multiply to $2 \times (-3) = -6$ and add to $+5$. That pair is $+6$ and $-1$.

Split the middle term using them:

$$2x^2 + 6x - x - 3$$

Factor in pairs:

$$2x(x + 3) - 1(x + 3) = (x + 3)(2x - 1)$$

So the factors are $x + 3$ and $2x - 1$. Of the options, $2x - 1$ is there.

Check by expanding: $(x+3)(2x-1) = 2x^2 - x + 6x - 3 = 2x^2 + 5x - 3$. Correct.

Why each wrong option is wrong:

• A, $x - 3$ — used factors of 3 alone, ignoring the leading 2.
• B, $2 x + 1$ — right numbers with the signs swapped; $(x-3)(2x+1)$ expands to $2x^2 - 5x - 3$, with the middle sign wrong.
• D, $x + 1$ — not a factor at all.

Takeaway: with a leading coefficient $a$, look for two numbers multiplying to $a \times c$ and adding to $b$, then split the middle term and factor in pairs.

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Answer: F

Explanation

$9x^2 - 49$ is a difference of two squares: $9x^2 = (3x)^2$ and $49 = 7^2$, with a minus between them and nothing else.

$$a^2 - b^2 = (a - b)(a + b)$$

with $a = 3x$ and $b = 7$:

$$9x^2 - 49 = (3x - 7)(3x + 7)$$

Check by expanding: $9x^2 + 21x - 21x - 49$, and the middle terms cancel — which is exactly why the pattern has no middle term.

Why each wrong option is wrong:

• G, $\left(3 x - 7\right)^{2}$ — $(3x-7)^2$ expands to $9x^2 - 42x + 49$, which has a middle term and the wrong constant sign.
• H, $\left(x + 7\right) \left(9 x - 7\right)$ and J, $\left(3 x - 49\right) \left(3 x + 1\right)$ — took a square root of only one part.

Takeaway: two perfect squares with a minus between them and no middle term factors as $(a-b)(a+b)$. Take the square root of both parts.

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Section 4 — Quadratic Equations

A quadratic equation is one with an $x^2$ and no higher power. Three ways to solve one, in the order you should try them:

1. Factor, if it factors easily. Fastest by far.
2. Take square roots, if there is no middle term.
3. The quadratic formula, which always works:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The ACT does not print that formula. Learn it.

The part under the root, $b^2 - 4ac$, is the discriminant, and it tells you how many real solutions there are before you find any of them:

discriminant	real solutions
positive	two
zero	exactly one
negative	none

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Answer: D

Explanation

Find two numbers multiplying to $-14$ and adding to $-5$. They are $-7$ and $+2$:

$$x^2 - 5x - 14 = (x - 7)(x + 2) = 0$$

A product is zero when one of its factors is zero, so $x = 7$ or $x = -2$.

Time cannot be negative, so the answer is $x = 7$.

Why each wrong option is wrong:

• A, $-2$ — gave the negative root, which the context rules out. This is why the question says the solution must be positive.
• B, $14$ and C, $5$ — read numbers straight off the equation.

Takeaway: factor, set each bracket to zero, then check the context. A length, a time or a count cannot be negative, and the ACT relies on you noticing.

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Answer: J

Explanation

Exactly one real solution means the discriminant is zero:

$$b^2 - 4ac = 0$$

Here $a = 1$, $b = k$ and $c = 16$:

$$k^2 - 4(1)(16) = 0$$ $$k^2 = 64$$ $$k = \pm 8$$

The question says $k$ is positive, so $k = 8$.

Check: $x^2 + 8x + 16 = (x + 4)^2$, which is zero only at $x = -4$. Exactly one solution. Correct.

Why each wrong option is wrong:

• F, $4$ — took the square root of 16 rather than of $4ac$.
• G, $16$ — copied the constant.
• H, $64$ — computed $4ac = 64$ and stopped before the square root.

Takeaway: one solution means $b^2 = 4ac$. And when a squared quantity is solved, there are two answers — read the question to see which sign it wants.

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Answer: B

Explanation

Take the square root of both sides, remembering both signs:

$$x - 3 = \pm 5$$

That is two separate equations:

$$x - 3 = 5 \Rightarrow x = 8 \qquad x - 3 = -5 \Rightarrow x = -2$$

Check both: $(8-3)^2 = 25$ and $(-2-3)^2 = (-5)^2 = 25$. Both correct.

Why each wrong option is wrong:

• A, $\left( -8, \ 2\right)$ — subtracted the 3 rather than adding it when isolating $x$.
• C, $\left( 5, \ 5\right)$ — used only $+5$, giving one solution twice. A squared equation almost always has two.
• D, $\left( 2, \ 8\right)$ — got $+8$ right and slipped the sign on the other case.

Takeaway: taking a square root gives $\pm$. Write out both equations separately — trying to hold both in your head is where the sign errors come from.

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Section 5 — Radical, Rational and Exponential Equations

Three equation types that each carry one specific hazard.

Radical — square both sides to clear the root, then check every answer in the original equation. Squaring can manufacture solutions that were never there.

Rational (a variable in a denominator) — multiply through by the denominator. Any value that would make a denominator zero is excluded.

Exponential — if both sides can be written with the same base, the exponents must be equal.

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Answer: J

Explanation

Square both sides:

$$x + 6 = x^2$$

Bring everything to one side and factor:

$$x^2 - x - 6 = 0 \quad\Longrightarrow\quad (x - 3)(x + 2) = 0$$

So the squared equation gives $x = 3$ and $x = -2$. Both must now be tested in the original, because squaring turns a false statement into a true one: $-2 \ne 2$, but $(-2)^2 = 2^2$.

At $x = 3$: $\sqrt{9} = 3$. ✓

At $x = -2$: $\sqrt{4} = 2$, and $2 \ne -2$. ✗

The square-root symbol means the non-negative root, so it can never equal a negative number. $x = -2$ is extraneous — manufactured by the squaring.

Why each wrong option is wrong:

• F, $-2$ — kept the extraneous root without testing it.
• G, $1$ — added the two roots of the squared equation.
• H, $6$ — doubled the right-hand side instead of squaring the left.

Takeaway: after squaring, every solution is only a candidate. Substitute each back into the original equation. A square root is never negative.

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Answer: D

Explanation

Both 9 and 27 are powers of 3:

$$9 = 3^2 \qquad 27 = 3^3$$

Rewrite both sides:

$$(3^2)^x = (3^3)^4$$ $$3^{2x} = 3^{12}$$

With the same base, the exponents must be equal:

$$2x = 12 \qquad x = 6$$

Check: $9^6 = 531441$ and $27^4 = 531441$. Correct.

Why each wrong option is wrong:

• B, $4$ — copied the 4 from the right-hand side.
• C, $12$ — reached $2x = 12$ and gave 12 instead of dividing.
• A, $\frac{4}{3}$ — divided the exponents without converting to a common base.

Takeaway: rewrite both sides as powers of the same base, then set the exponents equal. Look for the small base hiding inside both numbers — usually 2 or 3.

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Answer: H

Explanation

Gather the fractions on one side and the numbers on the other:

$$\frac{12}{x} - \frac{8}{x} = 3 - 1$$

$$\frac{4}{x} = 2$$

Multiply both sides by $x$ and divide by 2:

$$4 = 2x \qquad x = 2$$

Check in the original: $\frac{12}{2} + 1 = 7$ and $\frac{8}{2} + 3 = 7$. Correct. And $x = 2$ does not make any denominator zero, so it is allowed.

Why each wrong option is wrong:

• F, $4$ — subtracted the numerators and stopped, ignoring the constants.
• G, $-2$ — sign slip while gathering terms.
• J, $20$ — added the numerators.

Takeaway: with the same denominator on both sides, combine the fractions first — it is quicker than multiplying everything out. Then check the answer does not make a denominator zero.

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Section 6 — Rearranging a Formula

"Solve for $R$ in terms of the others" is a standard ACT question and it appeared on both official enhanced forms. There is no new mathematics in it: treat every other letter as if it were a number, and isolate the one you want.

The only real difficulty is that the answer is an expression rather than a number, so there is nothing to substitute back — which makes it worth being slow.

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Answer: B

Explanation

Isolate the term containing $w$:

$$P - 2l = 2w$$

Divide both sides by 2:

$$w = \frac{P - 2l}{2}$$

Test it with numbers, which is the only real check available. Take $l = 3$ and $w = 4$; then $P = 2(3) + 2(4) = 14$. Put $P = 14$ and $l = 3$ into the answer: $\frac{14 - 6}{2} = 4$. Correct.

Why each wrong option is wrong:

• A, $\frac{P}{2} - 2 l$ — halved $P$ before subtracting, so the $2l$ never got halved.
• C, $\frac{P}{2} - \frac{l}{2}$ — subtracted $l$ rather than $2l$.
• D, $P - 2 l$ — subtracted correctly and never divided.

Takeaway: isolate the term with your letter in it, then divide by everything multiplying it. Test with easy numbers — it catches almost every rearranging error in ten seconds.

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Answer: G

Explanation

Multiply both sides by $t$ to bring it up out of the denominator:

$$Ft = mv$$

Now divide by $F$:

$$t = \frac{mv}{F}$$

Test with numbers: let $m = 2$, $v = 3$, $t = 6$. Then $F = \frac{6}{6} = 1$, and the answer gives $t = \frac{2 \times 3}{1} = 6$. Correct.

Why each wrong option is wrong:

• F, $F m v$ — multiplied by $F$ instead of dividing.
• H, $\frac{F}{m v}$ — the reciprocal of the right answer, which is what you get by dividing in the wrong direction.
• J, $\frac{F m}{v}$ — moved $m$ but not $v$.

Takeaway: to free a variable from a denominator, multiply both sides by it first. Then test with three small numbers; an inverted answer shows up immediately.

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Answer: D

Explanation

Square both sides:

$$2x + 7 = (x - 4)^2 = x^2 - 8x + 16$$

$$x^2 - 10x + 9 = 0 \quad\Longrightarrow\quad (x - 9)(x - 1) = 0$$

Test both. At $x = 9$: $\sqrt{25} = 5$ and $9 - 4 = 5$. ✓ At $x = 1$: $\sqrt{9} = 3$ but $1 - 4 = -3$. ✗

A square root cannot be negative, so $x = 1$ is extraneous.

Why each wrong option is wrong:

• B, $1$ — kept the extraneous root without testing it.
• C, $10$ — added the two roots of the squared equation.
• A, $-1$ — a sign slip expanding $(x-4)^2$.

Takeaway: every solution after squaring is a candidate until substituted back into the original equation.

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Answer: H

Explanation

Gather fractions on one side and numbers on the other:

$$\frac{20}{x} - \frac{8}{x} = 7 - 3 \quad\Longrightarrow\quad \frac{12}{x} = 4$$

$$x = 3$$

Check: $\frac{20}{3} + 3 = \frac{29}{3}$ and $\frac{8}{3} + 7 = \frac{29}{3}$. ✓

Why each wrong option is wrong:

• F, $12$ and G, $4$ — did one half of the gathering and stopped.
• J, $28$ — added the numerators.

Takeaway: with a shared denominator, combine the fractions before doing anything else, and check the answer does not make the bottom zero.

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Answer: C

Explanation

$$500 \times 1.2 \times 1.2 = 500 \times 1.44 = 720$$

The second hour's growth is calculated on 600, not on 500 — which is why the answer exceeds $500 + 100 + 100$.

Why each wrong option is wrong:

• A, $600$ — stopped after one hour.
• B, $700$ — added 20% of the original twice, which is simple growth, not compound.
• D, $620$ — grew the wrong base.

Takeaway: compound growth multiplies by $(1 + r)$ once per period. Each period grows from the new amount.

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Answer: G

Explanation

"Decays to 85%" states the multiplier directly: $0.85$.

$$M = 400(0.85)^h$$

Check one hour: $400 \times 0.85 = 340$, a loss of 60, which is 15% of 400.

Why each wrong option is wrong:

• F, $400(0.15)^h$ — used $0.15$, which would leave only 15% each hour.
• H, $400 - 0.15h$ — linear decay, losing a fixed 0.15 grams per hour.
• J, $400(1.15)^h$ — a multiplier above 1 grows the sample.

Takeaway: the multiplier is what remains. Decay is below 1, growth above.

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Answer: B

Explanation

Multiply both sides by $(x - 2)$:

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

$$3x = 12 \qquad x = 4$$

Check: $\frac{6}{4 - 2} = \frac{6}{2} = 3$. ✓ And $x = 4$ does not make the denominator zero, so it is allowed.

Why each wrong option is wrong:

• A, $2$ — $x = 2$ is precisely the value that is not allowed: it makes the denominator zero and the expression undefined.
• C, $20$ — multiplied 6 by 3 rather than dividing.
• D, $-4$ — a sign slip.

Takeaway: clear the denominator by multiplying, solve, then confirm the answer does not make any denominator zero.

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Answer: G

Explanation

Find two numbers multiplying to $-40$ and adding to $-3$. They are $-8$ and $+5$:

$$x^2 - 3x - 40 = (x - 8)(x + 5) = 0$$

A product is zero when one factor is zero, so $x = 8$ or $x = -5$. A length cannot be negative, so $x = 8$.

Why each wrong option is wrong:

• F, $-5$ — gave the root the context rules out.
• H, $40$ and J, $3$ — read numbers straight off the equation without factoring.

Takeaway: factor, set each bracket to zero, then check which root the situation allows.

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Answer: C

Explanation

With $a = 3$, $b = -5$, $c = -2$:

$$x = \frac{5 \pm \sqrt{25 - 4(3)(-2)}}{6} = \frac{5 \pm \sqrt{49}}{6} = \frac{5 \pm 7}{6}$$

$$x = 2 \qquad \text{or} \qquad x = -\tfrac{1}{3}$$

Watch the $-4ac$ term: $c$ is negative, so $-4(3)(-2) = +24$, not $-24$. That double negative is where the formula usually goes wrong.

Why each wrong option is wrong:

• A, $- \frac{1}{3}$ — the smaller root; the question asked for the larger.
• B, $\frac{5}{6}$ — computed $\frac{5}{6}$, forgetting the $\pm\sqrt{\;}$ part entirely.
• D, $3$ — read the leading coefficient off the equation.

Takeaway: substitute $a$, $b$ and $c$ with their signs, and expect $-4ac$ to be positive whenever $c$ is negative.

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Answer: H

Explanation

Move the constant across and divide by 4:

$$4x^2 = 49 \qquad x^2 = \frac{49}{4}$$

Take the square root of both sides, remembering both signs:

$$x = \pm\frac{7}{2}$$

Check: $4 \times \frac{49}{4} = 49$. ✓

Why each wrong option is wrong:

• F, $\left( - \frac{49}{4}, \ \frac{49}{4}\right)$ — divided by 4 and stopped before the square root.
• G, $\left( -7, \ 7\right)$ — rooted the 49 and ignored the 4.
• J, $\left( - \frac{49}{2}, \ \frac{49}{2}\right)$ — halved 49 instead of rooting.

Takeaway: no middle term means take square roots. Isolate $x^2$ first, and never lose the $\pm$.

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Answer: D

Explanation

One squaring is not enough here, because a root sign survives it:

$$x + 5 = \left(\sqrt{x} + 1\right)^2 = x + 2\sqrt{x} + 1$$

The $x$ terms cancel, which leaves the remaining root on its own:

$$4 = 2\sqrt{x} \quad\Longrightarrow\quad \sqrt{x} = 2$$

Square a second time: $x = 4$.

Check in the original: $\sqrt{4 + 5} = 3$ and $\sqrt{4} + 1 = 3$. ✓

Why each wrong option is wrong:

• A, $2$ — found $\sqrt{x} = 2$ and stopped before the second squaring.
• B, $8$ — squared $2\sqrt{x}$ as $2x$ instead of $4x$, so $16 = 2x$.
• C, $16$ — dropped the 2 and read $\sqrt{x} = 4$.

Takeaway: $\left(\sqrt{x} + 1\right)^2$ is $x + 2\sqrt{x} + 1$, not $x + 1$. The middle term keeps a root alive, so isolate it and square again, then check the answer in the original equation.

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Answer: F

Explanation

Cross-multiply:

$$3(x + 6) = 12x$$ $$3x + 18 = 12x$$ $$18 = 9x \qquad x = 2$$

Check: $\frac{3}{2} = 1.5$ and $\frac{12}{8} = 1.5$. ✓ Neither denominator is zero at $x = 2$, so it is allowed.

Why each wrong option is wrong:

• G, $6$ — copied the 6 out of the denominator.
• H, $-6$ — $x = -6$ makes the second denominator zero, which is exactly the value the equation excludes.
• J, $18$ — added the two numbers in the question.

Takeaway: cross-multiply, solve, then confirm the answer does not make any denominator zero.

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Answer: D

Explanation

Multiply every term by $x$:

$$x^2 + 6 = 5x \quad\Longrightarrow\quad x^2 - 5x + 6 = 0$$

$$(x - 2)(x - 3) = 0 \quad\Longrightarrow\quad x = 2 \text{ or } x = 3$$

The smaller is 2. Check: $2 + \frac{6}{2} = 5$. ✓

Why each wrong option is wrong:

• B, $3$ — the larger of the two solutions.
• C, $5$ and A, $6$ — copied numbers from the equation.

Takeaway: clearing a denominator that contains $x$ usually produces a quadratic. Expect two answers and read which one is wanted.

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Answer: G

Explanation

A 6% rise multiplies by $1.06$ — the whole original, plus another 6% of it:

$$V = 2000(1.06)^{n}$$

Check one year: $2000 \times 1.06 = 2120$, which is $2000 + 120$, and 120 is 6% of 2000. ✓

Why each wrong option is wrong:

• F, $2000(0.06)^n$ — $0.06$ would leave 6% of the money after a year, a 94% loss.
• H, $2000 + 0.06n$ — linear growth, adding 6 cents a year.
• J, $2000(6)^n$ — multiplies the money sixfold every year.

Takeaway: a $p\%$ rise per period is a multiplier of $1 + \frac{p}{100}$. Test one period against plain arithmetic.

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Answer: C

Explanation

Both 8 and 32 are powers of 2:

$$8 = 2^3 \qquad 32 = 2^5$$

$$(2^3)^{x+1} = 2^5 \quad\Longrightarrow\quad 2^{3(x+1)} = 2^5$$

Same base, so the exponents match:

$$3(x + 1) = 5 \quad\Longrightarrow\quad 3x = 2 \quad\Longrightarrow\quad x = \tfrac{2}{3}$$

Why each wrong option is wrong:

• A, $4$ — divided 32 by 8, which is not how exponents combine.
• B, $\frac{5}{3}$ — solved $3x = 5$, forgetting the $+1$ inside the bracket.
• D, $2$ — a whole-number guess; $8^3 = 512$, not 32.

Takeaway: find the small base hiding inside both numbers — usually 2 or 3 — then set the exponents equal. Do not forget what is inside the bracket.

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Answer: H

Explanation

Keeping 80% each hour means multiplying by $0.8$ three times:

$$500 \times 0.8 \times 0.8 \times 0.8 = 500 \times 0.512 = 256$$

Note it is not $500 - 300$. Each hour's loss is 20% of what is left, which shrinks: 100, then 80, then 64.

Why each wrong option is wrong:

• F, $200$ — took 20% of the original three times, which is simple decay, not compound.
• G, $400$ — stopped after one hour.
• J, $140$ — subtracted a fixed amount three times.

Takeaway: decay multiplies by $1 - \frac{p}{100}$ once per period. Each period works from the new amount, so the loss gets smaller every time.

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