Linear Equations and Inequalities

Dividing both sides by $-3$ reverses the inequality: $-3x < 12$ becomes $x > -4$. Forgetting to flip the sign when dividing by a negative is the classic error.

Move the variables to one side: $5x - 2x = 11 + 4$, so $3x = 15$ and $x = 5$. ($x = \dfrac{15}{7}$ added the variables ($5x + 2x$); $x = 15$ forgot to divide by 3.)

Expanding gives $2x + 6 = 2x + 7$. Subtracting $2x$ from both sides leaves $6 = 7$, which is never true, so there is no solution.

The gap between the cars grows at $60 + 80 = 140$ km/h. Let $t$ be the time: $140t = 350$, so $t = 2.5$ hours. ($5$ hours used only one of the two speeds.)

Distribute $-2$ across $(x - 1)$: $-2 \times x = -2x$ and $-2 \times -1 = +2$, so $4 - 2x + 2 = 8$, giving $6 - 2x = 8$, then $-2x = 2$ and $x = -1$. ($x = -3$ treated $-2 \times -1$ as $-2$; $x = 1$ did not flip the sign when dividing by $-2$.)

Cross-multiply: $6(2x - 1) = 4(x + 3)$, so $12x - 6 = 4x + 12$, giving $8x = 18$ and $x = \dfrac{9}{4}$.

Set the costs equal: $20 + 0.10n = 30 + 0.05n$. Then $0.05n = 10$, so $n = 200$ texts. (At that point both plans cost $\$40$.)

Let the smaller angle be $x$; the other is $x + 20$. They add to 90: $x + (x + 20) = 90$, so $2x = 70$ and $x = 35^\circ$. ($55^\circ$ is the larger angle.)

Cross-multiply: $4(x + 1) = 3(x - 1)$; distribute: $4x + 4 = 3x - 3$; move variables: $4x - 3x = -3 - 4$, so $x = -7$. ($x = 7$ cross-multiplied in the wrong order; $x = -1$ treated $-1$ as $+1$.)

There are $3$ parts boys and $2$ parts girls, so $3k + 2k = 30$, giving $5k = 30$ and $k = 6$. The number of boys is $3 \times 6 = 18$. ($12$ is the number of girls.)

Move variables: $3x - 5x < 10 - 4$, so $-2x < 6$. Divide by $-2$ and flip the inequality (dividing by a negative): $x > -3$. ($x < -3$ is the canonical error: it did not flip the inequality when dividing by a negative.)

Add 2: $3x \le 12$; divide by 3 (positive, so no flip): $x \le 4$. ($x \ge 4$ flipped the inequality for no reason; $x \le \dfrac{8}{3}$ subtracted 2 instead of adding.)

Divide both sides by $-2$ and flip the inequality (dividing by a negative): $x < -4$. ($x > -4$ did not flip the inequality; $x > 4$ also lost the negative sign.)

The cost must satisfy $4 + 2d \le 20$. Subtract 4: $2d \le 16$; divide by 2: $d \le 8$. The greatest whole number is $8$ km. ($10$ km would cost $\$24$, over budget.)

The perimeter satisfies $2(5 + L) < 40$. Divide by 2: $5 + L < 20$; subtract 5: $L < 15$. The greatest whole number below 15 is $14$ cm. ($15$ cm gives a perimeter of exactly 40 cm, which is not less than 40.)

Let the smallest be $x$; the integers are $x$, $x + 1$ and $x + 2$. Their sum: $3x + 3 = 48$, so $3x = 45$ and $x = 15$ (the integers are 15, 16, 17). (16 is the middle integer; 17 is the largest; 48 is the sum.)

Let the sister be $x$; Sam is $x + 4$. Together: $x + (x + 4) = 22$, so $2x + 4 = 22$, $2x = 18$ and $x = 9$. Sam is $9 + 4 = 13$. (9 is the sister's age; 11 splits 22 equally, ignoring the 4-year difference.)

Apply inverse operations in reverse order: subtract 3 ($y - 3 = 2x$), then divide by 2: $x = \dfrac{y - 3}{2}$. ($x = \dfrac{y + 3}{2}$ added 3 instead of subtracting; $x = 2(y - 3)$ multiplied instead of dividing.)

Divide both sides by $2\pi$: $r = \dfrac{C}{2\pi}$. ($r = \dfrac{2\pi}{C}$ took the reciprocal; $r = \dfrac{C}{\pi}$ forgot the 2; $r = C - 2\pi$ subtracted, but $2\pi$ is multiplied by $r$, not added.)

Divide both sides by $\pi r^2$: $h = \dfrac{V}{\pi r^2}$. ($h = \dfrac{\pi r^2}{V}$ took the reciprocal; $h = \dfrac{V}{\pi r}$ forgot to square the $r$.)

Multiply every term by the LCM of 3 and 2, which is 6: $2(2x + 1) + 12 = 3(x + 5)$; distribute: $4x + 2 + 12 = 3x + 15$; combine: $4x + 14 = 3x + 15$; solve: $x = 1$. ($x = 11$ forgot to multiply the constant 2 by 6.)

Multiply both sides by 2 (positive, so no flip): $x + 4 > 2(x - 1) = 2x - 2$; move terms: $4 + 2 > 2x - x$, so $6 > x$, that is $x < 6$. ($x < 5$ treated $2(x - 1)$ as $2x - 1$; $x > 6$ made a sign error at the final step.)

Let the width be $w$; the length is $w + 3$. Perimeter: $2(w + w + 3) = 4w + 6 = 26$, so $4w = 20$ and $w = 5$ cm (the length is 8 cm). (8 cm is the length; 6.5 cm divided 26 by 4 as if it were a square.)

Multiply both sides by 2 to clear the half: $2A = bh$; divide by $b$: $h = \dfrac{2A}{b}$. ($h = \dfrac{A}{2b}$ divided by 2 instead of multiplying; $h = 2Ab$ multiplied by $b$ instead of dividing.)

Distribute both brackets, especially $-2$ across $(x + 4)$: $6x - 3 - 2x - 8 = 5$; combine: $4x - 11 = 5$; solve: $4x = 16$, $x = 4$. ($x = 0$ kept $+8$ instead of $-8$, failing to distribute the $-2$ to the $+4$; $x = 0.5$ did not distribute either bracket.)