NUMBER

$$15+8=23$$

$$6\times7-4 = 42-4=38$$

$$|-8|+|8| = 8+8=16$$

$$|x - 120| \le 0.5$$

Minimum: $119.5$ mm. Maximum: $120.5$ mm.

$|120.7-120|=0.7$. Since $0.7 > 0.5$, the component is rejected.

$a=5,b=3$: $|5+3|=8$ and $|5|+|3|=8$ — equal. $a=-5,b=-3$: $|-8|=8$ and $|-5|+|-3|=8$ — equal.

$|5+(-3)|=|2|=2$. $|5|+|-3|=5+3=8$. These are NOT equal ($2 \ne 8$).

It holds true when $a$ and $b$ have the same sign (both positive or both negative). It does not hold when $a$ and $b$ have opposite signs.

$x=10$: actual $|10-7|=3$, classmate's version gives $10-7=3$ — matches. $x=3$: actual $|3-7|=|-4|=4$, classmate's version gives $3-7=-4$ — does NOT match.

The classmate forgot that absolute value always gives a non-negative result — when $x-7$ is negative (i.e. $x<7$), the absolute value flips its sign. Correct definition: $|x-7| = x-7$ when $x\ge7$, and $|x-7|=7-x$ when $x<7$.

$|-3|=3$, $|-1|=1$, $|2|=2$, $|4|=4$, $|1|=1$ km, respectively.

Changes: $-1-(-3)=2$, $2-(-1)=3$, $4-2=2$, $1-4=-3$. Total distance $=|2|+|3|+|2|+|-3|=2+3+2+3=10$ km.

$$49.7 \text{ mm} \le d \le 50.3 \text{ mm}$$

Checking all: $49.8$?, $50.4$? ($|50.4-50|=0.4>0.3$), $49.6$? ($|49.6-50|=0.4>0.3$), $50.1$?, $50.35$? ($|50.35-50|=0.35>0.3$), $49.9$?. Rejected: $50.4, 49.6, 50.35$. Accepted: $49.8, 50.1, 49.9$.

$|6-3|+|6+2|=3+8=11\ne9$. Does not satisfy.

$x=5$: $|5-3|+|5+2|=2+7=9$ ?. $x=-4$: $|-4-3|+|-4+2|=7+2=9$ ?. Both satisfy!

$|x-3|+|x+2|$ represents the SUM of distances from $x$ to the two fixed points 3 and $-2$ (which are themselves 5 units apart). For any $x$ BETWEEN $-2$ and 3, this sum of distances equals exactly 5 (the distance between the two fixed points) — but we need it to equal 9, which is MORE than 5, so $x$ must lie OUTSIDE the interval $[-2,3]$. In fact, checking further would reveal an entire continuous RANGE of solutions exists on each side (not just the two isolated points found), since moving further from either fixed point increases the total distance sum smoothly.

By definition, $|x|$ represents the distance of $x$ from zero on a number line — and a distance can NEVER be negative (the smallest possible value of $|x|$ is 0, when $x=0$ itself). Since $-5$ is negative, there is NO value of $x$ for which $|x|$ could ever equal $-5$; the equation has no solution.

$|x|=5$ has solutions $x=5$ and $x=-5$ (both are exactly 5 units from zero). The key difference: $5$ is a valid (non-negative) value for a distance, while $-5$ is not — this is precisely why $|x|=5$ is solvable but $|x|=-5$ is not.

$40, 25, 15, 60, 10$ metres respectively.

Checkpoint 4 (position $60$) violates the regulation, since its distance (60m) exceeds the 50m limit, by $60-50=10$ metres. All other checkpoints (40, 25, 15, 10m) are within the safe range.

$$49.7 \text{ mm} \le d \le 50.3 \text{ mm}$$

Checking all: $49.8$?, $50.4$? ($|50.4-50|=0.4>0.3$), $49.6$? ($|49.6-50|=0.4>0.3$), $50.1$?, $50.35$? ($|50.35-50|=0.35>0.3$), $49.9$?. Rejected: $50.4, 49.6, 50.35$. Accepted: $49.8, 50.1, 49.9$.

$|6-3|+|6+2|=3+8=11\ne9$. Does not satisfy.

$x=5$: $|5-3|+|5+2|=2+7=9$ ?. $x=-4$: $|-4-3|+|-4+2|=7+2=9$ ?. Both satisfy!

$|x-3|+|x+2|$ represents the SUM of distances from $x$ to the two fixed points 3 and $-2$ (which are themselves 5 units apart). For any $x$ BETWEEN $-2$ and 3, this sum of distances equals exactly 5 (the distance between the two fixed points) — but we need it to equal 9, which is MORE than 5, so $x$ must lie OUTSIDE the interval $[-2,3]$. In fact, checking further would reveal an entire continuous RANGE of solutions exists on each side (not just the two isolated points found), since moving further from either fixed point increases the total distance sum smoothly.

By definition, $|x|$ represents the distance of $x$ from zero on a number line — and a distance can NEVER be negative (the smallest possible value of $|x|$ is 0, when $x=0$ itself). Since $-5$ is negative, there is NO value of $x$ for which $|x|$ could ever equal $-5$; the equation has no solution.

$|x|=5$ has solutions $x=5$ and $x=-5$ (both are exactly 5 units from zero). The key difference: $5$ is a valid (non-negative) value for a distance, while $-5$ is not — this is precisely why $|x|=5$ is solvable but $|x|=-5$ is not.

$40, 25, 15, 60, 10$ metres respectively.

Checkpoint 4 (position $60$) violates the regulation, since its distance (60m) exceeds the 50m limit, by $60-50=10$ metres. All other checkpoints (40, 25, 15, 10m) are within the safe range.