OMPT inequalities practice: lesson and 10 problems

Answer: 4

I subtract 4 from both sides: $-3u>-15$. Dividing by $-3$ flips the inequality sign, so $u<5$. The largest integer strictly below 5 is 4. Check with $u=4$: $4-12=-8>-11$, while $u=5$ gives $-11>-11$, which is false.

Answer: A — $r\ge -3$

I divide both sides by $-2$ first, flipping the sign: $r-1\ge -4$. Adding 1 gives $r\ge -3$. A spot check with $r=0$: $-2(-1)=2\le 8$, true, and 0 is indeed in $r\ge -3$.

Answer: Step 1

Step 1 multiplies by $x$ without knowing its sign. If $x$ is negative, the inequality must flip. Splitting into cases: for $x>0$ we get $x>3$; for $x<0$ the left side $6/x$ is already negative, so it is automatically less than 2. The full solution is $x<0$ or $x>3$ — the student lost the entire negative branch. Try $x=-1$: $6/(-1)=-6<2$, clearly true.

Answer: A — $-3<k<3$

I factor: $(k-3)(k+3)<0$. A product of two factors is negative exactly when the factors have opposite signs, which happens between the roots. So $-3<k<3$. A parabola picture confirms it: $k^2-9$ opens upward and dips below zero only between its zeros $\pm 3$.

Answer: 8

The absolute value inequality unpacks to $-7\le 2d-1\le 7$. Adding 1 everywhere gives $-6\le 2d\le 8$, and dividing by 2 gives $-3\le d\le 4$. The integers in that range are $-3,-2,-1,0,1,2,3,4$ — eight of them.

Answer: 9

With $x$ kilometres the cost is $3.50+1.90x\le 22$, so $1.90x\le 18.50$ and $x\le \mathrm{9.73...}$. Since I need a whole number of kilometres, the answer is 9. Checking: 9 km costs $3.50+17.10=\text{€}20.60$, within budget, while 10 km would cost $\text{€}22.50$.

Answer: 3

Expand the left side: $3n-6-7=3n-13$, so the inequality is $3n-13>5n-20$. I move the $n$-terms to the right, where the coefficient stays positive: $-13+20>2n$, so $7>2n$ and $n<3.5$. The largest integer strictly below 3.5 is 3. Check $n=3$: left $3-7=-4$, right $15-20=-5$, and $-4>-5$ holds. Check $n=4$: left $6-7=-1$, right $0$, and $-1>0$ fails.

Answer: A — $-2<x<4$

The $x^2$ coefficient is negative, so I multiply by $-1$ and flip: $x^2-2x-8<0$. Factor: $(x-4)(x+2)<0$, boundary points $x=4$ and $x=-2$. An upward parabola is below zero between its roots, so $-2<x<4$. Check $x=0$ in the original: $8>0$, true. Check $x=5$: $-25+10+8=-7$, false. Without the flip you can also read it from the downward parabola directly: it is above the axis between its roots.

Answer: 6

First condition: $|x+2|\le 6$ unpacks to $-6\le x+2\le 6$, so $-8\le x\le 4$. Second condition: $x^2>9$ means $x<-3$ or $x>3$, the outside region of an upward parabola with roots $\pm 3$. Now intersect. From $-8$ to $4$, the integers below $-3$ are $-8,-7,-6,-5,-4$, five of them; the integers above 3 are just $4$, one. Total 6. Check the edge $x=-3$: $9>9$ is false, so it is correctly left out. Check $x=4$: $|6|\le 6$ and $16>9$, both true.

Answer: A — $p>4$

Positive for every $x$ needs two things: the parabola opens upward, so $p>0$, and it never reaches the axis, so $D<0$. Here $D=16-4p$, and $16-4p<0$ gives $p>4$, which already forces $p>0$. So the answer is $p>4$. Check $p=5$: the minimum sits at $x=\frac{4}{10}=\frac{2}{5}$ and equals $5\cdot \frac{4}{25}-\frac{8}{5}+1=\frac{1}{5}>0$. Check $p=4$: $4x^2-4x+1=(2x-1{)}^2$, which is 0 at $x=\frac{1}{2}$, so the strict inequality fails there.