OMPT drill — Algebra

Inequalities

One habit decides everything here: flip the sign when you multiply or divide by a negative. Quadratic inequalities add a sketch-the-parabola step most students skip at their peril. Work through all six below before moving on; algebra slips compound into every other strand.

Tested inOMPT-A OMPT-B OMPT-C OMPT-D OMPT-E OMPT-F OMPT-G

Lesson

An inequality is solved almost exactly like an equation, same legal moves, same goal of isolating $x$ , with one exception that decides everything: when you multiply or divide both sides by a negative number, the inequality sign flips. $- 2 x < 6$ becomes $x > - 3$ , not $x < - 3$ . Why? Multiplying by a negative mirrors the number line, and mirroring swaps left and right. If $3 < 5$ , then $- 3 > - 5$ .

That single rule accounts for most lost points on linear inequalities, and it fails silently: the wrong answer looks just as tidy as the right one. The second half of this topic, quadratic inequalities, needs a different habit entirely: you stop pushing symbols around and draw a two-second sketch of a parabola instead. Students who skip the sketch and try to solve quadratics by symbol-pushing alone get the boundary points right and the region wrong, which scores zero.

Problem 1Numeric answer

What is the largest integer $u$ that satisfies $4 - 3 u > - 11$ ?

Show the worked solution

Answer: 4

I subtract 4 from both sides: $- 3 u > - 15$ . 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.

Problem 2Multiple choice

Solve $- 2 (r - 1) \leq 8$ .

A $r \geq - 3$
B $r \leq - 3$
C $r \geq 3$
D $r \leq 5$

Show the worked solution

Answer: A — $r \geq - 3$

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

Problem 3Spot the error

A student solves $\frac{6}{x} < 2$ like this. Step 1: multiply both sides by $x$ to get $6 < 2 x$ . Step 2: divide by 2 to get $3 < x$ . Step 3: conclusion, the solution is $x > 3$ . Which step contains the error?

Show the worked solution

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.

Problem 4Multiple choice

Solve $k^{2} - 9 < 0$ .

A $- 3 < k < 3$
B $k < - 3$ or $k > 3$
C $k < 3$
D $k < 9$

Show the worked solution

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 $\pm 3$ .

Problem 5Numeric answer

How many integers $d$ satisfy $∣2 d - 1∣ \leq 7$ ?

Show the worked solution

Answer: 8

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

Problem 6Numeric answer

A taxi ride costs $€ 3.50$ plus $€ 1.90$ per kilometre. You have $€ 22$ . What is the largest whole number of kilometres you can afford?

Show the worked solution

Answer: 9

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