OMPT drill — Algebra

Linear equations

Isolate the unknown, keep both sides honest. Sounds trivial until a fraction or a bracketed negative shows up; that is where I watch students drop easy points. 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

A linear equation is a balance with one unknown in it. Whatever sits on the left equals whatever sits on the right, and every legal move keeps that true: add the same thing to both sides, subtract it, multiply or divide both sides by the same non-zero number. That is the entire theory. Solving $3 x - 5 = 7$ means undoing what was done to $x$ , in reverse order: the 5 was subtracted last, so add it back first ( $3 x = 12$ ), then divide by 3 ( $x = 4$ ).

Because the theory is so small, the OMPT does not test whether you know it. It tests whether you can run it without slipping when the equation is dressed up: brackets to expand, fractions to clear, the unknown on both sides at once. Nearly every wrong answer I mark on this topic is a sign error or a step done to one side only. So the skill you are actually training here is bookkeeping under light time pressure.

Problem 1Numeric answer

Solve for $k$ : $3 (2 k - 5) = k + 10$ .

Show the worked solution

Answer: 5

I start by expanding the left side: $6 k - 15 = k + 10$ . Then I move the $k$ -terms to one side, $5 k = 25$ , and divide by 5 to get $k = 5$ . A quick check: $3 (10 - 5) = 15$ and $5 + 10 = 15$ , so it holds.

Problem 2Multiple choice

For which value of $p$ does the equation $p x + 4 = 2 x + 7$ have no solution at all?

A $p = 2$
B $p = - 2$
C $p = \frac{7}{4}$
DNo such value exists

Show the worked solution

Answer: A — $p = 2$

I rewrite the equation as $(p - 2) x = 3$ . If $p \neq = 2$ I can divide and get exactly one solution. But when $p = 2$ the left side collapses to $0 = 3$ , which is false for every $x$ — no solution. So $p = 2$ is the answer.

Problem 3Numeric answer

A phone plan costs a fixed $€ 12$ per month plus $€ 0.08$ per minute of calling. In one month the bill came to $€ 19.20$ . How many minutes were called?

Show the worked solution

Answer: 90

I let $m$ be the minutes and write the bill as $12 + 0.08 m = 19.20$ . Subtracting 12 gives $0.08 m = 7.20$ , and dividing by 0.08 gives $m = 90$ minutes. I always subtract the fixed part first — the per-minute rate only applies to the variable part of the bill.

Problem 4Multiple choice

Solve for $m$ : $\frac{m - 3}{4} = \frac{2 m + 1}{6}$ .

A $m = - 11$
B $m = 11$
C $m = - 7$
D $m = \frac{7}{2}$

Show the worked solution

Answer: A — $m = - 11$

I clear both denominators by multiplying through by 12: $3 (m - 3) = 2 (2 m + 1)$ , so $3 m - 9 = 4 m + 2$ . Collecting terms gives $- 11 = m$ . Checking: $(- 11 - 3) /4 = - 3.5$ and $(- 22 + 1) /6 = - 3.5$ . It matches.

Problem 5Spot the error

A student solves $5 - 2 t = 11$ like this. Step 1: $- 2 t = 11 - 5$ . Step 2: $- 2 t = 6$ . Step 3: $t = 3$ . Which step contains the error?

Show the worked solution

Answer: Step 3

Steps 1 and 2 are fine: subtracting 5 from both sides gives $- 2 t = 6$ . The slip is in Step 3. Dividing 6 by $- 2$ gives $t = - 3$ , not $t = 3$ — the student dropped the minus sign on the coefficient. Substituting $t = - 3$ back in: $5 - 2 (- 3) = 5 + 6 = 11$ . Correct.

Problem 6Numeric answer

Two numbers add up to 26. The larger number is 5 more than twice the smaller. What is the larger number?

Show the worked solution

Answer: 19

I call the smaller number $x$ , so the larger is $2 x + 5$ . Their sum gives $x + 2 x + 5 = 26$ , so $3 x = 21$ and $x = 7$ . The larger number is then $2 \cdot 7 + 5 = 19$ . I check the sum: $7 + 19 = 26$ .