Linear equations in one variable — solving, transposition method: Exercises

Level 1 — Recognition

Can you spot a linear equation in one variable and identify its pieces?

Recall Solution L1.1

A linear equation in one variable needs: one letter, that letter to the first power only, and an equals sign with the variable actually present.

(a) $3 x + 5 = 11$ — one variable $x$ , highest power $1$ . ✅ This is the answer.
(b) $x^{2} - 4 = 0$ — has $x^{2}$ , power $2$ . Not linear (it's quadratic).
(c) $2 x + 3 y = 7$ — two variables $x$ and $y$ . That's two variables, different game.
(d) $7 = 7$ — true, but no variable to solve for.

Answer: (a).

Recall Solution L1.2

Compare to the standard shape $a x + b = c$ from the parent note.

Variable: $x$ (the unknown — see 2.1.01-Variables-and-constants).
Coefficient of $x$ : $a = - 4$ (the number multiplying $x$ ; keep its minus sign).
Constants: $b = 9$ (on the left) and $c = 1$ (on the right).

Level 2 — Application

Run the transposition algorithm on clean equations.

Recall Solution L2.1

Step 1 — move the constant. The $+ 5$ is stuck to $6 x$ ; transpose it across the equals sign, where $+ 5$ becomes $- 5$ : $6 x = 29 - 5 = 24$ Why: subtracting $5$ from both sides is the inverse of $+ 5$ , isolating the $x$ -term. Step 2 — undo the coefficient. $6$ multiplies $x$ , so transpose as division: $x = \frac{24}{6} = 4$ Check: $6 (4) + 5 = 24 + 5 = 29$ ✓. Answer: $x = 4$ .

Recall Solution L2.1

Step 1 — gather variables left, constants right. Transpose $5 x$ (crosses as $- 5 x$ ) and $- 4$ (crosses as $+ 4$ ): $9 x - 5 x = 12 + 4$ Step 2 — simplify each side: $4 x = 16$ Step 3 — divide by the coefficient: $x = \frac{16}{4} = 4$ Check: Left $= 9 (4) - 4 = 32$ ; Right $= 5 (4) + 12 = 32$ ✓. Answer: $x = 4$ .

Recall Solution L2.3

Step 1 — transpose the $15$ : it becomes $- 15$ on the right: $- 3 x = 6 - 15 = - 9$ Step 2 — divide both sides by $- 3$ : $x = \frac{- 9}{- 3} = 3$ Why positive: a negative divided by a negative is positive. Check: $15 - 3 (3) = 15 - 9 = 6$ ✓. Answer: $x = 3$ .

Level 3 — Analysis

Fractions, brackets, and equations that need cleaning before transposition.

Recall Solution L3.1

Step 1 — clear the fraction. Division by $4$ blocks us; multiply both whole sides by $4$ (the denominator crosses as multiplication): $3 x - 2 = 28$ Why the whole side: $4 \times 7 = 28$ , applied to the entire right side, not just part of it. Step 2 — transpose $- 2$ : $3 x = 28 + 2 = 30$ Step 3 — divide by $3$ : $x = \frac{30}{3} = 10$ Check: $\frac{3 ( 10 ) - 2}{4} = \frac{28}{4} = 7$ ✓. Answer: $x = 10$ .

Recall Solution L3.2

Step 1 — expand the bracket (simplify each side first): $2 x + 6 = 5 x - 9$ Step 2 — gather variables and constants. Transpose $5 x$ (→ $- 5 x$ ) and $6$ (→ $- 6$ ): $2 x - 5 x = - 9 - 6$ Step 3 — simplify: $- 3 x = - 15$ Step 4 — divide by $- 3$ : $x = \frac{- 15}{- 3} = 5$ Check: Left $= 2 (5 + 3) = 16$ ; Right $= 5 (5) - 9 = 16$ ✓. Answer: $x = 5$ .

Recall Solution L3.3

Step 1 — combine the fractions. The lowest common denominator of $2$ and $3$ is $6$ : $\frac{3 x}{6} + \frac{2 x}{6} = \frac{5 x}{6} = 10$ Why LCD: to add fractions the denominators must match; $6$ is the smallest that both $2$ and $3$ divide. Step 2 — clear the denominator (multiply both sides by $6$ ): $5 x = 60$ Step 3 — divide by $5$ : $x = \frac{60}{5} = 12$ Check: $\frac{12}{2} + \frac{12}{3} = 6 + 4 = 10$ ✓. Answer: $x = 12$ .

Level 4 — Synthesis

Build the equation yourself, then solve it.

Recall Solution L4.1

Step 1 — form the equation (this is the skill from 2.1.06-Forming-equations-from-word-problems). Let the smallest number be $x$ . Consecutive numbers step up by $1$ : $x + (x + 1) + (x + 2) = 72$ Step 2 — simplify the left side: $3 x + 3 = 72$ Step 3 — transpose $+ 3$ (→ $- 3$ ): $3 x = 69$ Step 4 — divide by $3$ : $x = 23$ Step 5 — report all three: $23, 24, 25$ . Check: $23 + 24 + 25 = 72$ ✓. Answer: $23, 24, 25$ .

Recall Solution L4.2

Step 1 — name the unknown. Let the son's present age be $x$ . Then the father is $3 x$ . Step 2 — translate "in $12$ years, father is twice the son." Add $12$ to each age: $3 x + 12 = 2 (x + 12)$ Step 3 — expand the bracket: $3 x + 12 = 2 x + 24$ Step 4 — transpose $2 x$ (→ $- 2 x$ ) and $12$ (→ $- 12$ ): $3 x - 2 x = 24 - 12$ $x = 12$ Step 5 — report both: son $= 12$ , father $= 3 (12) = 36$ . Check: in $12$ years, son $= 24$ , father $= 48 = 2 \times 24$ ✓. Answer: son $12$ , father $36$ .

Recall Solution L4.3

Step 1 — match to $a x + b = c x + d$ : here $a = 8$ , $b = 1$ , $c = 3$ , $d = 16$ . Step 2 — plug into the formula: $x = \frac{d - b}{a - c} = \frac{16 - 1}{8 - 3} = \frac{15}{5} = 3$ Step 3 — verify by ordinary transposition: $8 x - 3 x = 16 - 1 \Rightarrow 5 x = 15 \Rightarrow x = 3$ . Same answer — the formula is just the algorithm done symbolically. Check: $8 (3) + 1 = 25$ ; $3 (3) + 16 = 25$ ✓. Answer: $x = 3$ .

Level 5 — Mastery

Reason about equations that are degenerate, parameterised, or trap-laden.

Recall Solution L5.1

Compare to $a x + b = c x + d$ and use the parent's special cases.

(a) $a = c = 4$ and $b = d = 7$ . Transpose: $4 x - 4 x = 7 - 7 \Rightarrow 0 = 0$ , always true. Infinitely many solutions (an identity — every $x$ works).
(b) $a = c = 4$ but $b = 7 \neq = d = 2$ . Transpose: $0 = 2 - 7 = - 5$ , never true. No solution (a contradiction).
(c) $a = 4 \neq = c = 3$ . Genuine: $4 x - 3 x = 2 - 7 \Rightarrow x = - 5$ . One solution, $x = - 5$ . Check (c): $4 (- 5) + 7 = - 13$ ; $3 (- 5) + 2 = - 13$ ✓.

Recall Solution L5.2

This is not linear (it has $x^{2}$ ), but it teaches the deepest transposition trap. Wrong path: divide by $x$ to get $x = 5$ — this silently assumes $x \neq = 0$ and throws away a real solution. Right path — transpose everything to one side, then factor: $x^{2} - 5 x = 0 \Rightarrow x (x - 5) = 0$ A product is zero only if a factor is zero, so $x = 0$ or $x = 5$ . Check: $0^{2} = 5 (0) = 0$ ✓ and $5^{2} = 25 = 5 (5)$ ✓. Answer: $x = 0$ or $x = 5$ . Moral: dividing by a variable can secretly divide by zero and delete solutions — exactly why the parent note says never divide by the variable.

Recall Solution L5.3

Match to $a x + b = c x + d$ : here $a = k$ , $b = 3$ , $c = 2$ , $d = 3$ . Note $b = d = 3$ already.

Infinitely many: need $a = c$ and $b = d$ . Since $b = d$ holds, we need $k = 2$ . Then the equation is $2 x + 3 = 2 x + 3$ , true for all $x$ . $k = 2$ .
No solution: need $a = c$ and $b \neq = d$ . But here $b = d$ , so equal coefficients can never give a contradiction — no value of $k$ produces "no solution."
For any $k \neq = 2$ : transpose $\Rightarrow (k - 2) x = 0 \Rightarrow x = 0$ , a single solution. Answer: infinitely many at $k = 2$ ; no value of $k$ gives no solution.

Connections

Linear equations in one variable — solving, transposition method — the parent note these exercises drill.
2.1.01-Variables-and-constants — naming the unknown in L1–L4.
2.1.06-Forming-equations-from-word-problems — the skill powering L4.
2.1.08-Linear-equations-in-two-variables — where L1.1(c) leads next.
2.3.01-Linear-inequalities — the same transposing with $<$ or $>$ .
3.2.04-Graphical-solution-of-linear-equations — the geometry behind L5.1 and L5.3.

Recall Answer key at a glance

L2.1 $x = 4$ · L2.2 $x = 4$ · L2.3 $x = 3$ · L3.1 $x = 10$ · L3.2 $x = 5$ · L3.3 $x = 12$ · L4.1 $23, 24, 25$ · L4.2 son $12$ , father $36$ · L4.3 $x = 3$ · L5.1 (a) infinite (b) none (c) $x = - 5$ · L5.2 $x = 0$ or $5$ · L5.3 $k = 2$ / no $k$ .