Simultaneous equations — substitution, elimination, cross-multiplication: Exercises

This page is a self-test ladder. Each problem lives above a collapsible solution — try it first, then reveal. The levels climb from "can you spot the method" (L1) up to "can you invent a strategy" (L5). Return to the parent topic whenever a step feels unfamiliar.

Figure — Simultaneous equations — substitution, elimination, cross-multiplication

Level 1 — Recognition

Here you only need to see which variable is easiest to grab and follow the machinery.

Exercise 1.1

Solve by substitution: ${y = 3 x - 1 ...(1) 2 x + y = 9 ...(2)$

Recall Solution 1.1

Equation (1) already tells us what $y$ is: $y = 3 x - 1$ . That is a ready-made "definition of $y$ ".

Substitute it into (2), replacing the letter $y$ with the whole expression: $2 x + (3 x - 1) = 9$ Why? Now every symbol in the equation is $x$ — one letter, one equation, solvable. $5 x - 1 = 9 \Rightarrow 5 x = 10 \Rightarrow x = 2$ Put $x = 2$ back into (1): $y = 3 (2) - 1 = 5$ . Answer: $x = 2, y = 5$ . Check (2): $2 (2) + 5 = 9$ ✓

Exercise 1.2

Solve by elimination: ${x + y = 10 ...(1) x - y = 4 ...(2)$

Recall Solution 1.2

Look at the $y$ terms: $+ y$ and $- y$ . They are already exact opposites, so adding the equations kills $y$ with no extra multiplying. $(x + y) + (x - y) = 10 + 4 \Rightarrow 2 x = 14 \Rightarrow x = 7$ Sub into (1): $7 + y = 10 \Rightarrow y = 3$ . Answer: $x = 7, y = 3$ . Check (2): $7 - 3 = 4$ ✓

Level 2 — Application

Now coefficients are not gift-wrapped; you must set up the method yourself.

Exercise 2.1

Solve by substitution: ${2 x + y = 7 ...(1) 3 x - 2 y = 4 ...(2)$

Recall Solution 2.1

In (1), $y$ has coefficient $1$ , so isolating it costs no fractions: $y = 7 - 2 x$ . Substitute into (2): $3 x - 2 (7 - 2 x) = 4$ Why the brackets? We replace the single letter $y$ by an expression, so it must travel inside parentheses to keep the $- 2$ multiplying all of it. $3 x - 14 + 4 x = 4 \Rightarrow 7 x = 18 \Rightarrow x = \frac{18}{7}$ Back-substitute: $y = 7 - 2 \cdot \frac{18}{7} = \frac{49 - 36}{7} = \frac{13}{7}$ . Answer: $x = \frac{18}{7}, y = \frac{13}{7}$ . Check (2): $3 \cdot \frac{18}{7} - 2 \cdot \frac{13}{7} = \frac{54 - 26}{7} = \frac{28}{7} = 4$ ✓

Exercise 2.2

Solve by elimination: ${3 x + 4 y = 10 ...(1) 5 x - 2 y = 4 ...(2)$

Recall Solution 2.2

Target the $y$ terms: $+ 4 y$ and $- 2 y$ . Multiply (2) by $2$ so it becomes $- 4 y$ — then $y$ cancels on adding. $(1) : 3 x + 4 y = 10, 2 \times (2) : 10 x - 4 y = 8$ Add: $13 x = 18 \Rightarrow x = \frac{18}{13}$ Sub into (1): $3 \cdot \frac{18}{13} + 4 y = 10 \Rightarrow \frac{54}{13} + 4 y = 10$ $4 y = \frac{130 - 54}{13} = \frac{76}{13} \Rightarrow y = \frac{19}{13}$ Answer: $x = \frac{18}{13}, y = \frac{19}{13}$ . Check (2): $5 \cdot \frac{18}{13} - 2 \cdot \frac{19}{13} = \frac{90 - 38}{13} = \frac{52}{13} = 4$ ✓

Exercise 2.3

Solve by cross-multiplication: ${2 x + 3 y = 8 5 x - 4 y = 3$

Recall Solution 2.3

First move everything to the form $a x + b y - c = 0$ is optional; here just read off with the RHS as $c$ : $a_{1} = 2, b_{1} = 3, c_{1} = 8, a_{2} = 5, b_{2} = - 4, c_{2} = 3.$ Denominator (cross-multiply the $a$ and $b$ columns): $a_{1} b_{2} - a_{2} b_{1} = 2 (- 4) - 5 (3) = - 8 - 15 = - 23.$ For $x$ (cross-multiply the $b$ and $c$ columns): $x = \frac{b _{1} c _{2} - b _{2} c _{1}}{a _{1} b _{2} - a _{2} b _{1}} = \frac{3 ( 3 ) - ( - 4 ) ( 8 )}{- 23} = \frac{9 + 32}{- 23} = \frac{41}{- 23} = \frac{41}{23} \cdot (- 1)?$ Careful with the double negative. Both numerator and denominator carry signs. Compute cleanly: $x = \frac{41}{- 23} = - \frac{41}{23} — but let us verify, because signs bite here.$ Verify with elimination: $4 \times$ (eqn 1) $= 8 x + 12 y = 32$ ; $3 \times$ (eqn 2) $= 15 x - 12 y = 9$ ; add $\Rightarrow 23 x = 41 \Rightarrow x = \frac{41}{23}$ (positive!). The mismatch means the arrangement of the formula matters: with the constant on the right, the correct cross-multiplication reads $x = \frac{b _{1} c _{2} - b _{2} c _{1}}{a _{1} b _{2} - a _{2} b _{1}}$ only when the system is written as $a x + b y = c$ consistently. Recomputing carefully: numerator $= 3 (3) - (- 4) (8) = 41$ , denominator $= - 23$ , giving $- \frac{41}{23}$ — which contradicts elimination. So the safe value from elimination is $x = \frac{41}{23}$ . Lesson: the cross-multiplication sign convention flips depending on whether you write $a x + b y = c$ or $a x + b y - c = 0$ . Always sanity-check one variable against elimination. Now $y$ from elimination: sub $x = \frac{41}{23}$ into $2 x + 3 y = 8$ : $3 y = 8 - \frac{82}{23} = \frac{184 - 82}{23} = \frac{102}{23} \Rightarrow y = \frac{34}{23} .$ Answer: $x = \frac{41}{23}, y = \frac{34}{23}$ . Check (eqn 2): $5 \cdot \frac{41}{23} - 4 \cdot \frac{34}{23} = \frac{205 - 136}{23} = \frac{69}{23} = 3$ ✓

Level 3 — Analysis

Now we ask what kind of system we have before solving.

Exercise 3.1

How many solutions does this system have? Justify using the determinant. ${2 x + 3 y = 5 4 x + 6 y = 11$

Recall Solution 3.1

Compute the determinant $D = a_{1} b_{2} - a_{2} b_{1} = 2 (6) - 4 (3) = 12 - 12 = 0.$ $D = 0$ means the lines are parallel or identical — never a single crossing. To decide which, compare ratios: $\frac{a _{1}}{a _{2}} = \frac{2}{4} = \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{3}{6} = \frac{1}{2}$ , but $\frac{c _{1}}{c _{2}} = \frac{5}{11} \neq = \frac{1}{2}$ . Since the $x, y$ ratios match but the constant ratio does not, the lines are parallel and distinct — no solution (inconsistent). Answer: No solution.

Exercise 3.2

For what value of $k$ does the system have no unique solution? ${3 x + k y = 6 6 x + 4 y = 12$

Recall Solution 3.2

A unique solution fails exactly when $D = 0$ : $D = 3 (4) - 6 (k) = 12 - 6 k = 0 \Rightarrow k = 2.$ What does $k = 2$ look like? Then eqn 1 is $3 x + 2 y = 6$ ; doubling gives $6 x + 4 y = 12$ — the same line as eqn 2. So at $k = 2$ the two lines coincide: infinitely many solutions. Answer: $k = 2$ .

Level 4 — Synthesis

Here the equations are disguised. You must transform them into linear form first.

Exercise 4.1

Solve: $⎩ ⎨ ⎧ \frac{2}{x} + \frac{3}{y} = 13 \frac{5}{x} - \frac{4}{y} = - 2 (x, y \neq = 0)$

Recall Solution 4.1

These are not linear in $x, y$ — but they ARE linear in $\frac{1}{x}$ and $\frac{1}{y}$ . Substitute $u = \frac{1}{x}, v = \frac{1}{y}$ : ${2 u + 3 v = 13 5 u - 4 v = - 2$ Eliminate $v$ : multiply row 1 by $4$ , row 2 by $3$ : $8 u + 12 v = 52$ and $15 u - 12 v = - 6$ . Add: $23 u = 46 \Rightarrow u = 2.$ Then $2 (2) + 3 v = 13 \Rightarrow 3 v = 9 \Rightarrow v = 3.$ Undo the substitution: $u = \frac{1}{x} = 2 \Rightarrow x = \frac{1}{2}$ ; $v = \frac{1}{y} = 3 \Rightarrow y = \frac{1}{3}$ . Answer: $x = \frac{1}{2}, y = \frac{1}{3}$ . Check (row 2): $\frac{5}{1/2} - \frac{4}{1/3} = 10 - 12 = - 2$ ✓

Exercise 4.2

The sum of two numbers is $25$ . Twice the larger exceeds three times the smaller by $5$ . Find them.

Recall Solution 4.2

Translate words to symbols. Let the larger be $x$ , smaller be $y$ .

"Sum is 25": $x + y = 25$ ...(1)
"Twice the larger exceeds three times the smaller by 5": $2 x - 3 y = 5$ ...(2)

Substitution: from (1), $x = 25 - y$ . Into (2): $2 (25 - y) - 3 y = 5 \Rightarrow 50 - 2 y - 3 y = 5 \Rightarrow - 5 y = - 45 \Rightarrow y = 9.$ Then $x = 25 - 9 = 16$ . Answer: larger $= 16$ , smaller $= 9$ . Check: $16 + 9 = 25$ ✓ and $2 (16) - 3 (9) = 32 - 27 = 5$ ✓

Level 5 — Mastery

Now you choose the method and handle a parameter or a proof.

Exercise 5.1

Solve for $x, y$ in terms of the constants $a, b$ (assume $a \neq = b$ ): ${a x + b y = a^{2} + b^{2} x + y = a + b$

Recall Solution 5.1

The second equation is cleanest, so isolate $x = (a + b) - y$ and substitute into the first: $a [(a + b) - y] + b y = a^{2} + b^{2}$ $a^{2} + ab - a y + b y = a^{2} + b^{2}$ Cancel $a^{2}$ from both sides: $ab - a y + b y = b^{2} \Rightarrow ab - b^{2} = a y - b y \Rightarrow b (a - b) = y (a - b) .$ Since $a \neq = b$ , divide by $(a - b)$ : $y = b$ . Then $x = (a + b) - b = a$ . Answer: $x = a, y = b$ . Check (eqn 1): $a (a) + b (b) = a^{2} + b^{2}$ ✓

Exercise 5.2

A boat goes $30$ km downstream and $18$ km upstream in the same total time it takes to go $24$ km down and $24$ km up. If the boat's still-water speed is $10$ km/h, find the stream speed $s$ . (Use downstream speed $10 + s$ , upstream $10 - s$ ; time $= \frac{distance}{speed}$ .)

Recall Solution 5.2

Downstream speed $= 10 + s$ , upstream $= 10 - s$ . Equal-time condition: $\frac{30}{10 + s} + \frac{18}{10 - s} = \frac{24}{10 + s} + \frac{24}{10 - s} .$ Group the $\frac{1}{10 + s}$ and $\frac{1}{10 - s}$ terms: $\frac{30 - 24}{10 + s} = \frac{24 - 18}{10 - s} \Rightarrow \frac{6}{10 + s} = \frac{6}{10 - s} .$ Cancel the $6$ : $\frac{1}{10 + s} = \frac{1}{10 - s} \Rightarrow 10 - s = 10 + s \Rightarrow 2 s = 0 \Rightarrow s = 0.$ Interpretation: the only way those two journeys take equal time is if there is no current ( $s = 0$ ). A genuine, non-trivial answer — the algebra proves the stream must be still. Answer: $s = 0$ km/h (still water).

Recall Self-test roll-up (answers only)

Which value satisfies Ex 1.1? ::: $x = 2, y = 5$ Ex 2.1 answer? ::: $x = \frac{18}{7}, y = \frac{13}{7}$ Ex 2.3 answer (verified)? ::: $x = \frac{41}{23}, y = \frac{34}{23}$ Ex 3.1 — how many solutions? ::: None (parallel distinct lines, $D = 0$ ) Ex 3.2 — value of $k$ ? ::: $k = 2$ Ex 4.1 answer? ::: $x = \frac{1}{2}, y = \frac{1}{3}$ Ex 5.1 answer? ::: $x = a, y = b$ Ex 5.2 stream speed? ::: $s = 0$ km/h