EQUATIONS

QUESTION 1 [6 marks] — Criterion A Medium

Solve each equation, where $x$ appears more than once:

a. $5x+3=2x+18$

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$3x=15 \Rightarrow x=5$

b. $7x-4=3x+20$

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$4x=24 \Rightarrow x=6$

c. $9-2x=x+3$

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$6=3x \Rightarrow x=2$

QUESTION 2 [3 marks] — Criterion A Medium

Solve $3(x+2)=x+14$, where $x$ appears on both sides (one inside brackets).

a. Expand the left side first.

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$3x+6=x+14$

b. Collect $x$ terms and solve.

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$3x-x=14-6 \Rightarrow 2x=8 \Rightarrow x=4$

QUESTION 3 [5 marks] — Criterion B Medium

Investigate whether it matters WHICH side you move the $x$ terms to when solving an equation with $x$ on both sides.

a. Solve $6x+2=2x+18$ by moving the $x$ terms to the LEFT side (subtracting $2x$ from both sides).

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$4x+2=18 \Rightarrow 4x=16 \Rightarrow x=4$

b. Now solve the SAME equation by instead moving the $x$ terms to the RIGHT side (subtracting $6x$ from both sides).

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$2=-4x+18 \Rightarrow -16=-4x \Rightarrow x=4$

c. Compare your two answers, and explain why the choice of which side to collect $x$ terms on doesn't affect the final solution.

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Both give $x=4$ — the choice is just a matter of preference/convenience (usually choosing the side that avoids negative coefficients), since either path represents valid, balanced operations leading to the same true solution.

QUESTION 4 [5 marks] — Criterion C Medium

A student solving $8x-3=3x+22$ moves terms around and writes '$8x-3x=22-3$', getting $5x=19$.

a. Identify the student's error (a sign mistake when moving terms), and give the correct working.

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When moving $-3$ from the left side to the right, it should become $+3$ (not stay as $-3$ to subtract), and when moving $3x$ from the right to the left, it becomes $-3x$. Correct: $8x-3x=22+3 \Rightarrow 5x=25 \Rightarrow x=5$.

b. Verify $x=5$ by substituting into the original equation.

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$8(5)-3=37$ and $3(5)+22=37$ — both sides equal 37 ?.

QUESTION 5 [5 marks] — Criterion D Medium

Two mobile phone plans are being compared. Plan A costs \$20 flat plus \$0.10 per minute. Plan B costs \$35 flat plus \$0.05 per minute.

a. Write an equation setting the two plans' total costs equal (using $m$ for minutes), and solve to find the number of minutes at which both plans cost the same.

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$20+0.10m=35+0.05m \Rightarrow 0.05m=15 \Rightarrow m=300$ minutes.

b. For a customer who typically uses 500 minutes per month, which plan is cheaper? Justify your answer using your equation's result.

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Since the break-even point is 300 minutes, and 500 minutes is MORE than this, the plan with the lower per-minute rate (Plan B) becomes cheaper beyond 300 minutes — so Plan B is the better choice for a 500-minute user.

QUESTION 6 [3 marks] — Criterion A Medium

Solve $9x-4=5x+20$, where $x$ appears on both sides.

a. Collect the $x$ terms and constants, then solve.

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$9x-5x=20+4 \Rightarrow 4x=24 \Rightarrow x=6$

QUESTION 7 [7 marks] — Criterion B Hard

Investigate what happens when solving an equation with $x$ on both sides leads to the $x$ terms CANCELLING OUT completely, using $6x+12=2x+12$ vs $4x+7=4x+9$.

a. Solve $6x+12=2x+12$ using standard balance operations, and interpret the result.

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$6x-2x=12-12 \Rightarrow 4x=0 \Rightarrow x=0$. This has exactly ONE valid solution ($x=0$).

b. Now attempt to solve $4x+7=4x+9$. What happens when you try to collect the $x$ terms, and what does this tell you about how many solutions this equation has?

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Subtracting $4x$ from both sides: $7=9$ — this is a FALSE statement (7 does not equal 9), regardless of what $x$ is. This means the equation $4x+7=4x+9$ has NO SOLUTION at all — there is no value of $x$ that could ever make both sides equal, since the $x$ terms cancel identically but leave behind two DIFFERENT constants.

QUESTION 8 [6 marks] — Criterion B Hard

Investigate the equation $3(x+2)=3x+6$, where the $x$ terms ALSO cancel — but compare the result to the 'no solution' case above.

a. Expand the left side, then attempt to collect $x$ terms as usual.

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$3x+6=3x+6$. Subtracting $3x$ from both sides: $6=6$ — this is a TRUE statement, for any value of $x$.

b. Explain what this result means: how many solutions does $3(x+2)=3x+6$ actually have, and why is this a fundamentally DIFFERENT outcome from the 'no solution' case investigated earlier, even though both involved the $x$ terms cancelling?

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This equation has INFINITELY MANY solutions — EVERY value of $x$ satisfies it, since $3(x+2)$ is simply the EXPANDED (equivalent) form of $3x+6$; they're not two different expressions that happen to be equal, they're the SAME expression written two ways. This is fundamentally different from the 'no solution' case: there, the $x$ terms cancelled leaving a FALSE constant statement (impossible for any $x$); here, they cancel leaving a TRUE constant statement (true for every $x$) — the key difference is whether the leftover constant equation is true or false.

QUESTION 9 [3 marks] — Criterion C Medium

A student solving $7x-3=7x+5$ doesn't notice the $x$ terms are identical, and after subtracting $7x$ from both sides gets '$-3=5$', then writes 'x = -3/5 or something, I'm confused' without recognizing what this actually means.

a. Explain to the confused student what '$-3=5$' (a statement with NO variable left at all) actually indicates about the ORIGINAL equation's solution set, correcting their attempt to still solve for a numerical $x$.

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When ALL the variable terms cancel and you're left with a FALSE numerical statement (like $-3=5$, which is never true), this means the original equation has NO SOLUTION whatsoever — there is no value of $x$, no matter how it's calculated, that could make the original equation true. The student's instinct to 'solve for $x$' from this point is misguided, since there's no $x$ left in the statement to solve for; the correct response is to conclude 'no solution', not attempt further algebra.

QUESTION 10 [3 marks] — Criterion C Medium

A classmate claims: 'if solving an equation ever gives a statement with no $x$ left in it, that ALWAYS means there's no solution.'

a. Explain why this claim is only PARTIALLY correct, using the earlier finding that $3(x+2)=3x+6$ (which also loses its $x$ terms during solving) actually has INFINITELY MANY solutions, not zero.

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The classmate's claim is incomplete — losing the $x$ terms during solving can lead to TWO very different outcomes: if the remaining constant statement is FALSE (like $-3=5$), there's no solution; but if the remaining constant statement is TRUE (like $6=6$, from $3(x+2)=3x+6$), the equation actually has INFINITELY MANY solutions (every value of $x$ works). The key isn't just 'the $x$ disappeared' — it's whether the LEFTOVER constant statement is true or false that determines the outcome.

QUESTION 11 [6 marks] — Criterion D Hard

Two ride-share drivers compare pay structures. Driver A earns a flat \$80 per shift plus \$1.50 per km driven. Driver B earns a flat \$50 per shift plus \$2.10 per km driven.

a. Write an equation setting the two drivers' total pay equal (using $k$ for km driven), and solve to find the break-even distance.

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$80+1.50k=50+2.10k \Rightarrow 30=0.60k \Rightarrow k=50$km.

b. For a shift involving 80km of driving, determine which driver earns MORE, and by how much, justifying your answer using the break-even point found in part (a).

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Since 80km EXCEEDS the 50km break-even point, and Driver B has the HIGHER per-km rate (\$2.10 vs \$1.50), Driver B earns more beyond the break-even distance. Driver A: $80+1.50(80)=80+120=\$200$. Driver B: $50+2.10(80)=50+168=\$218$. Driver B earns \$18 more.

QUESTION 12 [7 marks] — Criterion D Hard

A gym offers two membership plans. Plan X: \$30/month plus \$8 per visit. Plan Y: \$60/month plus \$4 per visit.

a. Find the number of visits per month at which BOTH plans cost the same.

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$30+8v=60+4v \Rightarrow 4v=30 \Rightarrow v=7.5$ visits.

b. Since visits must be a WHOLE number, explain what this non-whole break-even point (7.5) means practically for someone deciding between the two plans — specifically, which plan is cheaper for someone visiting 7 times per month, and which is cheaper for 8 times per month?

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At 7 visits: Plan X $=30+56=\$86$; Plan Y $=60+28=\$88$ — Plan X is cheaper. At 8 visits: Plan X $=30+64=\$94$; Plan Y $=60+32=\$92$ — Plan Y is cheaper. The non-whole break-even (7.5) means the CHEAPER plan actually SWITCHES between 7 and 8 visits — Plan X is better for LIGHTER gym use (?7 visits/month), while Plan Y becomes better once visits reach 8 or more per month.

QUESTION 13 [7 marks] — Criterion D Hard

Two internet providers offer plans. Provider A: \$600 setup fee plus \$9 per month. Provider B: \$900 setup fee plus \$15 per month, but includes a free premium router (value irrelevant to the cost comparison here).

a. Write an equation setting the total costs equal after $m$ months, and solve for $m$.

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$$600+9m=900+15m$$ Solving: $600-900=15m-9m \Rightarrow -300=6m \Rightarrow m=-50$ — a NEGATIVE result.

b. Explain what a NEGATIVE 'break-even time' means in this context, and what it tells a customer about which plan is cheaper for EVERY realistic (positive) number of months.

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A negative break-even point means the two costs NEVER become equal for any realistic (positive) number of months — since Provider A has BOTH a lower setup fee AND a lower monthly rate, Provider A is cheaper for every possible positive value of $m$, with the cost gap only growing larger over time. The negative solution is a mathematical signal that the 'equal cost' scenario would only occur in a nonsensical negative-time scenario, confirming Provider A dominates entirely in this case.

Equations with a repeated unknown