Chapter 3 · Pair of Linear Equations in Two Variables

3.3 The Elimination Method and Word Problems

Add or subtract the two equations so that one variable cancels out, then use the method to crack word problems about prices, ages, digits, fractions and speeds.

Back at the sports-day stall from Section 3.1: $2$ samosas and $3$ cups of tea cost ₹ $80$, and $3$ samosas and $2$ cups of tea cost ₹ $95$. You can solve $2x+3y=80$, $3x+2y=95$ by substitution, but no coefficient is $1$, so you would have to carry the fraction $x=\dfrac{80-3y}{2}$ through the working. There is a neater way. Compare the two bills directly: if one bill contained exactly as many samosas as the other, the difference in price would be caused by the tea alone, and you could find the price of a cup of tea at once.

That is the elimination method: multiply the equations so that one variable has the same coefficient (or opposite coefficients) in both, then subtract (or add) the equations so that this variable disappears. For example, adding $x+y=7$ and $x-y=1$ gives $2x=8$ in one line, because $+y$ and $-y$ cancel. In the second half of this section you will use substitution and elimination to solve word problems on prices, ages, digits, fractions and speeds.

The elimination method

Elimination rests on one simple fact: if two equations are both true, then their sum and their difference are true as well. And multiplying both sides of an equation by the same non-zero number does not change which points satisfy it.

📐 Theorem 3.3 — Operations that keep the solutions
Start with a pair of linear equations in $x$ and $y$. Each of these operations produces a new pair with exactly the same solutions as the original pair:
  1. multiplying both sides of one equation by a non-zero number $k$;
  2. replacing one equation by its sum with (or its difference from) the other equation, or with a multiple of the other.

Why it works. If $(x_0, y_0)$ makes both equations true, it makes any sum, difference or multiple of them true, so every old solution is a new solution. Each operation can also be undone (divide by $k\ne0$; subtract what you added), so every new solution is an old one too. Note that $k$ must not be $0$: multiplying by $0$ turns any equation into $0=0$ and throws its information away.

📘 Definition 3.5 — The elimination method
To solve a pair of linear equations by elimination, multiply one or both equations by suitable non-zero numbers so that the coefficients of one variable become equal (or opposites). Then subtract (or add) the equations to eliminate that variable, solve the resulting equation in one variable, and substitute the value into either original equation to find the other variable.
🧭 Problem-solving strategy — solving a pair by elimination
  1. Line up both equations as $ax+by=c$, with the $x$-terms, $y$-terms and constants in columns.
  2. Choose the variable to eliminate. Multiply the equations (every term!) so its coefficients match. A safe choice: multiply each equation by the other equation's coefficient of that variable, or use their LCM.
  3. Same signs → subtract; opposite signs → add. The chosen variable cancels.
  4. Solve the one-variable equation, then substitute into the simpler original equation for the other variable.
  5. Check in both original equations. If both variables cancel, use Theorem 3.2: a false statement means no solution, a true one means infinitely many.
✏️ Example 1 — The samosa-and-tea problem
Solve $2x+3y=80$ and $3x+2y=95$, where ₹ $x$ is the price of a samosa and ₹ $y$ the price of a cup of tea.
Match x Make the samosa counts equal: multiply the first equation by $3$ and the second by $2$: $$6x+9y=240$$ $$6x+4y=190$$
Subtract Same sign on $6x$, so subtract: $(6x+9y)-(6x+4y)=240-190$, so $5y=50$ and $y=10$. (Six samosas cost the same on both "bills", so the extra ₹ $50$ pays for $5$ extra cups of tea.)
Substitute In $2x+3y=80$: $2x+30=80$, $2x=50$, $x=25$.
Check $2(25)+3(10)=80$ ✓ and $3(25)+2(10)=95$ ✓.
Answer A samosa costs ₹ $25$ and a cup of tea ₹ $10$, as promised in Section 3.1.
A neat shortcut Because the coefficients are swapped ($2, 3$ and $3, 2$), you can also add: $5x+5y=175$, so $x+y=35$; and subtract: $-x+y=-15$, so $x-y=15$. Adding these gives $2x=50$: the same answer. Watch for this pattern in exam questions.
✏️ Example 2 — Opposite signs: add
Solve by elimination: $5x-3y=11$ and $3x+2y=18$.
Choose The $y$-coefficients $-3$ and $+2$ have opposite signs, so eliminating $y$ needs an addition. Multiply the first equation by $2$ and the second by $3$: $$10x-6y=22$$ $$9x+6y=54$$
Add $19x=76$, so $x=4$.
Substitute In $3x+2y=18$: $12+2y=18$, $y=3$.
Check $5(4)-3(3)=20-9=11$ ✓ and $3(4)+2(3)=12+6=18$ ✓. Solution: $x=4$, $y=3$.
Interactive 1 — What Elimination Does to the Lines

The purple line is the combination "equation 1 $+\ t\times$ equation 2". By Theorem 3.3 it passes through the same crossing point for every $t$, so as you move $t$ it swings around the solution. When the $y$-terms cancel, the purple line is vertical ($x=$ number); when the $x$-terms cancel, it is horizontal ($y=$ number). That is elimination, seen on a graph. Press Play, or jump straight to a cancelling value.

1/2
combined equation—
$x$-coefficient—
$y$-coefficient—

Choosing a method, and spotting the special cases

Both algebraic methods always give the same answer, so choose the one that keeps the arithmetic easy:

With elimination, as with substitution, both variables may cancel. Theorem 3.2 still tells you what that means: a false statement such as $0=3$ means no solution; a true statement such as $0=0$ means infinitely many.

✏️ Example 3 — When both variables cancel
Use elimination to decide how many solutions each pair has.
(a) $3x-2y=4$ and $6x-4y=5$    (b) $2x+5y=7$ and $6x+15y=21$
(a) Match x Multiply the first equation by $2$: $6x-4y=8$. Subtract the second equation, $6x-4y=5$: $0=3$.
(a) Conclude $0=3$ is false, so the pair has no solution: the lines are parallel. (Ratio check: $\tfrac36=\tfrac{-2}{-4}=\tfrac12\ne\tfrac{-4}{-5}=\tfrac45$.)
(b) Match x Multiply the first equation by $3$: $6x+15y=21$. Subtract the second: $0=0$.
(b) Conclude $0=0$ is always true: the equations describe the same line, so there are infinitely many solutions, for example $(1, 1)$, $(6, -1)$, $\left(0, \tfrac75\right)$.
⚠️ Common mistakes with elimination

1. Multiplying only the left side. $2\times(2x+3y=80)$ is $4x+6y=160$, not $4x+6y=80$. Multiply every term, including the constant.

2. Sign slips when subtracting. $(5x-3y)-(2x-7y)=3x+4y$, because $-(-7y)=+7y$. Write the subtraction in brackets, or multiply one equation by $-1$ and add instead.

3. Adding when you should subtract. Equal coefficients with the same sign cancel by subtraction; opposite signs cancel by addition. Check that the variable really disappears before going on.

4. Not checking. Substitute both values into both original equations, not into the multiplied versions: an error in the multiplying step will not show up if you check against your own working.

Word problems

Most real problems arrive in words. The algebra is the easy part; the skill is translating carefully. Here are the standard ways to write the common quantities:

SituationWrite it as
a two-digit number with tens digit $x$ and units digit $y$$10x+y$; with the digits reversed, $10y+x$
a fraction with numerator $x$ and denominator $y$$\dfrac{x}{y}$; "add $2$ to both" gives $\dfrac{x+2}{y+2}$
ages: now $x$ years$n$ years ago: $x-n$; in $n$ years: $x+n$
boat speed $u$ in still water, stream speed $v$downstream: $u+v$; upstream: $u-v$ (needs $u\gt v$)
distance, speed, timedistance $=$ speed $\times$ time
🧭 Problem-solving strategy — word problems with two unknowns
  1. Name the two unknowns, with units: "let the boat's speed in still water be $u$ km/h".
  2. Translate each sentence of information into one equation.
  3. Solve by substitution or elimination, whichever is easier.
  4. Check in the words of the problem, not just in your equations: a slip in step 2 is caught only this way. Make sure the answer is sensible (digits from $0$ to $9$, ages and speeds positive).
✏️ Example 4 — Digits
The sum of the digits of a two-digit number is $9$. If the digits are reversed, the new number is $27$ more than the original number. Find the number.
Name Let the tens digit be $x$ and the units digit be $y$. The number is $10x+y$; reversed, it is $10y+x$.
Translate $x+y=9$ and $(10y+x)-(10x+y)=27$, i.e. $9y-9x=27$, which simplifies to $y-x=3$.
Eliminate Write the equations as $x+y=9$ and $-x+y=3$, and add: $2y=12$, so $y=6$ and $x=3$.
Check in words The number is $36$: $3+6=9$ ✓, and $63-36=27$ ✓.
Answer The number is $\mathbf{36}$.
✏️ Example 5 — A fraction
If $2$ is added to both the numerator and the denominator of a fraction, it becomes $\dfrac23$. If $3$ is subtracted from both, it becomes $\dfrac12$. Find the fraction.
Name Let the fraction be $\dfrac xy$.
Translate $\dfrac{x+2}{y+2}=\dfrac23$ gives $3(x+2)=2(y+2)$, i.e. $3x-2y=-2$. And $\dfrac{x-3}{y-3}=\dfrac12$ gives $2(x-3)=y-3$, i.e. $2x-y=3$. Both equations are linear once you clear the denominators.
Eliminate y Multiply the second by $2$: $4x-2y=6$. Subtract the first: $(4x-2y)-(3x-2y)=6-(-2)$, so $x=8$. Then $2(8)-y=3$ gives $y=13$.
Check in words $\dfrac{8+2}{13+2}=\dfrac{10}{15}=\dfrac23$ ✓ and $\dfrac{8-3}{13-3}=\dfrac{5}{10}=\dfrac12$ ✓.
Answer The fraction is $\dfrac{8}{13}$.
✏️ Example 6 — A boat on a river
A motorboat travels $36$ km downstream in $2$ hours, and the same $36$ km back upstream in $3$ hours. Find the speed of the boat in still water and the speed of the stream.
Name Let the boat's speed in still water be $u$ km/h and the stream's speed $v$ km/h. Downstream the current helps: speed $u+v$. Upstream it holds the boat back: speed $u-v$.
Translate Speed $=\dfrac{\text{distance}}{\text{time}}$: downstream $u+v=\dfrac{36}{2}=18$; upstream $u-v=\dfrac{36}{3}=12$.
Eliminate Add: $2u=30$, so $u=15$. Subtract: $2v=6$, so $v=3$.
Check in words Downstream $15+3=18$ km/h covers $36$ km in $2$ h ✓; upstream $15-3=12$ km/h covers $36$ km in $3$ h ✓.
Answer Boat: $15$ km/h in still water; stream: $3$ km/h. Try it in Interactive 2.
Interactive 2 — Downstream and Upstream

A boat must cover $36$ km downstream in $2$ h and $36$ km upstream in $3$ h. Set the boat's speed $u$ in still water and the stream's speed $v$. Press Play to race the two trips on the distance–time graph: each trip is a straight line whose steepness is its speed. Both boats must reach the $36$ km line exactly at their target times (the rings). Then press Solve by elimination.

122
downstream speed $u+v$—
upstream speed $u-v$—
time for 36 km down—
time for 36 km up—

Exercises

I
Solve each pair by the elimination method, and check your answer in both equations.
  • 1 $x+y=12$ and $x-y=2$
  • 2 $3x+2y=16$ and $x-2y=0$
  • 3 $2x+3y=12$ and $2x-y=4$
  • 4 $4x+3y=24$ and $2x+5y=26$
  • 5 $3x-5y=-1$ and $2x+3y=12$
  • 6 $7x+2y=11$ and $3x-4y=29$
  • 7 $6x+5y=7$ and $4x+3y=5$
  • 8 $11x+15y=-23$ and $7x-2y=20$
  • 9 $\dfrac x3+\dfrac y2=4$ and $\dfrac x2-\dfrac y4=2$
  • 10 $0.4x+0.3y=1.7$ and $0.7x-0.2y=0.8$
  • 11 $47x+31y=63$ and $31x+47y=15$ (hint: add the equations, then subtract them)
1 Add: $2x=14$, $x=7$; then $y=5$. Check: $7+5=12$, $7-5=2$ ✓.
2 Add: $4x=16$, $x=4$; then $4-2y=0$, $y=2$. Check: $12+4=16$ ✓.
3 Subtract: $4y=8$, $y=2$; then $2x-2=4$, $x=3$. Check: $6+6=12$ ✓.
4 Second $\times2$: $4x+10y=52$. Subtract the first: $7y=28$, $y=4$; then $4x+12=24$, $x=3$. Check: $6+20=26$ ✓.
5 First $\times3$: $9x-15y=-3$; second $\times5$: $10x+15y=60$. Add: $19x=57$, $x=3$; then $6+3y=12$, $y=2$. Check: $9-10=-1$ ✓.
6 First $\times2$: $14x+4y=22$. Add the second: $17x=51$, $x=3$; then $21+2y=11$, $y=-5$. Check: $9+20=29$ ✓.
7 First $\times2$: $12x+10y=14$; second $\times3$: $12x+9y=15$. Subtract: $y=-1$; then $4x-3=5$, $x=2$. Check: $12-5=7$ ✓.
8 Eliminate $y$ (its coefficients $15$ and $-2$ have opposite signs). First $\times2$: $22x+30y=-46$, second $\times15$: $105x-30y=300$. Add: $127x=254$, $x=2$; then $14-2y=20$, $y=-3$. Check: $22-45=-23$ ✓.
9 Clear fractions: $2x+3y=24$ and $2x-y=8$. Subtract: $4y=16$, $y=4$; then $2x=12$, $x=6$. Check: $2+2=4$, $3-1=2$ ✓.
10 $\times10$: $4x+3y=17$, $7x-2y=8$. Then $\times2$ and $\times3$: $8x+6y=34$, $21x-6y=24$. Add: $29x=58$, $x=2$; then $8+3y=17$, $y=3$. Check: $0.8+0.9=1.7$, $1.4-0.6=0.8$ ✓.
11 Add: $78x+78y=78$, so $x+y=1$. Subtract: $16x-16y=48$, so $x-y=3$. Then $x=2$, $y=-1$. Check: $94-31=63$, $62-47=15$ ✓.
II
Special cases and unknown coefficients.
  • 1 Use elimination to show that $2x-3y=5$ and $4x-6y=7$ have no solution.
  • 2 Use elimination to show that $3x+y=4$ and $9x+3y=12$ have infinitely many solutions. Give two of them.
  • 3 The point $(2, 1)$ lies on both lines $ax+by=7$ and $bx-ay=4$. Find $a$ and $b$.
  • 4 Find $p$ and $q$ so that the pair $2x+3y=7$ and $(p+q)x+(2p-q)y=21$ has infinitely many solutions.
  • 5 For the pair $3x+ky=9$ and $6x+4y=18$, eliminate $x$. For which value of $k$ are there infinitely many solutions? Solve the pair when $k\ne2$.
  • 6 If $2x+y=23$ and $4x-y=19$, find the values of $x-3y$ and $5y-2x$.
1 First $\times2$: $4x-6y=10$. Subtract the second: $0=3$, which is false. No solution (parallel lines).
2 First $\times3$: $9x+3y=12$. Subtract the second: $0=0$, always true. Infinitely many solutions, e.g. $(0,4)$ and $(1,1)$.
3 $2a+b=7$ and $2b-a=4$. Second $\times2$: $-2a+4b=8$; add the first: $5b=15$, $b=3$, so $a=2$. Check: $4+3=7$, $6-2=4$ ✓.
4 All three ratios must be equal: $\dfrac{p+q}{2}=\dfrac{2p-q}{3}=\dfrac{21}{7}=3$. So $p+q=6$ and $2p-q=9$. Add: $3p=15$, $p=5$, $q=1$. Check: the second equation becomes $6x+9y=21=3(2x+3y)$ ✓.
5 First $\times2$: $6x+2ky=18$. Subtract the second: $(2k-4)y=0$. If $k=2$ this is $0=0$: infinitely many solutions. If $k\ne2$ then $y=0$ and $3x=9$, so the unique solution is $x=3$, $y=0$.
6 Add: $6x=42$, $x=7$; then $y=23-14=9$. So $x-3y=7-27=-20$ and $5y-2x=45-14=31$.
III
Money, ages, digits and fractions. Form a pair of equations, solve by any algebraic method, and check in words.
  • 1 The sum of two numbers is $75$ and their difference is $21$. Find the numbers.
  • 2 Five years ago a mother was three times as old as her daughter. Ten years from now she will be twice as old as her daughter. Find their present ages.
  • 3 $2$ kg of apples and $3$ kg of bananas cost ₹ $440$, while $3$ kg of apples and $1$ kg of bananas cost ₹ $520$. Find the price per kg of each.
  • 4 A gym charges a one-time joining fee plus a fixed monthly fee. Kavya paid ₹ $3{,}400$ in all for $4$ months, and Arjun paid ₹ $7{,}000$ in all for $10$ months. Find the joining fee and the monthly fee.
  • 5 The digits of a two-digit number add up to $11$. Reversing the digits gives a number $45$ greater than the original. Find the number.
  • 6 A two-digit number is $7$ times the sum of its digits. The number formed by reversing its digits is $27$ less than the original number. Find the number.
  • 7 If $1$ is added to both the numerator and the denominator of a fraction, it becomes $\dfrac45$. If $5$ is subtracted from both, it becomes $\dfrac12$. Find the fraction.
  • 8 The denominator of a fraction is $3$ more than its numerator. If $2$ is added to both, the fraction becomes $\dfrac{7}{10}$. Find the fraction.
Check: a digit explorer for two-digit numbers
number $10x+y$—
reversed $10y+x$—
sum of digits—
reversed − original—
1 $x+y=75$, $x-y=21$. Add: $2x=96$, $x=48$, $y=27$. $48$ and $27$.
2 Mother $m$, daughter $d$ (years): $m-5=3(d-5)$, i.e. $m-3d=-10$; $m+10=2(d+10)$, i.e. $m-2d=10$. Subtract: $d=20$, then $m=50$. Check: five years ago $45=3\times15$ ✓; in ten years $60=2\times30$ ✓. Mother $50$ years, daughter $20$ years.
3 Apples ₹ $a$/kg, bananas ₹ $b$/kg: $2a+3b=440$, $3a+b=520$. Second $\times3$: $9a+3b=1560$; subtract the first: $7a=1120$, $a=160$, $b=40$. Apples ₹ $160$ per kg, bananas ₹ $40$ per kg.
4 Joining fee ₹ $j$, monthly fee ₹ $f$: $j+4f=3400$, $j+10f=7000$. Subtract: $6f=3600$, $f=600$, $j=1000$. Joining fee ₹ $1{,}000$, monthly fee ₹ $600$.
5 Tens digit $x$, units digit $y$: $x+y=11$ and $(10y+x)-(10x+y)=45$, i.e. $y-x=5$. Add: $2y=16$, $y=8$, $x=3$. Check: $83-38=45$ ✓. The number is $38$.
6 $10x+y=7(x+y)$ gives $3x=6y$, i.e. $x=2y$. $(10x+y)-(10y+x)=27$ gives $x-y=3$. Substitute: $2y-y=3$, $y=3$, $x=6$. Check: $63=7\times9$ ✓, $63-36=27$ ✓. The number is $63$.
7 $\dfrac{x+1}{y+1}=\dfrac45$ gives $5x-4y=-1$; $\dfrac{x-5}{y-5}=\dfrac12$ gives $2x-y=5$. Second $\times4$: $8x-4y=20$; subtract the first: $3x=21$, $x=7$, $y=9$. Check: $\tfrac{8}{10}=\tfrac45$, $\tfrac24=\tfrac12$ ✓. The fraction is $\dfrac79$.
8 $y=x+3$ and $10(x+2)=7(y+2)$, i.e. $10x-7y=-6$. Substitute: $10x-7x-21=-6$, $3x=15$, $x=5$, $y=8$. Check: $\tfrac{7}{10}$ ✓. The fraction is $\dfrac58$.
IV
Speed, geometry and mixtures. Give units in your answers.
  • 1 A boat goes $48$ km downstream in $3$ h and returns upstream in $4$ h. Find the speed of the boat in still water and the speed of the current.
  • 2 An aircraft flies $1800$ km with a steady tail wind in $2$ h, and returns the same distance against the same wind in $2.5$ h. Find the speed of the aircraft in still air and the speed of the wind.
  • 3 A swimmer covers $1.2$ km downstream in $10$ minutes and the same distance upstream in $20$ minutes. Find, in km/h, her speed in still water and the speed of the stream.
  • 4 If the length of a rectangular hall is increased by $2$ m and its breadth decreased by $1$ m, the area does not change. If the length is decreased by $3$ m and the breadth increased by $2$ m, the area increases by $6$ m². Find the length and breadth of the hall.
  • 5 In $\triangle ABC$, $\angle A=(x+y)^\circ$, $\angle B=(2x-y)^\circ$ and $\angle C=(3y+30)^\circ$. If $\angle B$ is $20^\circ$ more than $\angle A$, find $x$, $y$ and the three angles.
  • 6 A dairy mixes milk with $6\%$ fat and milk with $3\%$ fat to make $90$ L of milk with $4\%$ fat. How many litres of each does it use?
  • 7 All $500$ tickets for a concert were sold: front-row tickets at ₹ $800$ and others at ₹ $300$, for a total of ₹ $2{,}00{,}000$. How many front-row tickets were sold?
1 $u+v=\tfrac{48}{3}=16$, $u-v=\tfrac{48}{4}=12$. Add: $u=14$; then $v=2$. Boat $14$ km/h, current $2$ km/h.
2 Plane $p$ km/h, wind $w$ km/h: $p+w=\tfrac{1800}{2}=900$, $p-w=\tfrac{1800}{2.5}=720$. Add: $2p=1620$, $p=810$, $w=90$. Aircraft $810$ km/h, wind $90$ km/h.
3 Convert times to hours: $10$ min $=\tfrac16$ h, $20$ min $=\tfrac13$ h. So $u+v=1.2\div\tfrac16=7.2$ and $u-v=1.2\div\tfrac13=3.6$. Add: $2u=10.8$, $u=5.4$; $v=1.8$. Swimmer $5.4$ km/h, stream $1.8$ km/h. (Using minutes would give speeds in km per minute — always say which unit you use.)
4 $(l+2)(b-1)=lb$ expands to $lb-l+2b-2=lb$, i.e. $l-2b=-2$. $(l-3)(b+2)=lb+6$ expands to $lb+2l-3b-6=lb+6$, i.e. $2l-3b=12$. The $lb$ terms cancel, leaving a linear pair. First $\times2$: $2l-4b=-4$; subtract from the second: $b=16$, $l=30$. Check: $32\times15=480=30\times16$ ✓, $27\times18=486=480+6$ ✓. Length $30$ m, breadth $16$ m.
5 Angle sum: $(x+y)+(2x-y)+(3y+30)=180$ gives $3x+3y=150$, i.e. $x+y=50$. And $(2x-y)-(x+y)=20$ gives $x-2y=20$. Subtract: $3y=30$, $y=10$, $x=40$. $\angle A=50^\circ$, $\angle B=70^\circ$, $\angle C=60^\circ$ (sum $180^\circ$ ✓).
6 $x$ L of $6\%$ milk, $y$ L of $3\%$ milk: $x+y=90$ and $0.06x+0.03y=0.04\times90=3.6$, i.e. $2x+y=120$. Subtract: $x=30$, $y=60$. $30$ L of $6\%$ milk and $60$ L of $3\%$ milk. Check: $1.8+1.8=3.6$ L of fat ✓.
7 $x$ front-row, $y$ other: $x+y=500$, $800x+300y=200000$, i.e. $8x+3y=2000$. First $\times3$: $3x+3y=1500$; subtract: $5x=500$, $x=100$, $y=400$. $100$ front-row tickets (and $400$ others).
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · 5 marksBoard · Case study

How this is examined: "solve by elimination" appears as a 2-mark question, and word problems (digits, fractions, ages, speeds, prices) are favourite 3- and 5-mark questions and case studies. Marks are given for naming the unknowns, forming both equations, solving, and answering in words with units.

E1. Board · 1 mark If $2x+3y=12$ and $3x+2y=13$, then $x+y$ equals  (A) $5$  (B) $25$  (C) $1$  (D) $6$

Add The coefficients are swapped, so add the equations: $5x+5y=25$, hence $x+y=5$. No need to find $x$ and $y$ separately (they are $3$ and $2$).
Trap (B) is $5x+5y$, forgetting to divide by $5$. (C) is $x-y$, which you get by subtracting.
Answer (A) $5$.

E2. Board · 1 mark Assertion (A): The pair $3x+4y=10$ and $6x+8y=25$ has no solution. Reason (R): Multiplying the first equation by $2$ and subtracting it from the second gives $0=5$.
(A) Both A and R are true and R is the correct explanation of A.  (B) Both A and R are true but R is not the correct explanation of A.  (C) A is true but R is false.  (D) A is false but R is true.

Check R $2(3x+4y)=6x+8y=20$. Subtracting from $6x+8y=25$: $0=5$. R is true.
Check A A false statement after elimination means no solution (Theorem 3.2). Ratio check: $\tfrac36=\tfrac48=\tfrac12$ but $\tfrac{10}{25}=\tfrac25$. A is true.
Answer (A): R is exactly the calculation that shows why A is true.

E3. Board · 2 marks Solve by the elimination method: $4x-3y=1$ and $3x+2y=22$.

Eliminate y First $\times2$: $8x-6y=2$; second $\times3$: $9x+6y=66$. Add: $17x=68$, $x=4$. (1 mark)
Substitute $3(4)+2y=22$, $y=5$. Check: $16-15=1$ ✓. $x=4$, $y=5$. (1 mark)

E4. Board · 3 marks The sum of a two-digit number and the number obtained by reversing its digits is $121$. The two digits differ by $3$. Find the number.

Form Tens digit $x$, units digit $y$: $(10x+y)+(10y+x)=121$, so $11x+11y=121$ and $x+y=11$. (1 mark)
Two cases "Differ by $3$" means $x-y=3$ or $y-x=3$. Case 1: add $x+y=11$ and $x-y=3$: $x=7$, $y=4$, number $74$. Case 2: $y-x=3$ gives $y=7$, $x=4$, number $47$. (1 mark)
Answer The number is $74$ or $47$. Check: $74+47=121$ ✓, and in each the digits differ by $3$ ✓. Giving only one number loses a mark. (1 mark)

E5. Board · 5 marks A ferry travels $40$ km downstream in $2$ hours, and $24$ km upstream in $2$ hours. (i) Find the speed of the ferry in still water and the speed of the stream. (ii) How long would the ferry take to travel $66$ km downstream? (iii) Explain why the ferry could not travel upstream at all if the stream were faster than $16$ km/h.

Name and form Ferry $u$ km/h, stream $v$ km/h. Downstream $u+v=\tfrac{40}{2}=20$; upstream $u-v=\tfrac{24}{2}=12$. (2 marks)
(i) Solve Add: $2u=32$, $u=16$; subtract: $2v=8$, $v=4$. Check: $20\times2=40$ km ✓, $12\times2=24$ km ✓. Ferry $16$ km/h, stream $4$ km/h. (1 mark)
(ii) Downstream speed $20$ km/h: time $=\tfrac{66}{20}=3.3$ h $=$ $3$ h $18$ min. (1 mark)
(iii) Upstream speed is $u-v=16-v$. If $v\gt16$ this is negative: the current carries the ferry backwards faster than its engine can push it forwards. (1 mark)

E6. Board · Case study The juice stall. At a school fete, a juice stall sells mango juice and lemon juice. One group pays ₹ $270$ for $4$ mango and $3$ lemon juices; another pays ₹ $240$ for $2$ mango and $5$ lemon juices. Let a mango juice cost ₹ $x$ and a lemon juice ₹ $y$.
(i) Write the pair of linear equations. (1 mark)
(ii) Use elimination to find the price of each juice. (2 marks)
(iii) Ravi has ₹ $300$. Can he buy $4$ mango and $4$ lemon juices? (1 mark)

(i) $4x+3y=270$ and $2x+5y=240$.
(ii) Second $\times2$: $4x+10y=480$. Subtract the first: $7y=210$, $y=30$. Then $4x+90=270$, $x=45$. Check: $90+150=240$ ✓. Mango ₹ $45$, lemon ₹ $30$.
(iii) $4(45)+4(30)=180+120=$ ₹ $300$. Yes, exactly, with nothing left over.
🔬 Research–Teaching NexusMini project · optional, about 1 hour

Research question. How fast is the wind? Cycle or run a measured stretch of straight road once with the wind and once against it, and use a pair of linear equations to separate your own speed from the wind's effect.

Why it matters. This is the "upstream–downstream" idea at work in the real world. Airline pilots plan fuel using ground speed $=$ air speed $\pm$ wind speed; rowing and kayaking coaches separate an athlete's effort from the river's current; ocean scientists measure currents by timing floats in both directions. In every case, two trips with and against the flow give two equations, and elimination splits them into "your speed" and "the medium's speed".

What to do.

  1. With an adult, choose a safe, straight, flat path (a school ground or park path) on a steady windy day. Measure a distance $d$ of $100$–$200$ m with a measuring tape or a phone map app.
  2. Ride or run the stretch with the wind at a steady, comfortable effort, and time it with a phone stopwatch. Rest, then do the same against the wind with the same effort. Repeat each direction three times and average the times.
  3. Compute the two speeds in m/s: $s_1=\dfrac{d}{t_1}$ and $s_2=\dfrac{d}{t_2}$. Write $u+w=s_1$ and $u-w=s_2$, where $u$ is your own speed and $w$ is the wind's effect on your speed.
  4. Solve by elimination. Convert your answers to km/h (multiply by $3.6$).
  5. Compare $w$ with the wind speed in a weather app for your town at that time.

Tools: Phone stopwatch and weather app, measuring tape or map app, a notebook or spreadsheet.

Hand in: A table of your times, the two speeds, the pair of equations and its solution, the weather-app comparison, and a few sentences on why your $w$ is smaller than the actual wind speed (hint: the wind pushes on you, but you are not a sail).

Going further. Try it on a still day: what should $w$ be, and what do you get? Then try a different effort level. Does your $w$ stay the same? What does that tell you about the "speeds simply add" model?

Teacher/parent note: safety first — use a traffic-free path and supervise. The model "speed with wind $=u+w$, against $=u-w$" is only approximately true for a cyclist or runner, because air resistance grows with speed; discussing why it is not exact is part of the learning.

🧪 Self-check quiz

0/5 correct

Q1. To eliminate $y$ from $3x+2y=7$ and $5x-2y=9$, you should

Q2. To eliminate $x$ from $2x+3y=8$ and $3x-4y=5$, a good first step is to

Q3. A two-digit number has tens digit $a$ and units digit $b$. The number is

Q4. A boat's speed in still water is $u$ km/h and the stream flows at $v$ km/h, with $u\gt v$. Going upstream, the boat's speed is

Q5. Elimination on a pair ends with $0=0$. Which statement is correct?

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