Chapter 3 · Pair of Linear Equations in Two Variables

3.2 The Substitution Method

Turn two unknowns into one: make one variable the subject of an equation, put that expression into the other equation, and solve. The answer is exact, even when it is an awkward fraction no graph could show.

In Section 3.1 you solved pairs of equations by drawing two lines and reading the point where they cross. That works beautifully when the answer is a whole number like $(4, 3)$. But try the pair $3x-y=4$ and $x+2y=5$. Its lines cross at $x=\tfrac{13}{7}$, $y=\tfrac{11}{7}$, that is, at about $(1.857\ldots,\ 1.571\ldots)$. No ruler and pencil will read that off graph paper. You might guess $(1.9, 1.6)$, which is close but wrong, and a board examiner wants the exact value.

The way out is to use algebra instead of a ruler. A pair of equations (also called simultaneous equations) has two unknowns, and you already know how to solve an equation with one unknown, such as $7x=13$. The substitution method turns the pair into exactly that kind of equation. In this section you will learn the method step by step, see what it tells you when the lines are parallel or coincide, and use it on word problems. Section 3.3 gives a second algebraic method, elimination.

The idea of substitution

Look at the pair $x+y=7$ and $x-y=1$. The first equation says that $y$ is always "$7$ minus $x$". So wherever you see $y$ in the second equation, you may write $(7-x)$ instead. The second equation then contains only $x$, and you can solve it.

📘 Definition 3.4 — The substitution method
To solve a pair of linear equations in $x$ and $y$ by substitution, use one equation to write one variable in terms of the other (for example $y=7-x$). This is called making that variable the subject of the equation. Then put this expression in place of that variable in the other equation, which leaves an equation in one variable. Solve it, and finally put the value you found back into the expression to get the second variable (back-substitution).
🧭 Problem-solving strategy — solving a pair by substitution
  1. Choose. Look for a variable whose coefficient is $1$ or $-1$; it is the easiest to make the subject. Write it in terms of the other variable, e.g. from $3x-y=4$ get $y=3x-4$.
  2. Substitute this expression, in brackets, into the other equation.
  3. Solve the resulting equation in one variable. Open the brackets carefully: a minus sign in front of a bracket changes every sign inside it.
  4. Back-substitute the value into the expression from step 1 to find the second variable.
  5. Check your pair of values in both original equations.
✏️ Example 1 — A first substitution
Solve by substitution: $x+y=7$ and $x-y=1$.
Choose Both variables in $x+y=7$ have coefficient $1$. Make $y$ the subject: $y=7-x$.
Substitute Put $(7-x)$ in place of $y$ in the other equation, $x-y=1$: $$x-(7-x)=1.$$
Solve Open the bracket: $x-7+x=1$, so $2x-7=1$, $2x=8$ and $x=4$.
Back-substitute $y=7-x=7-4=3$.
Check $4+3=7$ ✓ and $4-3=1$ ✓. The solution is $x=4$, $y=3$, the same point you found by drawing in Section 3.1, but with no graph paper at all.
✏️ Example 2 — An answer no graph can read
Solve by substitution: $3x-y=4$ and $x+2y=5$.
Choose In $3x-y=4$ the coefficient of $y$ is $-1$. Move $y$ to the right and $4$ to the left: $y=3x-4$.
Substitute Into $x+2y=5$: $$x+2(3x-4)=5.$$
Solve $x+6x-8=5$, so $7x=13$ and $x=\tfrac{13}{7}$.
Back-substitute $y=3\left(\tfrac{13}{7}\right)-4=\tfrac{39}{7}-\tfrac{28}{7}=\tfrac{11}{7}$.
Check $3\cdot\tfrac{13}{7}-\tfrac{11}{7}=\tfrac{39-11}{7}=\tfrac{28}{7}=4$ ✓ and $\tfrac{13}{7}+2\cdot\tfrac{11}{7}=\tfrac{13+22}{7}=\tfrac{35}{7}=5$ ✓. So $x=\tfrac{13}{7}$, $y=\tfrac{11}{7}$ exactly. Interactive 2 below zooms in on these two lines to show why a graph can only ever give an approximation.
Interactive 1 — Substitution Is a Walk Along One Line

Making $y$ the subject of the first equation means every point $P$ on the blue line has the form $(x,\ \text{expression in }x)$. Substituting into the second equation asks: for which $x$ does $P$ also satisfy it? Drag $x$ (or press Play) and watch the dashed gap between $P$ and the red line. The gap closes only where the equation in one variable is true.

1
point P on line 1—
2nd equation: left side—
needed—
after substituting—

Choosing what to substitute

When no coefficient is $1$ or $-1$, substitution still works, but the expression contains a fraction. Choose the variable with the smallest coefficient, and clear the fraction by multiplying the whole equation as soon as you can. If the equations themselves contain fractions or decimals, clear those before you start.

✏️ Example 3 — No coefficient equal to 1
Solve by substitution: $3x+4y=10$ and $2x-5y=-1$.
Choose The smallest coefficient is the $2$ in $2x-5y=-1$. Make $x$ the subject: $2x=5y-1$, so $x=\dfrac{5y-1}{2}$.
Substitute Into $3x+4y=10$: $$3\cdot\frac{5y-1}{2}+4y=10.$$
Clear the fraction Multiply every term by $2$: $3(5y-1)+8y=20$, so $15y-3+8y=20$, $23y=23$ and $y=1$.
Back-substitute $x=\dfrac{5(1)-1}{2}=\dfrac42=2$.
Check $3(2)+4(1)=10$ ✓ and $2(2)-5(1)=-1$ ✓. Solution: $x=2$, $y=1$.
✏️ Example 4 — Decimals first, then substitute
Solve: $0.2x+0.3y=1.3$ and $0.4x+0.5y=2.3$.
Clear decimals Multiply both equations by $10$: $2x+3y=13$ and $4x+5y=23$. Multiplying a whole equation by a non-zero number does not change its solutions.
Choose and substitute From the first, $x=\dfrac{13-3y}{2}$. Into the second: $4\cdot\dfrac{13-3y}{2}+5y=23$, i.e. $2(13-3y)+5y=23$.
Solve $26-6y+5y=23$, so $-y=-3$ and $y=3$. Then $x=\dfrac{13-9}{2}=2$.
Check In the original equations: $0.2(2)+0.3(3)=0.4+0.9=1.3$ ✓ and $0.4(2)+0.5(3)=0.8+1.5=2.3$ ✓. Solution: $x=2$, $y=3$.
⚠️ Common mistakes with substitution

1. Substituting back into the same equation. If you get $y=7-x$ from $x+y=7$ and put it into $x+y=7$ again, you get $x+7-x=7$, i.e. $7=7$. That tells you nothing new. The expression must go into the other equation.

2. Forgetting the brackets. $x-y=1$ with $y=7-x$ is $x-(7-x)=1$, which gives $2x-7=1$. Writing $x-7-x=1$ loses a sign and gives the impossible-looking $-7=1$.

3. Multiplying only part of an equation. When you clear a fraction, multiply every term on both sides, including the number on the right.

4. Stopping after one variable. "$x=4$" is only half an answer. A solution of a pair is a pair of values $(x, y)$.

When substitution gives no number

Sometimes the variable disappears completely after substitution, and you are left with a statement about numbers only, such as $12=9$ or $-4=-4$. That is not a mistake: it is the algebra telling you that the lines are parallel or coincide.

📐 Theorem 3.2 — What substitution tells you
Take one equation of a pair of linear equations in which a chosen variable has a non-zero coefficient, write that variable in terms of the other, substitute this expression into the other equation, and simplify.
  1. If you get an equation such as $7x=13$ with a non-zero coefficient, it has exactly one root, and the pair has a unique solution.
  2. If the variable cancels and you get a false statement such as $12=9$, the pair has no solution (inconsistent; the lines are parallel).
  3. If the variable cancels and you get a true statement such as $-4=-4$, the pair has infinitely many solutions (dependent; the lines coincide).

Why it works. A point $(x, y)$ with $y$ given by the expression lies on the first line. The equation in one variable is true exactly for those points of the first line that also lie on the second. If it is never true, no point is on both lines; if it is always true, every point of the first line is on the second. This agrees with the ratio test (Theorem 3.1), which you can use as a quick check.

✏️ Example 5 — Parallel or coincident?
Use substitution to decide how many solutions each pair has.
(a) $2x+3y=6$ and $4x+6y=9$    (b) $x-3y=2$ and $-2x+6y=-4$
(a) Substitute From the first equation, $y=\dfrac{6-2x}{3}$. Into the second: $4x+6\cdot\dfrac{6-2x}{3}=9$, i.e. $4x+2(6-2x)=9$.
(a) Simplify $4x+12-4x=9$, so $12=9$. This is false, whatever $x$ is. The pair has no solution. Ratio check: $\dfrac24=\dfrac36=\dfrac12$ but $\dfrac{-6}{-9}=\dfrac23$, so the lines are parallel ✓.
(b) Substitute From the first equation, $x=3y+2$. Into the second: $-2(3y+2)+6y=-4$, i.e. $-6y-4+6y=-4$.
(b) Simplify $-4=-4$, which is true for every $y$. The pair has infinitely many solutions: every point with $x=3y+2$, e.g. $(2, 0)$, $(5, 1)$, $(-1, -1)$. Ratio check: all three ratios equal $-\tfrac12$ ✓.
Interactive 2 — Zoom In: Why a Graph Gives Only an Estimate

Choose a pair and zoom in on the point where the first line and the second line cross. The amber dot is the best reading you could make from the grid at that zoom. Each zoom step makes the grid ten times finer, and the reading gets closer, but it never becomes the exact fraction that substitution gives in two lines of working. Press Play to zoom automatically.

×1
grid spacing—
graph reading—
exact (substitution)—
error in x—

Word problems by substitution

Many word problems hand you one unknown in terms of the other: "the mother is three times as old as her daughter" is already $m=3d$. Substitution is the natural method for these, because step 1 is done for you.

✏️ Example 6 — Ages
Asha's mother is three times as old as Asha. Five years ago, the mother was four times as old as Asha was then. Find their present ages.
Name the unknowns Let Asha be $a$ years old and her mother $m$ years old now.
Translate Now: $m=3a$. Five years ago both were $5$ years younger: $m-5=4(a-5)$.
Substitute Put $m=3a$ into the second equation: $3a-5=4a-20$, so $-a=-15$ and $a=15$. Then $m=3(15)=45$.
Check in words Now $45=3\times15$ ✓. Five years ago Asha was $10$ and her mother $40$, and $40=4\times10$ ✓.
Answer Asha is $15$ years old and her mother is $45$ years old.

Exercises

I
Solve each pair by the substitution method, and check your answer in both equations.
  • 1 $x+y=10$ and $x-y=4$
  • 2 $y=2x-3$ and $3x+y=12$
  • 3 $x=3y+1$ and $2x-5y=4$
  • 4 $2x+y=7$ and $4x-3y=-1$
  • 5 $x+4y=14$ and $3x-2y=0$
  • 6 $5x-y=13$ and $2x+3y=12$
  • 7 $3x-2y=1$ and $x+y=7$
  • 8 $2x+3y=4$ and $x-2y=-5$
  • 9 $4x+y=1$ and $3x-2y=9$
  • 10 $x-y=1$ and $2x+3y=5$ (the answer is not a whole number)
Check: enter a pair $ax+by=c$ and your answer — fractions like 13/7 are fine
equation 1—
equation 2—
1 $y=10-x$; $x-(10-x)=4$, so $2x=14$, $x=7$, $y=3$. Check: $7+3=10$, $7-3=4$ ✓.
2 $3x+(2x-3)=12$, so $5x=15$, $x=3$, $y=2(3)-3=3$. Check: $9+3=12$ ✓.
3 $2(3y+1)-5y=4$, so $6y+2-5y=4$, $y=2$, $x=3(2)+1=7$. Check: $14-10=4$ ✓.
4 $y=7-2x$; $4x-3(7-2x)=-1$, so $10x-21=-1$, $x=2$, $y=3$. Check: $8-9=-1$ ✓.
5 $x=14-4y$; $3(14-4y)-2y=0$, so $42-14y=0$, $y=3$, $x=2$. Check: $6-6=0$ ✓.
6 $y=5x-13$; $2x+3(5x-13)=12$, so $17x=51$, $x=3$, $y=2$. Check: $6+6=12$ ✓.
7 $y=7-x$; $3x-2(7-x)=1$, so $5x=15$, $x=3$, $y=4$. Check: $9-8=1$ ✓.
8 $x=2y-5$; $2(2y-5)+3y=4$, so $7y=14$, $y=2$, $x=-1$. Check: $-2+6=4$, $-1-4=-5$ ✓.
9 $y=1-4x$; $3x-2(1-4x)=9$, so $11x=11$, $x=1$, $y=-3$. Check: $4-3=1$, $3+6=9$ ✓.
10 $x=y+1$; $2(y+1)+3y=5$, so $5y=3$, $y=\tfrac35$, $x=\tfrac85$. Check: $\tfrac85-\tfrac35=1$, $\tfrac{16}{5}+\tfrac95=5$ ✓.
II
Fractions, decimals, and pairs that do not have a unique solution.
  • 1 $\dfrac x2+\dfrac y3=4$ and $x-y=3$
  • 2 $0.3x+0.2y=1.3$ and $0.5x-0.1y=1.3$
  • 3 $\dfrac x3-\dfrac y4=1$ and $\dfrac x2+\dfrac y2=5$
  • 4 $\dfrac{x+y}{2}=4$ and $\dfrac{x-y}{3}=2$
  • 5 Use substitution to show that $3x-y=4$ and $6x-2y=9$ have no solution.
  • 6 Use substitution to show that $x+2y=5$ and $3x+6y=15$ have infinitely many solutions, and write down three of them.
  • 7 Solve $2x+3y=13$ and $3x-y=3$, and hence find $m$ for which $y=mx+5$.
  • 8 Solve $x+y=5$ and $2x-3y=0$. Find $k$ if the solution also lies on the line $kx+y=7$.
  • 9 For the pair $px+2y=3$ and $3x+6y=9$, substitute $y=\dfrac{3-px}{2}$ into the second equation. For which value of $p$ is there no unique solution? Solve the pair for every other value of $p$.
  • 10 Show by substitution that the pair $2x-5y=3$ and $4x-10y=k$ has no solution unless $k=6$. What happens when $k=6$?
1 Multiply the first by $6$: $3x+2y=24$. With $x=y+3$: $3y+9+2y=24$, $y=3$, $x=6$. Check: $3+1=4$, $6-3=3$ ✓.
2 Multiply by $10$: $3x+2y=13$, $5x-y=13$. Then $y=5x-13$; $3x+2(5x-13)=13$, $13x=39$, $x=3$, $y=2$. Check: $0.9+0.4=1.3$, $1.5-0.2=1.3$ ✓.
3 Multiply by $12$ and by $2$: $4x-3y=12$, $x+y=10$. Then $y=10-x$; $4x-3(10-x)=12$, $7x=42$, $x=6$, $y=4$. Check: $2-1=1$, $3+2=5$ ✓.
4 $x+y=8$ and $x-y=6$. With $y=8-x$: $x-(8-x)=6$, $2x=14$, $x=7$, $y=1$. Check: $\tfrac82=4$, $\tfrac63=2$ ✓.
5 $y=3x-4$; $6x-2(3x-4)=9$ gives $6x-6x+8=9$, i.e. $8=9$: false, so no solution (parallel lines).
6 $x=5-2y$; $3(5-2y)+6y=15$ gives $15=15$: always true, so infinitely many solutions, e.g. $(5,0)$, $(3,1)$, $(1,2)$.
7 $y=3x-3$; $2x+3(3x-3)=13$, $11x=22$, $x=2$, $y=3$. Then $3=2m+5$, so $m=-1$.
8 $x=5-y$; $2(5-y)-3y=0$, $10=5y$, $y=2$, $x=3$. Then $3k+2=7$, so $k=\tfrac53$.
9 $3x+6\cdot\dfrac{3-px}{2}=9$ gives $3x+9-3px=9$, i.e. $3(1-p)x=0$. If $p=1$ this is $0=0$: infinitely many solutions (the first equation is then one-third of the second), so no unique solution. If $p\ne1$, then $x=0$ and $y=\tfrac32$: the unique solution is $\left(0, \tfrac32\right)$ for every $p\ne1$.
10 $x=\dfrac{3+5y}{2}$; $4\cdot\dfrac{3+5y}{2}-10y=k$ gives $6+10y-10y=k$, i.e. $6=k$. For $k\ne6$ this is false: no solution. For $k=6$ it is always true: infinitely many solutions (the second equation is twice the first).
III
Word problems. Name your unknowns with units, form a pair of equations, solve by substitution and check in the words of the problem.
  • 1 One number is four times another, and their difference is $15$. Find the numbers.
  • 2 A father is four times as old as his son. In $20$ years he will be twice as old as his son will be then. Find their present ages.
  • 3 A notebook costs ₹ $5$ more than a pen. Three pens and one notebook cost ₹ $85$. Find the cost of each.
  • 4 The length of a rectangular vegetable plot is $2$ m more than twice its breadth, and its perimeter is $46$ m. Find its length, breadth and area.
  • 5 Two angles are supplementary. The larger is $20^\circ$ more than three times the smaller. Find both angles.
  • 6 A box holds $40$ coins, all ₹ $5$ or ₹ $10$ coins, worth ₹ $290$ altogether. How many of each are there?
  • 7 A taxi charges a fixed amount plus a rate per km. A $10$ km ride costs ₹ $175$ and a $25$ km ride costs ₹ $370$. Find the fixed charge and the rate, and the cost of an $18$ km ride.
  • 8 In a T20 innings a batter hit $19$ boundaries (fours and sixes) worth $100$ runs in all. How many fours and how many sixes did she hit?
  • 9 A tea merchant blends tea costing ₹ $400$ per kg with tea costing ₹ $600$ per kg to make $20$ kg of a blend worth ₹ $520$ per kg. How many kg of each does she use?
1 $x=4y$ and $x-y=15$: $4y-y=15$, $y=5$, $x=20$. The numbers are $20$ and $5$.
2 Father $f$, son $s$ (years): $f=4s$ and $f+20=2(s+20)$. Then $4s+20=2s+40$, $s=10$, $f=40$. Check: in $20$ years $60=2\times30$ ✓. Father $40$ years, son $10$ years.
3 Pen ₹ $p$, notebook ₹ $n$: $n=p+5$ and $3p+n=85$. Then $4p+5=85$, $p=20$, $n=25$. Pen ₹ $20$, notebook ₹ $25$.
4 $l=2b+2$ and $2(l+b)=46$, i.e. $l+b=23$. Then $3b+2=23$, $b=7$, $l=16$. Length $16$ m, breadth $7$ m, area $16\times7=112$ m².
5 $x+y=180$ and $x=3y+20$: $4y+20=180$, $y=40$, $x=140$. $140^\circ$ and $40^\circ$.
6 $f$ five-rupee and $t$ ten-rupee coins: $f+t=40$, $5f+10t=290$. With $f=40-t$: $200-5t+10t=290$, $t=18$, $f=22$. Check: $110+180=290$ ✓. $22$ coins of ₹ $5$, $18$ coins of ₹ $10$.
7 Fixed charge ₹ $F$, rate ₹ $r$ per km: $F+10r=175$, $F+25r=370$. With $F=175-10r$: $175+15r=370$, $r=13$, $F=45$. ₹ $45$ fixed and ₹ $13$ per km; $18$ km costs $45+13\times18=$ ₹ $279$.
8 $f$ fours, $s$ sixes: $f+s=19$, $4f+6s=100$. With $f=19-s$: $76+2s=100$, $s=12$, $f=7$. Check: $28+72=100$ ✓. $7$ fours and $12$ sixes.
9 $x$ kg at ₹ $400$, $y$ kg at ₹ $600$: $x+y=20$, $400x+600y=520\times20=10400$. With $x=20-y$: $8000+200y=10400$, $y=12$, $x=8$. $8$ kg of the ₹ $400$ tea and $12$ kg of the ₹ $600$ tea.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case study

How this is examined: board papers often say "solve by substitution" in 2- and 3-mark questions, so the examiner looks for the expression you substitute, the equation in one variable, both values, and a check. A short MCQ or Assertion–Reason item may test what a false or true statement after substitution means.

E1. Board · 1 mark If $x=2y+1$ and $3x-4y=7$, then the value of $x+y$ is  (A) $7$  (B) $3$  (C) $5$  (D) $9$

Substitute $3(2y+1)-4y=7$ gives $2y+3=7$, so $y=2$ and $x=2(2)+1=5$. Check: $15-8=7$ ✓.
Trap (C) is the value of $x$ alone, and (B) is $x-y$. Read what the question asks for.
Answer (A) $x+y=5+2=7$.

E2. Board · 1 mark Assertion (A): The pair $2x-y=3$ and $6x-3y=9$ has infinitely many solutions. Reason (R): Substituting $y=2x-3$ from the first equation into the second gives the true statement $9=9$, in which no variable remains.
(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 $6x-3(2x-3)=6x-6x+9=9$, so the second equation becomes $9=9$: true, and no variable remains. R is true.
Check A By Theorem 3.2(3), a true statement with no variable left means every point of the first line lies on the second, so there are infinitely many solutions. A is true. (Ratio check: $\tfrac26=\tfrac{-1}{-3}=\tfrac{-3}{-9}=\tfrac13$.)
Answer (A): both are true, and the substitution result in R is exactly what proves A. Choosing (B) is the trap; ask yourself "does R give the reason why A happens?" Here it does.

E3. Board · 2 marks Solve by the substitution method: $2x+y=1$ and $3x-2y=12$.

Substitute From the first, $y=1-2x$. Then $3x-2(1-2x)=12$, so $7x-2=12$. (1 mark)
Solve $7x=14$, $x=2$, and $y=1-4=-3$. Check: $4-3=1$ ✓, $6+6=12$ ✓. $x=2$, $y=-3$. (1 mark)

E4. Board · 3 marks Solve $3x+y=5$ and $2x-3y=7$ by substitution, and hence find the value of $m$ for which $y=mx+4$.

Substitute $y=5-3x$; then $2x-3(5-3x)=7$, so $11x-15=7$. (1 mark)
Solve $11x=22$, $x=2$, $y=5-6=-1$. Check: $6-1=5$ ✓, $4+3=7$ ✓. (1 mark)
Find m $-1=2m+4$ gives $2m=-5$, so $m=-\tfrac52$. (1 mark)

E5. Board · Case study School science fair. A school sold $300$ tickets for its science fair. Adult tickets cost ₹ $50$ and student tickets ₹ $20$, and the total collection was ₹ $11{,}400$. Let $a$ be the number of adult tickets and $s$ the number of student tickets.
(i) Write a pair of linear equations for this situation. (1 mark)
(ii) Solve the pair by substitution to find $a$ and $s$. (2 marks)
(iii) Next year the student ticket will cost ₹ $25$. With the same numbers of tickets and the same adult price, how much more money would be collected? (1 mark)

(i) $a+s=300$ and $50a+20s=11400$.
(ii) $s=300-a$; $50a+20(300-a)=11400$, so $30a+6000=11400$, $30a=5400$, $a=180$ and $s=120$. Check: $9000+2400=11400$ ✓. $180$ adult and $120$ student tickets.
(iii) Each of the $120$ student tickets costs ₹ $5$ more: $120\times5=$ ₹ $600$ more (₹ $12{,}000$ in all).
🔬 Research–Teaching NexusMini project · optional, about 45 minutes

Research question. Can you work out the price of each item at your school canteen or a local snack shop using only the totals on two bills, and how sure can you be of the answer?

Why it matters. Finding hidden quantities from combined measurements is everyday work for scientists and businesses. A chemist finds how much of two substances is in a mixture from its total mass and total volume; an economist estimates two unknown prices from totals in shop records; a nutritionist works out the protein and fat in a meal from food labels. Each time, two facts about two unknowns become a pair of linear equations.

What to do.

  1. Choose two items that are often bought together (for example, idli and vada, or a juice and a sandwich). Collect at least three bills, or ask friends what they paid, each with different quantities of the two items. Note the quantities and the total.
  2. Take two of the bills and write a pair of equations, with ₹ $x$ and ₹ $y$ the prices of one of each item.
  3. Solve the pair by substitution. Then use the third bill as a check: does it fit your prices?
  4. If it does not fit exactly, find out why (a discount on a combo, a price change, a packing charge, a rounding of the total).
  5. Finally compare with the canteen's menu prices.

Tools: Bills or a short survey of friends, a calculator, a notebook (or a spreadsheet to check the third bill).

Hand in: A table of the bills, your pair of equations, the full substitution working, the check with the third bill, and two or three sentences on whether the "same price every time" model was good enough.

Going further. What happens if two of your bills have quantities in the same ratio (say $1$ idli + $2$ vadas and $2$ idlis + $4$ vadas)? Use Theorem 3.2 to explain why those two bills alone cannot give you the prices.

Teacher/parent note: real bills include taxes, discounts and rounding, so the "exact" solution may not quite fit a third bill. That is a valuable lesson about the difference between a mathematical model and the real world; encourage the student to state their assumptions clearly.

🧪 Self-check quiz

0/5 correct

Q1. You have made $y=7-x$ from the equation $x+y=7$ of a pair. Where should you substitute it?

Q2. After substitution, the working of a pair ends with $0=4$. The pair

Q3. The solution of $y=2x$ and $x+y=9$ is

Q4. For the pair $4x+3y=11$ and $5x-y=9$, which first step makes substitution easiest?

Q5. Substituting $y=2x-4$ from $2x-y=4$ into $4x-2y=8$ gives

← 3.1 Pairs of Linear Equations and the Graphical Method3.3 The Elimination Method and Word Problems →