Linear equation in two variables and its solutions
Two friends have 10 chocolates between them. If one has 3 the other has 7; if one has 5 the other has 5. Both numbers keep changing, so we use two letters, x and y, and write x + y = 10.
An equation with two variables, each appearing only once (power 1), is a linear equation in two variables, such as 3x − 4y − 15 = 0. The values of both variables written together form an ordered pair (value of x, value of y).
In the figure: Some solutions of x + y = 10 are shown as dots. All the dots lie on one straight line, and there are infinitely many such solutions (even more if negatives and fractions are allowed).
Rule 1: One equation has infinitely many solutions — Give x any value and find y; the two values together form one solution.
- Easy: In x + y = 10 take x = 4; then y = 10 − 4 = 6. The solution is (4, 6).
- Medium: 2x + y = 9: x = 3 ⇒ y = 3; x = 0 ⇒ y = 9; x = 5 ⇒ y = −1. Solutions: (3, 3), (0, 9), (5, −1).
- Exam type: 3x − 2y = 12: x = 4 ⇒ 12 − 2y = 12 ⇒ y = 0; x = 0 ⇒ −2y = 12 ⇒ y = −6; x = 2 ⇒ 6 − 2y = 12 ⇒ y = −3. Solutions: (4, 0), (0, −6), (2, −3).
Rule 2: The order in an ordered pair matters — Always write the pair as (x, y). If the order changes, the values of x and y change.
In the figure: In x − y = 4, (7, 3) gives 7 − 3 = 4, so it is a solution. (3, 7) gives 3 − 7 = −4, so it is not a solution.
- Easy: x − y = 4: (7, 3) ⇒ 7 − 3 = 4 ✔; (3, 7) ⇒ 3 − 7 = −4 ✘.
- Medium: 2x + y = 11: (4, 3) ⇒ 8 + 3 = 11 ✔; (3, 4) ⇒ 6 + 4 = 10 ✘.
- Exam type: Is (−1, 5) a solution of 3x + y = 2? 3 × (−1) + 5 = −3 + 5 = 2. Left side = right side, so yes.
Rule 3: Bring the equation to standard form, then substitute — The general form has every term on the left and 0 on the right: ax + by + c = 0. Moving the constant to the right changes its sign.
- Easy: 3x + y − 7 = 0 ⇒ 3x + y = 7.
- Medium: Is (2, 1) a solution of 3x + y − 7 = 0? 6 + 1 − 7 = 0 ✔. Yes.
- Exam type: 2x − 3y + 6 = 0 ⇒ 2x − 3y = −6. (3, 4): 6 − 12 = −6 ✔ and (−3, 0): −6 − 0 = −6 ✔. Both are solutions.
One equation in two variables has infinitely many solutions, not just one. So a single equation cannot fix one value of x and y.
Swapping the order of the pair: writing (7, 3) when it should be (3, 7). The first value always belongs to x.
To test a pair, replace x and y by its values and check that the left side equals the right side. Memory clue: An equation is a balance; if the pair fits, both pans are equal.
Spot the difference
| Equation | Variables | Number of solutions | How the solution is written |
|---|---|---|---|
| x + 5 = 9 | 1 | only one | x = 4 |
| x + y = 9 | 2 | infinitely many | (4, 5), (2, 7), (0, 9) ... |
| 2x − y = 3 | 2 | infinitely many | (2, 1), (3, 3), (0, −3) ... |
Practice questions
- 1In x + y = 12, if x = 5, what is y?
- 2Take x = 2 in 2x + y = 8 and write the solution.
- 3Is (3, 1) a solution of x + 2y = 5?
- 4Write the solutions of 3x + 2y = 12 for x = 0, 2, 4.
- 5Which of (6, 1) and (1, 6) is a solution of x − 2y = 4?
- 6Write 5x − 2y + 4 = 0 as 5x − 2y = −4. What is y when x = 0?
- 7If (2, 3) is a solution of 3x + ky = 18, find k.
- 8Find a solution of x + y = 12 in which y is double x.
Show answers (try them yourself first)
- 1y = 12 − 5 = 7. Solution (5, 7). 2) 4 + y = 8 ⇒ y = 4. Solution (2, 4). 3) 3 + 2 × 1 = 5 ✔. Yes.
- 4x = 0 ⇒ y = 6; x = 2 ⇒ 6 + 2y = 12 ⇒ y = 3; x = 4 ⇒ y = 0. Solutions (0, 6), (2, 3), (4, 0). 5) (6, 1): 6 − 2 = 4 ✔. (1, 6): 1 − 12 = −11 ✘. So (6, 1). 6) x = 0 ⇒ −2y = −4 ⇒ y = 2.
- 76 + 3k = 18 ⇒ 3k = 12 ⇒ k = 4. 8) y = 2x ⇒ x + 2x = 12 ⇒ x = 4, y = 8. Solution (4, 8); check: 4 + 8 = 12.
Simultaneous equations and the common solution
We need two numbers whose sum is 10 and whose difference is 4. "Sum 10" gives many pairs and "difference 4" also gives many pairs, but only one pair obeys both conditions at the same time.
When two linear equations are considered together they are called simultaneous equations. The pair that fits both equations is their common solution.
In the figure: All solutions of x + y = 10 lie on the blue line and all solutions of x − y = 4 lie on the orange line. The lines meet at (7, 3), which is the common solution.
Simultaneous equations and the common solution
In the app you can move this figure with your finger.
Open in the appRule 1: List the solutions of both equations and pick the common pair — Write a few pairs for each equation and choose the pair that appears in both lists.
- Easy: x + y = 10: (6, 4), (7, 3), (8, 2). x − y = 4: (6, 2), (7, 3), (8, 4). Common pair: (7, 3).
- Medium: x + y = 8 and x − y = 2: (5, 3) gives 5 + 3 = 8 and 5 − 3 = 2, both correct. Solution (5, 3).
- Exam type: Together they have 14 chocolates and one has 6 more than the other: x + y = 14, x − y = 6. (10, 4) gives 14 and 6. So they have 10 and 4 chocolates.
Rule 2: Check the answer in both equations — A common solution fits both equations. If it fits only one, it is not a common solution.
- Easy: (4, 1): x + y = 5 ⇒ 4 + 1 = 5 ✔; x − y = 3 ⇒ 4 − 1 = 3 ✔. So it is the common solution.
- Medium: (2, 3): 2x + y = 7 ⇒ 4 + 3 = 7 ✔; x + 2y = 8 ⇒ 2 + 6 = 8 ✔. Common solution.
- Exam type: (3, 2): x + y = 5 ⇒ 3 + 2 = 5 ✔; but 2x − y = 5 ⇒ 6 − 2 = 4 ✘. It fits only the first, so it is not a common solution.
Rule 3: If one equation is just a multiple of the other, all solutions are common — In x + y = 5 and 2x + 2y = 10 the second is double the first, so both have exactly the same list of solutions.
- Easy: x + y = 5: (1, 4), (2, 3). In 2x + 2y = 10: 2 + 8 = 10 ✔ and 4 + 6 = 10 ✔. Both pairs are common.
- Medium: x + y = 4 and 3x + 3y = 12: the second is 3 times the first. So all solutions are common (infinitely many).
- Exam type: x + y = 6 and x + y = 9: the same sum cannot have two different values. So there is no common solution.
The common solution must fit both equations. In the figure it is the point where the two lines meet.
Checking the answer in only one equation and calling it correct.
While checking, tick each equation separately. The answer is correct only if both get a ✔. Memory clue: Two equations are two conditions; the pair that obeys both is the answer.
Spot the difference
| One equation | Simultaneous equations | |
|---|---|---|
| Example | x + y = 10 | x + y = 10 and x − y = 4 |
| Solutions | infinitely many | one common pair (7, 3) |
| Figure | one line | two lines meeting at one point |
Practice questions
- 1Is (4, 2) the common solution of x + y = 6 and x − y = 2?
- 2Two numbers have sum 9 and difference 1. Which numbers are they?
- 3Is (1, 1) the common solution of x + y = 2 and x − y = 1?
- 4Find the common solution of x + y = 7 and x − y = 3 by listing pairs.
- 5Is (2, 5) the common solution of 3x − y = 1 and x + y = 7?
- 6What can you say about the common solutions of x + y = 4 and 3x + 3y = 12?
- 7Two numbers have sum 20 and difference 6. Find them.
- 8If (4, 3) is the common solution of x + y = a and x − y = b, find a and b.
Show answers (try them yourself first)
- 14 + 2 = 6 ✔ and 4 − 2 = 2 ✔. Yes. 2) (5, 4): 5 + 4 = 9 and 5 − 4 = 1. The numbers are 5 and 4. 3) 1 + 1 = 2 ✔ but 1 − 1 = 0 ≠ 1 ✘. No.
- 4x + y = 7: (4, 3), (5, 2), (6, 1). x − y = 3: (5, 2), (6, 3). Common (5, 2). 5) 3 × 2 − 5 = 1 ✔ and 2 + 5 = 7 ✔. Yes. 6) The second equation is 3 times the first, so every solution of the first also fits the second. The common solutions are infinitely many.
- 7(13, 7): 13 + 7 = 20 and 13 − 7 = 6. The numbers are 13 and 7. 8) a = 4 + 3 = 7 and b = 4 − 3 = 1.
Elimination method: adding and subtracting
If equal weights are added to both pans of a balance, it stays balanced. In the same way, if the left sides of two equations are added together and the right sides are added together, the equality still holds.
Elimination means removing one variable. Adding or subtracting the two equations makes one variable vanish and leaves an equation in one variable. Solve it, then put the value into an original equation to get the other variable.
In the figure: The equations 2x + y = 13 and x − y = 2 are written one below the other and added. The y coefficients +1 and −1 cancel, leaving 3x = 15, so x = 5.
Rule 1: If the coefficients are opposites, add — If one equation has +y and the other has −y, then y vanishes in the sum.
- Easy: x + y = 9 and x − y = 3: add ⇒ 2x = 12 ⇒ x = 6. 6 + y = 9 ⇒ y = 3. Solution (6, 3).
- Medium: 2x + y = 13 and x − y = 2: add ⇒ 3x = 15 ⇒ x = 5. 10 + y = 13 ⇒ y = 3. Check: 5 − 3 = 2 ✔.
- Exam type: 5x + 3y = 21 and 2x − 3y = 0: add ⇒ 7x = 21 ⇒ x = 3. 6 − 3y = 0 ⇒ y = 2. Check: 15 + 6 = 21 ✔.
Rule 2: If the coefficients are equal, subtract — If both equations have the same coefficient of y, subtract one equation from the other.
- Easy: 3x + y = 13 and x + y = 7: subtract ⇒ 2x = 6 ⇒ x = 3. 3 + y = 7 ⇒ y = 4.
- Medium: 4x + 3y = 25 and 4x − y = 13: subtract ⇒ 4y = 12 ⇒ y = 3. 4x − 3 = 13 ⇒ x = 4.
- Exam type: 7x + 2y = 29 and 3x + 2y = 17: subtract ⇒ 4x = 12 ⇒ x = 3. 21 + 2y = 29 ⇒ y = 4. Check: 9 + 8 = 17 ✔.
Rule 3: Change the signs carefully while subtracting — When you subtract, every term of the second equation changes its sign. −(−2y) = +2y.
- Easy: 5x − 2y = 11 and 3x − 2y = 5: subtract ⇒ 2x = 6 ⇒ x = 3. 15 − 2y = 11 ⇒ y = 2.
- Medium: 6x + 5y = 28 and 2x + 5y = 16: subtract ⇒ 4x = 12 ⇒ x = 3. 18 + 5y = 28 ⇒ y = 2.
- Exam type: x − 3y = −5 and 2x − 3y = −1: subtract the first from the second ⇒ x = 4. 4 − 3y = −5 ⇒ −3y = −9 ⇒ y = 3. Check: 8 − 9 = −1 ✔.
The variable you want to remove must have the same number as its coefficient. Opposite signs: add. Same signs: subtract.
Changing the sign of only the first term while subtracting and leaving the other terms as they are.
Put the answer into both original equations. If both come out correct, you are done. Memory clue: Elimination: "same signs, subtract; different signs, add".
Spot the difference
| Coefficients of y | Signs | Action |
|---|---|---|
| +3 and −3 | different | add |
| +4 and +4 | same | subtract |
| −2 and −2 | same | subtract |
| +5 and −5 | different | add |
Practice questions
- 1Solve x + y = 8 and x − y = 2.
- 2Solve 2x + y = 11 and x + y = 7.
- 3Solve 3x + y = 14 and 3x − y = 10.
- 4Solve 4x + 3y = 23 and 4x + y = 13.
- 5Solve 5x − 4y = 7 and 3x + 4y = 17.
- 6Solve x − 2y = 0 and 3x − 2y = 8.
- 72 notebooks and 1 pen cost ₹35. 1 notebook and 1 pen cost ₹20. Find the cost of each.
- 8Two numbers have sum 48 and difference 12. Find the numbers.
Show answers (try them yourself first)
- 1Add ⇒ 2x = 10 ⇒ x = 5; 5 + y = 8 ⇒ y = 3. (5, 3). 2) Subtract ⇒ x = 4; 4 + y = 7 ⇒ y = 3. (4, 3). 3) Add ⇒ 6x = 24 ⇒ x = 4; 12 + y = 14 ⇒ y = 2. (4, 2).
- 4Subtract ⇒ 2y = 10 ⇒ y = 5; 4x + 5 = 13 ⇒ x = 2. (2, 5). 5) Add ⇒ 8x = 24 ⇒ x = 3; 15 − 4y = 7 ⇒ y = 2. (3, 2). 6) Subtract ⇒ 2x = 8 ⇒ x = 4; 4 − 2y = 0 ⇒ y = 2. (4, 2).
- 72x + y = 35 and x + y = 20; subtract ⇒ x = 15; y = 5. Notebook ₹15, pen ₹5. Check: 30 + 5 = 35 ✔. 8) Add ⇒ 2x = 60 ⇒ x = 30; y = 18. The numbers are 30 and 18.
Elimination method: multiply first, then eliminate
If a bag with 2 bananas and 3 apples costs ₹26, then 3 such bags hold 6 bananas and 9 apples and cost ₹78. So if both sides of an equation are multiplied by the same number, the equation stays true.
If no variable has equal or opposite coefficients, multiply one (or both) equations by a suitable number. Then one variable gets equal coefficients and you can add or subtract as before.
In the figure: Multiplying 2x + 3y = 13 by 2 gives 4x + 6y = 26. Multiplying 5x − 2y = 4 by 3 gives 15x − 6y = 12. Now the y coefficients are +6 and −6, so adding removes y.
Rule 1: Multiply only one equation to match the coefficients — Pick a variable whose coefficient in one equation is a multiple of the other, and multiply just that equation. Do not forget the right side.
- Easy: 3x + y = 9 and 2x + 2y = 10: multiply the first by 2 ⇒ 6x + 2y = 18. Subtract the second ⇒ 4x = 8 ⇒ x = 2. 6 + y = 9 ⇒ y = 3.
- Medium: 5x + 2y = 22 and 3x + 4y = 16: multiply the first by 2 ⇒ 10x + 4y = 44. Subtract the second ⇒ 7x = 28 ⇒ x = 4. 20 + 2y = 22 ⇒ y = 1.
- Exam type: 2x + y = 4 and 5x − 3y = 21: multiply the first by 3 ⇒ 6x + 3y = 12. Add ⇒ 11x = 33 ⇒ x = 3. 6 + y = 4 ⇒ y = −2. Check: 15 + 6 = 21 ✔.
Rule 2: Multiply both equations — If no coefficient is a multiple of another, multiply both equations by different numbers so that the chosen variable gets the same coefficient.
- Easy: 2x + 3y = 12 and 3x + 2y = 13: first × 3 ⇒ 6x + 9y = 36; second × 2 ⇒ 6x + 4y = 26. Subtract ⇒ 5y = 10 ⇒ y = 2. x = 3.
- Medium: 3x + 5y = 25 and 4x − 3y = 14: first × 3 ⇒ 9x + 15y = 75; second × 5 ⇒ 20x − 15y = 70. Add ⇒ 29x = 145 ⇒ x = 5. 15 + 5y = 25 ⇒ y = 2.
- Exam type: 2x + 3y = 13 and 5x − 2y = 4: first × 2 ⇒ 4x + 6y = 26; second × 3 ⇒ 15x − 6y = 12. Add ⇒ 19x = 38 ⇒ x = 2. 4 + 3y = 13 ⇒ y = 3.
Rule 3: For standard-form equations, move the constant to the right first — If the equations look like ax + by + c = 0, move the constant to the right and change its sign. Then eliminate.
- Easy: x + y − 5 = 0 and x − y − 1 = 0 ⇒ x + y = 5 and x − y = 1. Add ⇒ 2x = 6 ⇒ x = 3, y = 2.
- Medium: 2x + 3y − 14 = 0 and 3x + y − 7 = 0 ⇒ 2x + 3y = 14 and 3x + y = 7. Second × 3 ⇒ 9x + 3y = 21. Subtract ⇒ 7x = 7 ⇒ x = 1, y = 4.
- Exam type: 4x − 3y − 24 = 0 and x + y + 1 = 0 ⇒ 4x − 3y = 24 and x + y = −1. Second × 3 ⇒ 3x + 3y = −3. Add ⇒ 7x = 21 ⇒ x = 3. 3 + y = −1 ⇒ y = −4.
When you multiply an equation, multiply both the left and the right side. Then the equation stays true.
Multiplying only the left side and leaving the right side unchanged: 3 × (2x + y = 4) ⇒ 6x + 3y = 4.
Take the lowest common multiple of the two coefficients to know the multipliers at once: 12 for 4 and 6. Memory clue: Multiply first, then eliminate: make one variable have the same coefficient in both equations.
Spot the difference
| Situation | Action | Example |
|---|---|---|
| One coefficient is a multiple of the other | Multiply one only | 3x + y and 2x + 2y: first × 2 |
| No coefficient is a multiple | Multiply both | 2y and 3y: make both 6y |
| Equation in ... = 0 form | Move the constant right | x + y − 5 = 0 ⇒ x + y = 5 |
Practice questions
- 1Solve 2x + y = 8 and x + 3y = 9. (Multiply the first by 3.)
- 2Solve x + 2y = 8 and 3x − y = 3. (Multiply the second by 2.)
- 3Solve 4x + y = 14 and 2x + 3y = 12.
- 4Solve 3x + 2y = 19 and 2x + 3y = 16.
- 5Solve 5x + 3y = 29 and 3x − 2y = 6.
- 6Solve 3x + 2y − 16 = 0 and x − 4y + 4 = 0.
- 73 notebooks and 2 pens cost ₹76, and 2 notebooks and 5 pens cost ₹80. Find the cost of one notebook and one pen.
- 8Solve 5x + 4y = 6 and 3x − 2y = 8.
Show answers (try them yourself first)
- 1First × 3 ⇒ 6x + 3y = 24. Subtract the second ⇒ 5x = 15 ⇒ x = 3; 6 + y = 8 ⇒ y = 2. (3, 2). 2) Second × 2 ⇒ 6x − 2y = 6. Add ⇒ 7x = 14 ⇒ x = 2; 2 + 2y = 8 ⇒ y = 3. (2, 3). 3) First × 3 ⇒ 12x + 3y = 42. Subtract the second ⇒ 10x = 30 ⇒ x = 3; 12 + y = 14 ⇒ y = 2. (3, 2).
- 4First × 3 ⇒ 9x + 6y = 57; second × 2 ⇒ 4x + 6y = 32. Subtract ⇒ 5x = 25 ⇒ x = 5; 15 + 2y = 19 ⇒ y = 2. (5, 2). 5) First × 2 ⇒ 10x + 6y = 58; second × 3 ⇒ 9x − 6y = 18. Add ⇒ 19x = 76 ⇒ x = 4; 20 + 3y = 29 ⇒ y = 3. (4, 3). 6) 3x + 2y = 16 and x − 4y = −4. First × 2 ⇒ 6x + 4y = 32. Add ⇒ 7x = 28 ⇒ x = 4; 4 − 4y = −4 ⇒ y = 2. (4, 2).
- 7Notebook x, pen y: 3x + 2y = 76, 2x + 5y = 80. First × 2 ⇒ 6x + 4y = 152; second × 3 ⇒ 6x + 15y = 240. Subtract ⇒ 11y = 88 ⇒ y = 8; x = 20. Notebook ₹20, pen ₹8. 8) Second × 2 ⇒ 6x − 4y = 16. Add ⇒ 11x = 22 ⇒ x = 2; 10 + 4y = 6 ⇒ y = −1. (2, −1); check: 6 + 2 = 8 ✔.
Substitution method
If 1 pen = 2 pencils, then wherever "3 pens" is written we can write "6 pencils". We write one thing in another form and put it in its place.
From one equation write one variable in terms of the other. Put that expression into the other equation (substitution). Now only one variable is left; solve it and then find the other variable.
In the figure: From (II) we get y = 3x − 6. Put it in place of y in (I) and only x is left: 7x − 12 = 9 ⇒ x = 3, then y = 3.
Rule 1: Isolate the variable whose coefficient is 1 — Keep the variable with coefficient 1 alone on one side. Then no fractions appear.
- Easy: x + 2y = 9 and 3x − y = 6: from the second, y = 3x − 6. Put in the first ⇒ x + 2(3x − 6) = 9 ⇒ 7x − 12 = 9 ⇒ x = 3. y = 9 − 6 = 3.
- Medium: 2x + 3y = 12 and x − y = 1: x = y + 1. Put in the first ⇒ 2(y + 1) + 3y = 12 ⇒ 5y + 2 = 12 ⇒ y = 2. x = 3.
- Exam type: 3x + y = 14 and 2x − 5y = −2: y = 14 − 3x. Put in the second ⇒ 2x − 5(14 − 3x) = −2 ⇒ 2x − 70 + 15x = −2 ⇒ 17x = 68 ⇒ x = 4. y = 14 − 12 = 2.
Rule 2: If a "variable = ..." form is already given, put it in directly — Use brackets, and apply the sign and multiplier in front of the bracket to every term inside.
- Easy: y = 2x and x + y = 12: x + 2x = 12 ⇒ 3x = 12 ⇒ x = 4. y = 8.
- Medium: x = 2y + 1 and 3x − 4y = 9: 3(2y + 1) − 4y = 9 ⇒ 6y + 3 − 4y = 9 ⇒ 2y = 6 ⇒ y = 3. x = 7.
- Exam type: y = 3x − 10 and 4x + 2y = 10: 4x + 2(3x − 10) = 10 ⇒ 10x − 20 = 10 ⇒ x = 3. y = 9 − 10 = −1. Check: 12 − 2 = 10 ✔.
Rule 3: Decide first which method is easier — Both methods give the same answer. Look at the coefficients and choose the easier one.
- Easy: x + y = 10 and x − y = 2: the y coefficients are +1 and −1, so elimination is easy. Add ⇒ 2x = 12 ⇒ x = 6, y = 4.
- Medium: y = 2x − 1 and 3x + y = 14: y is already alone, so substitution is easy. 3x + 2x − 1 = 14 ⇒ x = 3, y = 5.
- Exam type: 4x − 3y = 6 and 3x + 2y = 13: no coefficient is 1, so use elimination. First × 2 ⇒ 8x − 6y = 12; second × 3 ⇒ 9x + 6y = 39. Add ⇒ 17x = 51 ⇒ x = 3, y = 2.
Write the substituted expression inside brackets. Removing the bracket then handles the signs and multiplication correctly.
Writing 2 × 3x − 6 instead of 2y when substituting y = 3x − 6. The correct form is 2(3x − 6) = 6x − 12.
After finding x, use the simple "variable = ..." form to find y. You only need to put in the value. Memory clue: Substitution: "solve for one variable, put it in the other".
Spot the difference
| Method | When it is easy | Example |
|---|---|---|
| Elimination | coefficients equal or opposite | x + y = 10 and x − y = 2 |
| Substitution | one variable is already alone | y = 2x − 1 and 3x + y = 14 |
| Multiply, then eliminate | no coefficient is 1 | 4x − 3y = 6 and 3x + 2y = 13 |
Practice questions
- 1Solve x + y = 9 and y = 2x.
- 2Solve x = y + 2 and x + y = 10.
- 3Solve y = x − 1 and 2x + y = 11.
- 4Solve x + 3y = 11 and 2x − y = 1.
- 5Solve 2x + y = 7 and 3x − 2y = 7.
- 6Solve x = 3y + 5 and 2x + y = 17.
- 7Solve 4x − 3y = 7 and 2x + y = 11.
- 8If (2, 5) is the common solution of x + y = a and 3x − y = b, find a and b.
Show answers (try them yourself first)
- 1x + 2x = 9 ⇒ x = 3; y = 6. (3, 6). 2) (y + 2) + y = 10 ⇒ 2y = 8 ⇒ y = 4; x = 6. (6, 4). 3) 2x + x − 1 = 11 ⇒ 3x = 12 ⇒ x = 4; y = 3. (4, 3).
- 4x = 11 − 3y ⇒ 22 − 6y − y = 1 ⇒ 7y = 21 ⇒ y = 3; x = 2. (2, 3). 5) y = 7 − 2x ⇒ 3x − 14 + 4x = 7 ⇒ 7x = 21 ⇒ x = 3; y = 1. (3, 1). 6) 2(3y + 5) + y = 17 ⇒ 7y + 10 = 17 ⇒ y = 1; x = 8. (8, 1).
- 7y = 11 − 2x ⇒ 4x − 33 + 6x = 7 ⇒ 10x = 40 ⇒ x = 4; y = 3. (4, 3); check: 16 − 9 = 7 ✔. 8) a = 2 + 5 = 7 and b = 3 × 2 − 5 = 1.
Word problems: numbers, ages and digits
You buy 2 notebooks and 1 pen for ₹35, and 1 notebook and 1 pen for ₹20. The prices are unknown, so we call them x and y and write two equations.
When a word problem has two unknown quantities, forming two equations lets us solve it as simultaneous equations. Each sentence gives one equation.
In the figure: Use the same 6 steps for every word problem: read, choose variables, write two equations, solve, check, and write the answer in a sentence.
Rule 1: Problems on two numbers — Look for the words "sum", "difference", "times" and "more than" and write an equation for each sentence.
- Easy: Sum 30 and difference 8: x + y = 30, x − y = 8. Add ⇒ 2x = 38 ⇒ x = 19; y = 11. The numbers are 19 and 11.
- Medium: Three times the smaller number minus 4 gives the larger; the sum is 48. x = 3y − 4 and x + y = 48 ⇒ 4y − 4 = 48 ⇒ y = 13; x = 35. Check: 35 + 13 = 48 ✔.
- Exam type: Dividing the larger number by the smaller gives quotient 3 and remainder 4; the sum is 60. Larger = 3 × smaller + 4: x = 3y + 4 and x + y = 60 ⇒ 4y + 4 = 60 ⇒ y = 14; x = 46. Check: 3 × 14 + 4 = 46 ✔.
Rule 2: Problems on ages — Write separate equations for "today" and for "some years ago / later". The age n years ago is today's age − n.
- Easy: The mother's and daughter's ages add up to 40; the mother is 4 times as old. x = 4y and x + y = 40 ⇒ 5y = 40 ⇒ y = 8; x = 32. Mother 32 years, daughter 8 years.
- Medium: The father is 25 years older than the son; their ages add up to 61. x − y = 25 and x + y = 61 ⇒ 2x = 86 ⇒ x = 43; y = 18.
- Exam type: Today the father's and son's ages add up to 50. 5 years ago the father was 7 times as old as the son. x + y = 50 and x − 5 = 7(y − 5) ⇒ x − 7y = −30. Subtract ⇒ 8y = 80 ⇒ y = 10; x = 40. Check: 35 = 7 × 5 ✔.
Rule 3: Problems on two-digit numbers — If the tens digit is x and the units digit is y, the number is 10x + y. When the digits are reversed, the new number is 10y + x.
In the figure: In the number 47 the tens digit is 4 and the units digit is 7, so the number is 10 × 4 + 7. Reversing the digits gives 10 × 7 + 4 = 74.
- Easy: The digits add up to 11 and the units digit is 3 more than the tens digit. x + y = 11, y − x = 3 ⇒ 2y = 14 ⇒ y = 7; x = 4. The number is 47.
- Medium: The digits add up to 10; reversing them lowers the number by 18. (10x + y) − (10y + x) = 18 ⇒ x − y = 2. With x + y = 10 ⇒ x = 6, y = 4. The number is 64.
- Exam type: The number is 6 times the sum of its digits and the digits add up to 9. 10x + y = 6(x + y) ⇒ 4x − 5y = 0. With y = 9 − x: 4x − 45 + 5x = 0 ⇒ x = 5; y = 4. The number is 54. Check: 6 × 9 = 54 ✔.
A two-digit number is 10x + y. Do not write it as x + y; that is the sum of the digits.
Writing y + 5 for "5 years ago". The correct form is y − 5 (the earlier age is smaller).
Once you have the answer, test it against every sentence of the problem. Check in the words, not only in the equations. Memory clue: Word problem: read, choose the variables, write two equations.
Spot the difference
| Type of problem | Unknowns | First equation |
|---|---|---|
| Two numbers | larger x, smaller y | x + y = sum |
| Ages | today's ages x, y | x + y = sum of ages |
| Two-digit number | tens x, units y | number = 10x + y |
| 5 years ago | today's age − 5 | (x − 5) and (y − 5) |
Practice questions
- 1Two numbers have sum 25 and difference 7. Find them.
- 2Rahul is twice as old as Saili and their ages add up to 36. Find their ages.
- 3The digits of a two-digit number add up to 7 and the tens digit is 1 more than the units digit. Find the number.
- 4Today the father's and son's ages add up to 52. After 4 years the father will be 3 times as old as the son. Find their ages today.
- 5Twice the smaller number plus 5 gives the larger number, and the sum is 41. Find the numbers.
- 6The digits of a two-digit number add up to 12. Reversing the digits lowers the number by 36. Find the number.
- 7After 5 years a mother will be 3 times as old as her daughter, and 3 years ago she was 7 times as old. Find their present ages.
- 8A two-digit number is 7 times the sum of its digits, and reversing the digits lowers it by 18. Find the number.
Show answers (try them yourself first)
- 1x + y = 25, x − y = 7 ⇒ 2x = 32 ⇒ x = 16; y = 9. The numbers are 16 and 9. 2) x = 2y and x + y = 36 ⇒ 3y = 36 ⇒ y = 12; x = 24. Rahul 24, Saili 12. 3) x + y = 7, x − y = 1 ⇒ x = 4, y = 3. The number is 43.
- 4x + y = 52 and x + 4 = 3(y + 4) ⇒ x − 3y = 8. Subtract ⇒ 4y = 44 ⇒ y = 11; x = 41. Check: 45 = 3 × 15 ✔. 5) x = 2y + 5 and x + y = 41 ⇒ 3y + 5 = 41 ⇒ y = 12; x = 29. The numbers are 29 and 12. 6) x + y = 12 and 9(x − y) = 36 ⇒ x − y = 4 ⇒ x = 8, y = 4. The number is 84. Check: 84 − 48 = 36 ✔.
- 7Mother x, daughter y. x + 5 = 3(y + 5) ⇒ x − 3y = 10; x − 3 = 7(y − 3) ⇒ x − 7y = −18. Subtract ⇒ 4y = 28 ⇒ y = 7; x = 31. Mother 31, daughter 7. 8) 10x + y = 7(x + y) ⇒ x = 2y. 9(x − y) = 18 ⇒ x − y = 2 ⇒ y = 2, x = 4. The number is 42. Check: 7 × 6 = 42 and 42 − 24 = 18 ✔.
Word problems: money, fractions, geometry and speed
A shopkeeper has ₹10 and ₹20 notes. If the total number of notes and the total amount are known, two equations tell us how many notes of each kind there are.
The method is the same here: choose two unknowns, write two equations and solve. Only the underlying rule changes: amount = number × value, perimeter = 2 × (length + breadth), distance = speed × time.
In the figure: 10 notes of ₹10 (₹100) and 8 notes of ₹20 (₹160) make 18 notes and ₹260 together.
Rule 1: Money problems — The first equation is about the number of notes and the second about the total amount. Amount = number of notes × value of each note.
- Easy: 12 coins of ₹5 and ₹2 worth ₹42: x + y = 12, 5x + 2y = 42. First × 2 ⇒ 2x + 2y = 24. Subtract ⇒ 3x = 18 ⇒ x = 6; y = 6.
- Medium: 18 notes of ₹10 and ₹20 worth ₹260: x + y = 18, 10x + 20y = 260 ⇒ x + 2y = 26. Subtract ⇒ y = 8; x = 10.
- Exam type: 3 notebooks and 4 pens cost ₹130; 2 notebooks and 3 pens cost ₹90. First × 2 ⇒ 6x + 8y = 260; second × 3 ⇒ 6x + 9y = 270. Subtract ⇒ y = 10; x = 30. Notebook ₹30, pen ₹10.
Rule 2: Fraction problems — Take the numerator as x and the denominator as y. Apply the changes in the words (added, subtracted) to the numerator and denominator, then cross-multiply.
- Easy: The numerator and denominator add up to 17; the denominator is 3 more than the numerator. x + y = 17, y − x = 3 ⇒ y = 10, x = 7. The fraction is 7/10.
- Medium: The numerator and denominator add up to 10; adding 1 to both gives 1/2. (x + 1)/(y + 1) = 1/2 ⇒ 2x + 2 = y + 1 ⇒ 2x − y = −1. With x + y = 10 ⇒ add ⇒ 3x = 9 ⇒ x = 3, y = 7. The fraction is 3/7.
- Exam type: Subtracting 1 from the numerator gives 1/3, and adding 1 to the denominator gives 1/2. 3(x − 1) = y ⇒ 3x − y = 3; 2x = y + 1 ⇒ 2x − y = 1. Subtract ⇒ x = 2; y = 3. The fraction is 2/3. Check: 1/3 ✔ and 2/4 = 1/2 ✔.
Rule 3: Geometry and speed problems — Turn rules such as perimeter = 2(length + breadth), angles of a triangle add up to 180°, and distance = speed × time into equations.
In the figure: The length is twice the breadth (x = 2y) and the perimeter is 36 cm: 2(x + y) = 36 ⇒ x + y = 18. 2y + y = 18 ⇒ y = 6, x = 12.
In the figure: When vehicles approach each other their speeds add (x + y); when one chases the other, the difference of speeds (x − y) closes the gap.
- Easy: A rectangle's length is twice its breadth and its perimeter is 36 cm: x = 2y and x + y = 18 ⇒ 3y = 18 ⇒ y = 6; x = 12. Length 12 cm, breadth 6 cm.
- Medium: One angle of a triangle is 40°; the other two angles are x and y with difference 20°. x + y = 180 − 40 = 140 and x − y = 20 ⇒ 2x = 160 ⇒ x = 80; y = 60.
- Exam type: A and B are 180 km apart. Starting towards each other they meet after 2 hours; going in the same direction A catches B after 6 hours. 2(x + y) = 180 ⇒ x + y = 90; 6(x − y) = 180 ⇒ x − y = 30. Add ⇒ x = 60; y = 30. Speeds are 60 km/h and 30 km/h.
Remember the rule for each type: amount = number × value, perimeter = 2(length + breadth), distance = speed × time. Then the equations come easily.
Writing x + y = 36 when the perimeter is 36. The correct form is 2(x + y) = 36, that is x + y = 18.
Use the difference of speeds when moving in the same direction and the sum of speeds when moving towards each other. Memory clue: Speed: add when facing each other, subtract in the same direction; "face-to-face add, chase subtract".
Spot the difference
| Type of problem | Idea | Example equations |
|---|---|---|
| Notes / coins | number × value | x + y = 18 ; 10x + 20y = 260 |
| Fractions | numerator x, denominator y | y = x + 3 ; x + y = 17 |
| Rectangle | perimeter = 2(x + y) | 2(x + y) = 36 ; x = 2y |
| Speed and distance | distance = speed × time | 2(x + y) = 180 ; 6(x − y) = 180 |
Practice questions
- 1There are 14 notes of ₹10 and ₹5 worth ₹100 in all. How many notes of each kind?
- 2The length of a rectangle is 4 cm more than its breadth and its perimeter is 32 cm. Find the length and breadth.
- 3The numerator and denominator of a fraction add up to 13 and the denominator is 1 more than the numerator. Find the fraction.
- 4Two vehicles 240 km apart meet after 3 hours when going towards each other and after 12 hours when going the same way. Find their speeds.
- 52 cups and 3 saucers cost ₹130, and 3 cups and 2 saucers cost ₹120. Find the cost of a cup and a saucer.
- 6In an isosceles triangle the equal angles are x each and the third angle is y. y is 30° more than x. Find all three angles.
- 7Adding 3 to the numerator of a fraction makes it 1, and subtracting 1 from the denominator makes it 1/2. Find the fraction.
- 8An exam had 40 questions. A correct answer earns 3 marks and a wrong answer loses 1 mark. A student attempted all and got 100 marks. How many answers were wrong?
Show answers (try them yourself first)
- 1x + y = 14, 10x + 5y = 100 ⇒ 2x + y = 20. Subtract ⇒ x = 6; y = 8. 6 notes of ₹10 and 8 notes of ₹5. 2) x − y = 4 and x + y = 16 ⇒ x = 10; y = 6. Length 10 cm, breadth 6 cm. 3) x + y = 13, y − x = 1 ⇒ y = 7; x = 6. The fraction is 6/7.
- 43(x + y) = 240 ⇒ x + y = 80; 12(x − y) = 240 ⇒ x − y = 20. Add ⇒ x = 50; y = 30. Speeds 50 and 30 km/h. 5) 2x + 3y = 130, 3x + 2y = 120. Add ⇒ 5x + 5y = 250 ⇒ x + y = 50. Subtract ⇒ x − y = −10. x = 20; y = 30. Cup ₹20, saucer ₹30. 6) 2x + y = 180 and y = x + 30 ⇒ 3x + 30 = 180 ⇒ x = 50; y = 80. Angles 50°, 50°, 80°.
- 7x + 3 = y ⇒ y = x + 3; x/(y − 1) = 1/2 ⇒ 2x = y − 1. Put y ⇒ 2x = x + 2 ⇒ x = 2; y = 5. The fraction is 2/5. Check: 5/5 = 1 ✔ and 2/4 = 1/2 ✔. 8) Correct x, wrong y: x + y = 40, 3x − y = 100. Add ⇒ 4x = 140 ⇒ x = 35; y = 5. 5 answers were wrong. Check: 105 − 5 = 100 ✔.