Pair of Linear Equations in Two Variables studies two linear equations in x and y taken together, and the values that satisfy both. Each equation is a straight line, so a pair is two lines that intersect, run parallel or coincide, giving one solution, no solution or infinitely many. The chapter teaches the graphical method, the substitution method and the elimination method, and a ratio test on the coefficients that decides the case without drawing. Its word problems on ages, digits, fractions, prices and fixed charges make it a steady, checkable source of board marks.
Key Concepts
1. Linear Equations in Two Variables and a Pair of Them
A linear equation in two variables is an equation that can be written as ax + by + c = 0, where a, b and c are real numbers and a and b are not both zero. Each variable appears only to the power 1. There is no x2, no xy and no 1/x. So 3x + 4y = 20 is linear, while x2 + y = 5 and xy = 6 are not.
A pair of linear equations in two variables is two such equations in the same two variables, taken together. The general form of a pair is:
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
Here a1, b1, c1 are the coefficients of the first equation and a2, b2, c2 those of the second. A solution of the pair is a pair of values (x, y) that satisfies both equations at the same time. One equation alone has infinitely many solutions. The pair narrows this down to the values that work for both.
Where a pair comes from. The chapter opens with Akhila at a village fair. She rides the Giant Wheel and plays Hoopla. The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs ₹3, a game of Hoopla costs ₹4, and she spent ₹20. Let the number of rides be x and the number of Hoopla games be y. The two conditions give two equations:
- “Hoopla is half the rides”: y = x/2
- “She spent ₹20”: 3x + 4y = 20
Two unknowns need two separate conditions. Each condition in the story becomes one equation. Solving this pair gives x = 4 and y = 2 (see Competency Question 1).
2. Graphical Meaning: Three Possible Cases
The graph of every linear equation in two variables is a straight line. So a pair of linear equations is shown by two lines on the same axes. Every point on a line is a solution of that line’s equation. A point that lies on both lines is a solution of the pair. Two lines in a plane can sit in only three ways:
| How the lines behave | Number of solutions | Name of the pair |
|---|---|---|
| Intersect in a single point | Exactly one (unique) solution: the point of intersection | Consistent |
| Parallel (never meet) | No solution | Inconsistent |
| Coincident (one lies exactly on the other) | Infinitely many solutions: every point on the line | Dependent (and consistent) |
The textbook’s definitions are worth learning word for word:
- A pair of linear equations which has no solution is called an inconsistent pair.
- A pair of linear equations in two variables which has a solution is called a consistent pair.
- A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a dependent pair. A dependent pair is always consistent.
“Equivalent” means one equation is a non-zero multiple of the other. For example, 2x + 3y − 9 = 0 and 4x + 6y − 18 = 0 are equivalent, because the second is the first multiplied by 2. Both equations describe the same line, so every point of that line solves both.
3. Solving a Pair Graphically
To solve a pair by graph, draw both lines on the same graph paper and read off the point where they meet. For each equation:
- Rewrite it to give y in terms of x (or x in terms of y).
- Choose at least two convenient values of x and find the matching y. A third point is a useful check. Values that give whole numbers are easiest to plot. Putting x = 0 and y = 0 gives the points where the line cuts the axes.
- Plot the points and join them with a straight line, extended both ways.
Then look at the two lines. If they meet at a point, read its coordinates: that is the solution. If they are parallel, there is no solution. If they lie on top of each other, there are infinitely many.
Worked illustration (NCERT Example 1). Check graphically whether x + 3y = 6 and 2x − 3y = 12 is consistent, and if so, solve.
| x + 3y = 6, so y = (6 − x)/3 | 2x − 3y = 12, so y = (2x − 12)/3 | ||||
|---|---|---|---|---|---|
| x | 0 | 6 | x | 0 | 3 |
| y | 2 | 0 | y | −4 | −2 |
Plot A(0, 2) and B(6, 0) and join them to get line AB. Plot P(0, −4) and Q(3, −2) and join them to get line PQ. The point B(6, 0) is common to both lines. So the solution is x = 6, y = 0, and the pair is consistent. Check: 6 + 3(0) = 6 and 2(6) − 3(0) = 12.
Limitation of the graph. The graphical method is not convenient when the solution has non-integral coordinates such as (−1.75, 3.3) or (4/13, 1/19). Such points are hard to read exactly and mistakes are likely. That is why the algebraic methods come next.
4. Comparing Ratios of Coefficients
You can tell which of the three cases a pair falls into without drawing, by comparing three ratios: a1/a2, b1/b2 and c1/c2. The textbook builds this rule from three pairs whose graphs are known (Table 3.1):
| Pair | a1/a2 | b1/b2 | c1/c2 | Compare | Graph | Solutions |
|---|---|---|---|---|---|---|
| x − 2y = 0 3x + 4y − 20 = 0 | 1/3 | −2/4 | 0/(−20) | a1/a2 ≠ b1/b2 | Intersecting | Exactly one |
| 2x + 3y − 9 = 0 4x + 6y − 18 = 0 | 2/4 | 3/6 | −9/−18 | All three equal | Coincident | Infinitely many |
| x + 2y − 4 = 0 2x + 4y − 12 = 0 | 1/2 | 2/4 | −4/−12 | a1/a2 = b1/b2 ≠ c1/c2 | Parallel | No solution |
So for a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0:
- a1/a2 ≠ b1/b2: the lines intersect, one unique solution, consistent.
- a1/a2 = b1/b2 = c1/c2: the lines coincide, infinitely many solutions, dependent and consistent.
- a1/a2 = b1/b2 ≠ c1/c2: the lines are parallel, no solution, inconsistent.
The textbook adds that the converse is also true for any pair of lines. So if the first two ratios are equal and the third is different, the lines must be parallel, and the same holds for the other two cases.
Keep the signs. Always write the constant with its sign. In 3x + 4y − 20 = 0, c = −20. If both equations are written in the form ax + by = c instead, the constants both change sign, so the ratio c1/c2 stays the same. The comparison still works as long as both equations are written in the same form.
Finding an unknown coefficient. Many questions give a pair with a letter such as k and ask for the value of k that makes the pair consistent, inconsistent or dependent. Set up the right ratio condition and solve for k. For example, x + 2y = 3 and 5x + ky = 7 have a unique solution when 1/5 ≠ 2/k, that is when k ≠ 10.
5. Substitution Method
The substitution method removes one variable by writing it in terms of the other. The textbook sets out three steps:
- Step 1: From either equation, whichever is convenient, find the value of one variable, say y, in terms of the other, x.
- Step 2: Substitute this value of y in the other equation. It becomes an equation in x alone. Solve it.
- Step 3: Substitute the value of x from Step 2 into the equation used in Step 1 to get y.
Worked illustration (NCERT Example 4). Solve 7x − 15y = 2 and x + 2y = 3.
Step 1: from x + 2y = 3, x = 3 − 2y.
Step 2: put this in 7x − 15y = 2: 7(3 − 2y) − 15y = 2, so 21 − 14y − 15y = 2, so −29y = −19 and y = 19/29.
Step 3: x = 3 − 2(19/29) = 87/29 − 38/29 = 49/29.
Solution: x = 49/29, y = 19/29. Check in the first equation: 7(49/29) − 15(19/29) = (343 − 285)/29 = 58/29 = 2.
When the variable disappears. Sometimes Step 2 gives a statement with no variable left:
- If the statement is true (like 18 = 18), the pair has infinitely many solutions. In NCERT Example 6, 2x + 3y = 9 and 4x + 6y = 18 (the cost of pencils and erasers) give 18 − 6y + 6y = 18, that is 18 = 18. The two equations are the same, so no unique cost can be found.
- If the statement is false (like −4 = 0), the pair has no solution. In NCERT Example 7, the rails x + 2y − 4 = 0 and 2x + 4y − 12 = 0 give 2(4 − 2y) + 4y − 12 = 0, that is 8 − 12 = 0, or −4 = 0. So the rails will not cross each other.
6. Elimination Method
The elimination method removes (eliminates) one variable by making its coefficients equal in both equations and then adding or subtracting. It is sometimes more convenient than substitution, especially when no coefficient is 1. The textbook’s steps:
- Step 1: Multiply both equations by suitable non-zero constants so that the coefficients of one variable (x or y) become numerically equal.
- Step 2: Add or subtract one equation from the other so that this variable is eliminated. If you get an equation in one variable, go to Step 3. If you get a true statement with no variable, the pair has infinitely many solutions. If you get a false statement with no variable, the pair has no solution (inconsistent).
- Step 3: Solve the equation in one variable.
- Step 4: Substitute this value into either original equation to find the other variable.
Adding or subtracting. If the equal coefficients have the same sign, subtract. If they have opposite signs, add. For example, in 3x + 5y = 21 and 2x − 5y = −1, the y-terms are +5y and −5y, so adding gives 5x = 20 and x = 4.
Worked illustration (NCERT Example 8). The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. Each saves ₹2000 per month. Let the incomes be ₹9x and ₹7x and the expenditures ₹4y and ₹3y. Savings = income − expenditure, so:
9x − 4y = 2000 (1)
7x − 3y = 2000 (2)
Step 1: multiply (1) by 3 and (2) by 4 to make the y-coefficients equal: 27x − 12y = 6000 (3) and 28x − 12y = 8000 (4).
Step 2: subtract (3) from (4): x = 2000.
Step 3: substitute in (1): 9(2000) − 4y = 2000, so 4y = 16000 and y = 4000.
The monthly incomes are 9 × 2000 = ₹18,000 and 7 × 2000 = ₹14,000.
Verification: 18000 : 14000 = 9 : 7. Expenditures are 16000 and 12000, and 16000 : 12000 = 4 : 3.
7. Forming Equations from Word Problems
Many questions in this chapter are word problems. The method is always the same:
- Name the two unknowns clearly, with units. For example, “let the cost of one bat be ₹x and one ball be ₹y”.
- Turn each condition into one equation.
- Solve by substitution or elimination (or by graph if the question asks for it).
- Answer in words, with units, and verify against the original conditions.
The common types and how to write their equations:
| Type | How to set it up |
|---|---|
| Ages | Present ages x and y. “n years ago” means x − n and y − n. “n years hence” means x + n and y + n. Both people age by the same amount. |
| Two-digit numbers | Tens digit x, units digit y. Number = 10x + y. Reversed number = 10y + x. “Digits differ by 2” gives two cases: x − y = 2 or y − x = 2. |
| Fractions | Let the fraction be x/y. “Add 2 to both” gives (x + 2)/(y + 2). Cross-multiply to get a linear equation. |
| Fixed charge plus rate | Total = fixed charge + rate × quantity. For a taxi: x + 10y = 105 for 10 km. For a library with a fixed charge for the first three days: total = x + (days − 3)y. |
| Costs and notes | Number of items and total money give two equations: x + y = 25 notes, 50x + 100y = 2000 rupees. |
| Ratios | Write the quantities as 9x and 7x (common multiplier), then use the extra condition. |
| Geometry | Supplementary angles: x + y = 180. Half the perimeter of a rectangle: length + width. |
Which method to use. A question may name the method; then you must use it. If it does not, pick substitution when a variable has coefficient 1 or −1, and elimination when all coefficients are larger numbers. Use the graph only when asked, or when you must describe the lines or a region.
Formula and Theorem Sheet
| Topic | Rule or result | Remember |
|---|---|---|
| Linear equation in two variables | ax + by + c = 0, a and b not both zero | Powers of x and y are 1; no xy term |
| General form of a pair | a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 | Keep the sign of each coefficient |
| Solution of a pair | (x, y) satisfying both equations | Check by substituting in both |
| Intersecting lines | a1/a2 ≠ b1/b2 | Unique solution; consistent |
| Coincident lines | a1/a2 = b1/b2 = c1/c2 | Infinitely many; dependent (consistent) |
| Parallel lines | a1/a2 = b1/b2 ≠ c1/c2 | No solution; inconsistent |
| Substitution method | Express one variable in terms of the other, substitute in the other equation | Choose the variable with coefficient 1 or −1 |
| Elimination method | Equalise one coefficient, then add or subtract | Same signs: subtract. Opposite signs: add |
| No variable left, true statement | Example: 18 = 18 | Infinitely many solutions |
| No variable left, false statement | Example: 0 = 9 or −4 = 0 | No solution |
| Two-digit number | 10x + y; reversed 10y + x | x is the tens digit, y the units digit |
| Ages | n years ago: x − n; n years later: x + n | Apply the shift to both people |
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Important Definitions
- Linear equation in two variables: an equation that can be written as ax + by + c = 0, where a, b, c are real numbers and a and b are not both zero.
- Pair of linear equations in two variables: two linear equations in the same two variables, considered together.
- Solution of a pair: a pair of values of x and y that satisfies both equations at the same time. On a graph, it is a point common to both lines.
- Consistent pair: a pair of linear equations in two variables which has a solution. Its lines intersect or coincide.
- Inconsistent pair: a pair of linear equations which has no solution. Its lines are parallel.
- Dependent pair: a pair of linear equations which are equivalent, and so have infinitely many distinct common solutions. Its lines coincide. A dependent pair is always consistent.
- Unique solution: exactly one solution, which happens when the two lines intersect in a single point.
- Substitution method: solving a pair by expressing one variable in terms of the other from one equation and substituting it into the other equation.
- Elimination method: solving a pair by making the coefficients of one variable numerically equal and then adding or subtracting the equations to remove that variable.
Solved Examples (NCERT-Based)
Example 1: Pencils and pens, solved graphically (NCERT Exercise 3.1)
Question: Form the pair of linear equations in the following problem, and find its solution graphically: 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.
Solution: Let one pencil cost ₹x and one pen cost ₹y. The equations are 5x + 7y = 50 and 7x + 5y = 46.
| 5x + 7y = 50 | 7x + 5y = 46 | ||||||
|---|---|---|---|---|---|---|---|
| x | 3 | 10 | −4 | x | 3 | 8 | −2 |
| y | 5 | 0 | 10 | y | 5 | −2 | 12 |
Plot both sets of points and draw the two lines. They intersect at (3, 5). So one pencil costs ₹3 and one pen costs ₹5. Check: 5(3) + 7(5) = 15 + 35 = 50 and 7(3) + 5(5) = 21 + 25 = 46.
Example 2: Intersecting, parallel or coincident (NCERT Exercise 3.1)
Question: On comparing the ratios a1/a2, b1/b2 and c1/c2, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (i) 5x − 4y + 8 = 0, 7x + 6y − 9 = 0 (ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0 (iii) 6x − 3y + 10 = 0, 2x − y + 9 = 0
Solution:
- (i) a1/a2 = 5/7 and b1/b2 = −4/6 = −2/3. Since 5/7 ≠ −2/3, the lines intersect at a point.
- (ii) 9/18 = 1/2, 3/6 = 1/2, 12/24 = 1/2. All three ratios are equal, so the lines are coincident.
- (iii) 6/2 = 3, −3/−1 = 3, 10/9. The first two are equal but differ from the third, so the lines are parallel.
Example 3: Consistency check, then a graphical solution (NCERT Exercise 3.1)
Question: Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: 2x + y − 6 = 0, 4x − 2y − 4 = 0
Solution: a1/a2 = 2/4 = 1/2 and b1/b2 = 1/(−2) = −1/2. These are unequal, so the lines intersect and the pair is consistent.
For 2x + y = 6, y = 6 − 2x: points (0, 6), (3, 0), (2, 2). For 4x − 2y = 4, y = 2x − 2: points (0, −2), (1, 0), (2, 2). Both lists contain (2, 2), so the lines meet there. Solution: x = 2, y = 2. Check: 2(2) + 2 − 6 = 0 and 4(2) − 2(2) − 4 = 0.
Example 4: Writing a second equation for each case (NCERT Exercise 3.1)
Question: Given the linear equation 2x + 3y − 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is: (i) intersecting lines (ii) parallel lines (iii) coincident lines.
Solution: Here a1 = 2, b1 = 3, c1 = −8.
- (i) Intersecting: we need a1/a2 ≠ b1/b2. Take 3x + 2y − 7 = 0: 2/3 ≠ 3/2.
- (ii) Parallel: multiply the x and y coefficients by the same number but not the constant. Take 4x + 6y − 8 = 0: 2/4 = 3/6 = 1/2 but −8/−8 = 1.
- (iii) Coincident: multiply the whole equation by one number. Take 6x + 9y − 24 = 0: 2/6 = 3/9 = −8/−24 = 1/3.
Many answers are possible. Any equation that meets the ratio condition is correct.
Example 5: Triangle formed with the x-axis (NCERT Exercise 3.1)
Question: Draw the graphs of the equations x − y + 1 = 0 and 3x + 2y − 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution: For x − y + 1 = 0, y = x + 1: points (−1, 0), (0, 1), (2, 3). For 3x + 2y − 12 = 0, y = (12 − 3x)/2: points (4, 0), (0, 6), (2, 3).
The two lines meet at (2, 3), since (2, 3) is in both lists. Check: 2 − 3 + 1 = 0 and 6 + 6 − 12 = 0. The first line cuts the x-axis (where y = 0) at (−1, 0). The second cuts it at (4, 0).
The vertices of the triangle are (2, 3), (−1, 0) and (4, 0). Shade the region enclosed by the two lines and the x-axis between x = −1 and x = 4. Its base on the x-axis is 4 − (−1) = 5 units and its height is 3 units, so its area is (1/2) × 5 × 3 = 7.5 square units.
Example 6: Solve, then find m (NCERT Exercise 3.2)
Question: Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which y = mx + 3.
Solution: From the first equation, 2x = 11 − 3y. Substitute into the second: (11 − 3y) − 4y = −24, so 11 − 7y = −24, −7y = −35 and y = 5. Then 2x = 11 − 15 = −4, so x = −2.
Now y = mx + 3 gives 5 = m(−2) + 3, so −2m = 2 and m = −1.
Example 7: Taxi charges (NCERT Exercise 3.2)
Question: The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Solution: Let the fixed charge be ₹x and the charge per km be ₹y.
x + 10y = 105 (1)
x + 15y = 155 (2)
From (1), x = 105 − 10y. Substitute in (2): 105 − 10y + 15y = 155, so 5y = 50 and y = 10. Then x = 105 − 100 = 5.
The fixed charge is ₹5 and the charge per km is ₹10. For 25 km: 5 + 25 × 10 = ₹255.
Example 8: Finding the fraction (NCERT Exercise 3.2)
Question: A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
Solution: Let the fraction be x/y.
(x + 2)/(y + 2) = 9/11 gives 11x + 22 = 9y + 18, so 11x − 9y = −4 (1)
(x + 3)/(y + 3) = 5/6 gives 6x + 18 = 5y + 15, so 6x − 5y = −3 (2)
Multiply (1) by 5 and (2) by 9: 55x − 45y = −20 and 54x − 45y = −27. Subtract: x = 7. From (1): 77 − 9y = −4, so 9y = 81 and y = 9.
The fraction is 7/9. Check: 9/11 is correct, and 10/12 = 5/6.
Example 9: Jacob and his son (NCERT Exercise 3.2)
Question: Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution: Let Jacob’s present age be x years and his son’s be y years.
Five years hence: x + 5 = 3(y + 5), so x − 3y = 10 (1)
Five years ago: x − 5 = 7(y − 5), so x − 7y = −30 (2)
Subtract (2) from (1): 4y = 40, so y = 10. Then x = 10 + 3(10) = 40.
Jacob is 40 years old and his son is 10 years old. Check: in 5 years, 45 = 3 × 15. Five years ago, 35 = 7 × 5.
Example 10: Elimination with fractional answers (NCERT Exercise 3.3)
Question: Solve the following pair of linear equations by the elimination method and the substitution method: 3x − 5y − 4 = 0 and 9x = 2y + 7
Solution: Rewrite: 3x − 5y = 4 (1) and 9x − 2y = 7 (2).
Multiply (1) by 3: 9x − 15y = 12 (3). Subtract (3) from (2): (9x − 2y) − (9x − 15y) = 7 − 12, so 13y = −5 and y = −5/13.
From (1): 3x = 4 + 5y = 4 − 25/13 = 27/13, so x = 9/13.
By substitution: from (1), x = (4 + 5y)/3. Put this in (2): 3(4 + 5y) − 2y = 7, so 12 + 13y = 7 and y = −5/13, the same value.
Solution: x = 9/13, y = −5/13. Check in (2): 9(9/13) − 2(−5/13) = 81/13 + 10/13 = 91/13 = 7.
Example 11: The digit problem (NCERT Exercise 3.3)
Question: The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Solution: Let the tens digit be x and the units digit be y. The number is 10x + y and the reversed number is 10y + x.
x + y = 9 (1)
9(10x + y) = 2(10y + x), so 90x + 9y = 20y + 2x, so 88x − 11y = 0, that is y = 8x (2)
Substitute (2) in (1): x + 8x = 9, so x = 1 and y = 8. The number is 18. Check: 1 + 8 = 9, and 9 × 18 = 162 = 2 × 81.
Competency-Based Questions (with answers)
1. Case-based: Akhila at the fair
Akhila went to a fair. The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs ₹3 and a game of Hoopla costs ₹4. She spent ₹20 in all.
(a) Write the pair of equations, taking x rides and y Hoopla games. (b) Solve the pair. (c) If she had spent ₹30 with the same rule, how many rides would she have had?
Answer: (a) y = x/2 and 3x + 4y = 20. (b) Substituting, 3x + 2x = 20, so x = 4 and y = 2. (c) 3x + 2x = 30 gives x = 6 rides (and 3 Hoopla games). Check: 18 + 12 = 30.
2. Case-based: The gym membership
A gym charges a one-time joining fee plus a fixed fee for every month. Riya paid ₹4,100 in total for 3 months. Kabir paid ₹8,500 in total for 7 months.
(a) Form the pair of equations. (b) Find the joining fee and the monthly fee. (c) How much would a 12-month membership cost in total?
Answer: (a) Let the joining fee be ₹x and the monthly fee ₹y. Then x + 3y = 4100 and x + 7y = 8500. (b) Subtracting: 4y = 4400, so y = 1100. Then x = 4100 − 3300 = 800. The joining fee is ₹800 and the monthly fee is ₹1,100. (c) 800 + 12 × 1100 = ₹14,000.
3. Source-based: Will the rails cross?
In NCERT Example 7, two rails are represented by the equations x + 2y − 4 = 0 and 2x + 4y − 12 = 0. Substituting x = 4 − 2y into the second equation gives 8 − 12 = 0, that is −4 = 0.
(a) What does the false statement tell you about the rails? (b) Confirm the answer using the ratios of coefficients. (c) Change the constant in the second equation so that the two rails lie along the same line.
Answer: (a) A false statement with no variable means the pair has no common solution. The rails will not cross each other; they are parallel. (b) 1/2 = 2/4 = 1/2, but c1/c2 = −4/−12 = 1/3. So a1/a2 = b1/b2 ≠ c1/c2: parallel lines, inconsistent pair. (c) For coincident lines we need c1/c2 = 1/2, so c2 = −8. The second rail would be 2x + 4y − 8 = 0.
4. Assertion-Reason
Assertion (A): The pair x + 2y = 3 and 2x + 4y = 6 has infinitely many solutions.
Reason (R): If a1/a2 = b1/b2 = c1/c2, the lines are coincident and the pair has infinitely many solutions.
(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.
Answer: (a). The ratios are 1/2, 2/4 and 3/6, all equal to 1/2, so A is true. R is the correct rule, and it is exactly the reason A holds.
5. Assertion-Reason
Assertion (A): The pair 2x + 3y = 5 and 4x + 6y = 11 has a unique solution.
Reason (R): If a1/a2 = b1/b2 ≠ c1/c2, the lines representing the pair are parallel.
(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.
Answer: (d). Here 2/4 = 3/6 = 1/2 but 5/11 ≠ 1/2. The lines are parallel, so the pair has no solution and A is false. R is a true statement of the rule.
6. Assertion-Reason
Assertion (A): The graph of a dependent pair of linear equations is a pair of coincident lines.
Reason (R): A dependent pair of linear equations is always consistent.
(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.
Answer: (b). A is true: the equations of a dependent pair are equivalent, so they describe the same line. R is also true, as stated in the textbook. But being consistent only means having a solution; intersecting lines are consistent too. So R does not explain why the lines coincide.
7. Error analysis: The sign slip
A student solves 3x + 2y = 11 and 2x + 3y = 4 by elimination as follows. “Multiply the first by 3 and the second by 2: 9x + 6y = 33 and 4x + 6y = 8. Add them: 13x = 41, so x = 41/13.” Find the error and solve correctly.
Answer: The y-coefficients are both +6y, with the same sign, so the equations must be subtracted. Adding leaves 12y in the result, so the student’s line 13x = 41 is wrong as well. Correct step: (9x + 6y) − (4x + 6y) = 33 − 8, so 5x = 25 and x = 5. Then 3(5) + 2y = 11, so 2y = −4 and y = −2. Check: 2(5) + 3(−2) = 10 − 6 = 4.
8. Competency MCQ: Value of k for no solution
For which value of k does the pair kx + 3y = k − 3 and 12x + ky = k have no solution?
(A) 6 (B) −6 (C) ±6 (D) 0
Answer: (B) −6. No solution needs k/12 = 3/k ≠ (k − 3)/k. From k/12 = 3/k, k2 = 36, so k = 6 or k = −6. For k = 6: the ratios are 1/2, 1/2 and 3/6 = 1/2, all equal, so the lines coincide (infinitely many solutions). Reject 6. For k = −6: the ratios are −1/2, −1/2 and −9/−6 = 3/2, so the third is different and the lines are parallel. Only k = −6 works.
Important Questions for Board Exams
1-Mark Questions
- For what value of k is the pair 2x + 3y = 7 and 4x + ky = 14 dependent? Need 2/4 = 3/k = 7/14. So 3/k = 1/2 and k = 6.
- Is the pair x − 2y = 0 and 3x + 4y − 20 = 0 consistent? 1/3 ≠ −2/4, so the lines intersect. Yes, it is consistent with a unique solution.
- How many solutions does a pair of linear equations have if its lines are parallel? None. The pair is inconsistent.
2-Mark Questions
- Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden. Let the length be l m and the width w m. l = w + 4 and l + w = 36. Substituting: 2w + 4 = 36, so w = 16 and l = 20. The garden is 20 m long and 16 m wide.
- The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. x + y = 180 and x − y = 18. Adding: 2x = 198, x = 99, so y = 81. The angles are 99° and 81°.
- Solve the following pair of linear equations by the elimination method and the substitution method: 3x + 4y = 10 and 2x − 2y = 2. By substitution: the second gives x − y = 1, so x = y + 1. Then 3(y + 1) + 4y = 10, 7y = 7, y = 1 and x = 2. By elimination: multiply 2x − 2y = 2 by 2 to get 4x − 4y = 4. Add it to 3x + 4y = 10: 7x = 14, x = 2, and then 4y = 10 − 6 = 4, y = 1. Both methods give x = 2, y = 1.
- Find k for which kx + 2y = 5 and 3x + y = 1 have no solution. Need k/3 = 2/1 ≠ 5/1. So k = 6, and then 2 ≠ 5 holds. k = 6.
3-Mark Questions
- Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? Let their present ages be x and y. x − 5 = 3(y − 5) gives x − 3y = −10. x + 10 = 2(y + 10) gives x − 2y = 10. Subtract the first from the second: y = 20. Then x = 10 + 2(20) = 50. Nuri is 50 years and Sonu is 20 years. Check: 45 = 3 × 15 and 60 = 2 × 30.
- Find a and b for which 2x + 3y = 7 and (a − b)x + (a + b)y = 3a + b − 2 have infinitely many solutions. Need 2/(a − b) = 3/(a + b) = 7/(3a + b − 2). From the first two: 2a + 2b = 3a − 3b, so a = 5b. Then a + b = 6b and 3a + b − 2 = 16b − 2. From 3/(6b) = 7/(16b − 2): 3(16b − 2) = 42b, so 48b − 6 = 42b and b = 1. Then a = 5. Check: the ratios are 2/4, 3/6 and 7/14, all 1/2.
- A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. Let the fixed charge be ₹x and the extra charge ₹y per day. Seven days means 4 extra days: x + 4y = 27. Five days means 2 extra days: x + 2y = 21. Subtract: 2y = 6, y = 3. Then x = 21 − 6 = 15. Fixed charge ₹15, extra ₹3 per day.
5-Mark Questions
- Solve graphically: 2x + y = 6 and 2x − y + 2 = 0. Find the vertices of the triangle formed by these lines and the x-axis, and its area. For 2x + y = 6, y = 6 − 2x: points (0, 6), (3, 0), (1, 4). For 2x − y + 2 = 0, y = 2x + 2: points (0, 2), (−1, 0), (1, 4). The lines intersect at (1, 4), so the solution is x = 1, y = 4. Check: 2 + 4 = 6 and 2 − 4 + 2 = 0. The lines meet the x-axis at (3, 0) and (−1, 0). The vertices of the triangle are (1, 4), (3, 0) and (−1, 0). Base = 3 − (−1) = 4 units, height = 4 units, area = (1/2) × 4 × 4 = 8 square units.
- If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction? Also check the pair for consistency using ratios before solving. Let the fraction be x/y. (x + 1)/(y − 1) = 1 gives x + 1 = y − 1, so x − y = −2. x/(y + 1) = 1/2 gives 2x = y + 1, so 2x − y = 1. Ratios: 1/2 and −1/−1 = 1. Since 1/2 ≠ 1, the pair has a unique solution. Subtract the first equation from the second: x = 3. Then y = x + 2 = 5. The fraction is 3/5. Check: 4/4 = 1 and 3/6 = 1/2.
Common Mistakes and Examiner Tips
- Adding when you should subtract in elimination. When the equal coefficients have the same sign, subtract; when opposite, add. Look at the signs before you combine.
- Distributing a minus sign wrongly. In (9x − 2y) − (9x − 15y), the second bracket becomes −9x + 15y. Write the bracket out on its own line before simplifying.
- Dropping the sign of c in the ratio test. In x + 2y − 4 = 0, c1 = −4, not 4. Write both equations in the same form first and keep every sign.
- Comparing only two ratios for coincident lines. a1/a2 = b1/b2 only says the lines are parallel or coincident. You must check c1/c2 to decide which.
- Calling a coincident pair inconsistent. Coincident lines have infinitely many solutions, so the pair is dependent and consistent. Only parallel lines give an inconsistent pair.
- Shifting only one person’s age. “Five years ago” applies to both people: x − 5 and y − 5. The same goes for “ten years later”.
- Writing a two-digit number as x + y or xy. The number with tens digit x and units digit y is 10x + y. Its reverse is 10y + x.
- Missing the second case in digit problems. “The digits differ by 2” allows x − y = 2 and y − x = 2. NCERT Example 10 has two answers, 42 and 24.
- Counting all days in the library problem. The fixed charge covers the first three days, so 7 days means 4 extra days, not 7.
- Plotting only two points and misdrawing a line. Plot a third point as a check. Use whole-number points, label each line with its equation, and mark the point of intersection with its coordinates.
- Stopping at x and y. Answer the question asked, with units (“one pen costs ₹5”), and verify in the original words, not only in your equations. A mistake in forming an equation passes an equation-only check.
Quick Revision Points
- A linear equation in two variables has the form ax + by + c = 0, with a and b not both zero.
- A pair of linear equations: a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.
- A solution of the pair satisfies both equations at once.
- The graph of each equation is a straight line; the pair is two lines.
- Intersecting lines: unique solution, consistent pair.
- Parallel lines: no solution, inconsistent pair.
- Coincident lines: infinitely many solutions, dependent (consistent) pair.
- A dependent pair is always consistent.
- a1/a2 ≠ b1/b2: intersecting.
- a1/a2 = b1/b2 = c1/c2: coincident.
- a1/a2 = b1/b2 ≠ c1/c2: parallel.
- Graphical method: two or three points per line, read the point of intersection.
- The graph is inconvenient for non-integral solutions such as (4/13, 1/19).
- Substitution: express one variable in terms of the other, substitute, solve, back-substitute.
- Elimination: equalise one coefficient, add or subtract, solve, back-substitute.
- Same signs: subtract. Opposite signs: add.
- True statement with no variable: infinitely many solutions. False statement: no solution.
- Two-digit number: 10x + y; reversed: 10y + x.
- Ages: shift both people by the same number of years.
- For a parallel partner line, scale a and b but not c; for a coincident one, scale all three.
Weightage in Board Exams
Pair of Linear Equations in Two Variables belongs to the Algebra unit of Class 10 Maths, together with Polynomials, Quadratic Equations and Arithmetic Progressions. Check the current CBSE course structure for the marks given to the unit. Within it, this chapter is assessed in a predictable way.
| Question type | What is usually asked |
|---|---|
| MCQ and Assertion-Reason | Ratio test for intersecting, parallel or coincident lines; the value of k for a unique solution, no solution or infinitely many; whether a given pair of values is a solution |
| Short answers | Solving a pair by substitution or elimination; writing a second equation for a given type of line; a short word problem on angles, perimeter or prices |
| Long answer | A graphical solution with a triangle formed with an axis; word problems on ages, digits, fractions, fixed charges or notes |
| Case-based | A real situation such as fees, prices or a fair: form the pair, solve it and interpret the answer |
In every question, marks go to forming the equations correctly, showing each step of the method asked for, and stating the final answer in words. Practise every NCERT exercise question first, then the k-value and graph-with-triangle types above.
Class 10 Maths · Chapter 3 – swipe through all 10 cards to understand the whole chapter.
A pair of linear equations
Two linear equations in the same two variables, taken together. A solution satisfies both at once.
Each variable has power 1: no x2, no xy, no 1/x.
- Two unknowns need two separate conditions
- Check a solution by substituting into both equations
- Akhila: y = x/2 and 3x + 4y = 20 give 4 rides, 2 games
Three ways two lines can sit
Each equation is a straight line, so the pair is two lines on the same axes.
Consistent means at least one solution.
- Intersecting: one unique solution, consistent
- Parallel: no solution, inconsistent
- Coincident: infinitely many solutions, dependent and consistent
Decide without drawing
Compare a1/a2, b1/b2 and c1/c2 to know the type of lines.
Keep every sign and use the same form for both equations.
- a1/a2 = b1/b2 = c1/c2 → coincident, infinitely many
- a1/a2 = b1/b2 ≠ c1/c2 → parallel, no solution
- The converse is also true for any pair of lines
Solve by drawing both lines
Plot two or three points for each line and read the point where they meet.
Hard to use when the answer is a fraction like 4/13.
- x + 3y = 6: (0, 2), (6, 0); 2x − 3y = 12: (0, −4), (3, −2)
- The lines meet at (6, 0), so x = 6, y = 0
- Label each line and mark the intersection with coordinates
Put one variable in terms of the other
Express x or y from one equation and substitute it into the other.
Pick the variable whose coefficient is 1 or −1.
- Step 1: write one variable in terms of the other
- Step 2: substitute in the other equation and solve
- Step 3: back-substitute to get the second value
Equalise, then add or subtract
Multiply the equations so one variable has equal coefficients, then remove it.
Same signs: subtract. Opposite signs: add.
- 27x − 12y = 6000 and 28x − 12y = 8000 give x = 2000
- Then y = 4000, incomes ₹18,000 and ₹14,000
- Use the LCM of the coefficients as the target
When the variable vanishes
Sometimes substitution or elimination leaves no variable at all.
Always state the conclusion in words.
- True statement like 18 = 18: infinitely many solutions
- False statement like 0 = 9: no solution
- Rails x + 2y − 4 = 0 and 2x + 4y − 12 = 0 never cross
Two-digit numbers
Use place value to turn digits into a number.
x is the tens digit, y the units digit.
- Sum of number and reverse: 11(x + y)
- Digits differ by 2: try x − y = 2 and y − x = 2
- NCERT Example 10 gives two answers, 42 and 24
Ages, fractions and charges
Name the unknowns with units and turn each condition into one equation.
Verify with the original words of the question.
- Ages: shift both people by the same years
- Fraction x/y: cross-multiply after the change
- Taxi: x + 10y = 105, x + 15y = 155 give ₹5 fixed and ₹10 per km
Unknown coefficient questions
Write the ratio condition for the case asked, then solve for k.
Test each value of k in the third ratio.
- 2x + 3y = 7, 4x + ky = 14 dependent when k = 6
- kx + 2y = 5, 3x + y = 1 have no solution when k = 6
- x + 2y = 3, 5x + ky = 7 unique when k ≠ 10
📝 Practice Pair of Linear Equations in Two Variables - 10 board questions
CBSE previous-year and competency-based · with answers & explanations
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Why D: Coincident lines lie exactly on top of each other, so every point of one line is also a point of the other. Each common point is a solution, and a line has infinitely many points, so the pair has infinitely many solutions (a dependent, consistent pair).
Why not A: A unique solution comes from two lines that cross at exactly one point, that is intersecting lines.
Why not B: Two distinct straight lines can never share exactly two points; they share 0, 1 or infinitely many.
Why not C: No solution belongs to parallel lines, which never meet; coincident lines meet everywhere.
Remember: intersecting gives one solution, parallel gives none, coincident gives infinitely many.
Why B: Step 1 (inconsistent means parallel, so a₁/a₂ = b₁/b₂): a₁/a₂ = (1/2)/2 = 1/4 and b₁/b₂ = (1/3)/k = 1/(3k). Step 2 (equate the ratios): 1/(3k) = 1/4, so 3k = 4 and k = 4/3. Step 3 (check c₁/c₂ is different): c₁/c₂ = 5/7, which is not 1/4, so the lines are parallel and the system is inconsistent.
Why not A: k = 3/4 is the reciprocal of 4/3, from solving 3k = 4 as k = 3/4.
Why not C: k = 1/3 just copies the coefficient of y from the first equation, which makes b₁/b₂ = 1, not 1/4.
Why not D: k = 3 takes the denominator of y/3 as the coefficient of y, but that coefficient is 1/3.
Remember: parallel (no solution) needs a₁/a₂ = b₁/b₂ but c₁/c₂ different.
Why C: For A, a₁/a₂ = 3/(−6) = −1/2, b₁/b₂ = (−5)/10 = −1/2 and c₁/c₂ = 7/14 = 1/2. The first two ratios are equal but the third differs, so the lines are parallel and the system is inconsistent: A is true. R is false, because a pair without a unique solution may be parallel (no solution) or coincident (infinitely many solutions).
Why not A: R is false, so it cannot be true, let alone the correct explanation.
Why not B: This needs R to be true, but R ignores the coincident-lines case.
Why not D: A is true; the sign of c₁/c₂ (+1/2) differs from −1/2, so the lines do not coincide.
Remember: no unique solution means parallel OR coincident, never only one of them.
Why B: Step 1 (substitute x = 1, y = 2 in the first equation): 2 − 6 + a = 0, so a = 4. Step 2 (substitute in the second equation): 2 + 6 − b = 0, so b = 8. Step 3 (compare): b = 8 = 2 × 4 = 2a, so 2a = b.
Why not A: a = 2b would need a = 16, but a = 4; the roles of a and b are swapped.
Why not C: a + 2b = 4 + 16 = 20, not 0; this comes from a sign slip giving a = −4.
Why not D: 2a + b = 16, not 0; again a sign slip on a or b.
Remember: to use a known solution, put the values into EACH equation and solve for the unknown constants.
Why B: For A with p = 1, a₁/a₂ = 1/2, b₁/b₂ = 3/6 = 1/2 and c₁/c₂ = 59/118 = 1/2. All three ratios are equal, so the lines coincide: A is true. For R, a unique solution needs p/2 ≠ 3/6, that is p ≠ 1, so R is true. But R talks about a different pair and about a unique solution, so it does not explain why A’s pair has infinitely many solutions.
Why not A: R is true, but it is about a different pair and a different case, so it is not the reason behind A.
Why not C: R is true; a unique solution does require p/2 ≠ 1/2, which is p ≠ 1.
Why not D: A is true, since 1/2 = 3/6 = 59/118.
Remember: all three ratios equal means coincident; only a₁/a₂ ≠ b₁/b₂ is needed for a unique solution.
Why B: Step 1 (compare x and y coefficients): a₁/a₂ = 6/36 = 1/6 and b₁/b₂ = 1/6, so the lines are at least parallel. Step 2 (constant ratio must also be 1/6 for coincident lines): c₁/c₂ = 3k/3 = k, so k = 1/6.
Why not A: 6 is the reciprocal of the needed ratio, from writing 36/6 instead of 6/36.
Why not C: 1/2 comes from writing the constant ratio as k/3 instead of 3k/3, then solving k/3 = 1/6.
Why not D: 1/3 comes from setting 3k = 1, equating the constants directly instead of making their ratio 1/6.
Remember: infinitely many solutions means all three ratios are equal.
Why C: Step 1 (add the two equations): (a + b)x + (a + b)y = a² − b² + 0, so (a + b)(x + y) = a² − b². Step 2 (factorise and divide by a + b): a² − b² = (a − b)(a + b), so x + y = a − b.
Why not A: a² − b² is the right-hand side before dividing by (a + b).
Why not B: a + b is the factor we divide by, not the quotient.
Why not D: a² + b² comes from a sign slip in the right-hand side.
Remember: when coefficients swap between equations, adding (or subtracting) them gives x + y (or x − y) in one step.
Why D: Step 1 (coefficient ratios): with 5x − 3y − 2 = 0 and −15x + 9y − 5 = 0, a₁/a₂ = 5/(−15) = −1/3 and b₁/b₂ = (−3)/9 = −1/3, equal. Step 2 (constant ratio): c₁/c₂ = (−2)/(−5) = 2/5, not −1/3, so the lines are parallel and distinct.
Why not A: a₁/a₂ = −1/3 but b₁/b₂ = (−3)/(−9) = 1/3, so the lines intersect.
Why not B: a₁/a₂ = 1/3 but b₁/b₂ = −1/3, so the lines intersect.
Why not C: a₁/a₂ = 5/9 and b₁/b₂ = 3/15 = 1/5 are unequal; the x and y coefficients were swapped.
Remember: a parallel line keeps the SAME x : y coefficient ratio, signs included, with a different constant.
Why B: Step 1 (known ratios): a₁/a₂ = 3/6 = 1/2 and c₁/c₂ = 8/16 = 1/2. Step 2 (b-ratio must match): b₁/b₂ = (−1)/(−k) = 1/k, and 1/k = 1/2 gives k = 2.
Why not A: −2 comes from losing one of the two minus signs in (−1)/(−k).
Why not C: 1/2 is the common value of the ratios, not the value of k.
Why not D: −1/2 combines both slips: the ratio value taken as k and a lost sign.
Remember: infinitely many solutions means all three ratios are equal; keep the signs of the coefficients.
Why D: x = a is a vertical line (parallel to the y-axis) and y = b is a horizontal line (parallel to the x-axis). A vertical and a horizontal line always cross at one point, where x = a and y = b, which is (a, b).
Why not A: One line is vertical and the other horizontal, so they are perpendicular, not parallel.
Why not B: In an ordered pair the x-value comes first, so the point is (a, b), not (b, a).
Why not C: Coincident lines are the same line; a vertical line cannot coincide with a horizontal one.
Remember: x = a is vertical, y = b is horizontal, and they meet at (a, b).
Chapter Navigation
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Related Chapters in Class 10 Maths
- Real Numbers Class 10 Notes
- Polynomials Class 10 Notes
- Quadratic Equations Class 10 Notes
- Arithmetic Progressions Class 10 Notes
- Triangles Class 10 Notes
- Coordinate Geometry Class 10 Notes
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Frequently Asked Questions
It is two linear equations in the same two variables, such as a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, considered together. A solution of the pair is a pair of values of x and y that satisfies both equations at the same time. On a graph it is a point that lies on both lines.
A consistent pair has at least one solution: its lines intersect or coincide. An inconsistent pair has no solution: its lines are parallel. A dependent pair has equivalent equations, so its lines coincide and it has infinitely many solutions. A dependent pair is always consistent, so consistent covers both the intersecting and the coincident case.
If a₁/a₂ ≠ b₁/b₂, the lines intersect and there is exactly one solution. If all three ratios are equal, the lines coincide and there are infinitely many solutions. If a₁/a₂ = b₁/b₂ but c₁/c₂ is different, the lines are parallel and there is no solution. Keep the sign of every coefficient and write both equations in the same form.
If the question names a method, use that one. Otherwise, substitution is quickest when one variable has coefficient 1 or −1, as in x + 2y = 3, since x = 3 − 2y needs no fractions. Elimination is better when all coefficients are larger numbers, such as 9x − 4y = 2000 and 7x − 3y = 2000, where multiplying by 3 and 4 equalises the y terms.
Both mean the variables have cancelled out. A false statement such as 0 = 9 or −4 = 0 means the pair has no solution, so the lines are parallel and the pair is inconsistent. A true statement such as 18 = 18 means the two equations are the same line, so the pair has infinitely many solutions. Write the conclusion in words.
The graph works well when the solution has whole-number coordinates, such as (6, 0) or (2, 2). When the solution is non-integral, like x = 49/29 and y = 19/29, the point cannot be read accurately from graph paper. The substitution and elimination methods give exact answers in every case, so they are the standard tools.
Let the tens digit be x and the units digit be y. The number is 10x + y and the number with the digits reversed is 10y + x. Turn each condition into an equation. If the digits differ by a number, consider both x − y and y − x, because there may be two answers, as with 42 and 24 in NCERT Example 10.
Write the condition for parallel lines: a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Solve a₁/a₂ = b₁/b₂ for k, then test each value in c₁/c₂. Reject any value that makes all three ratios equal, because that gives coincident lines. For kx + 3y = k − 3 and 12x + ky = k, k² = 36, and only k = −6 gives no solution.
