Pair of Linear Equations in Two Variables
CBSE Class 10 · Mathematics · Notes, formulas and practice questions
Two linear equations in the same two variables define a pair; this chapter explains how to tell whether they have a common solution and how to find it by the graphical method and by algebraic substitution and elimination, including word problems.
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This chapter deals with two linear equations in the same two variables. Together such equations form a pair; an ordered pair (x, y) that satisfies both equations is called a solution of the pair. Graphically, each equation is a straight line, so solving the pair means finding the point or points common to the two lines. If the lines intersect once, there is a unique solution; if they are parallel and distinct, there is none; if they coincide, there are infinitely many. This graphical view lies behind everything in the chapter.
Before drawing graphs, you can predict which case you are in by comparing the coefficients. When a₁/a₂ ≠ b₁/b₂, the lines meet at one point. When a₁/a₂ = b₁/b₂ but the constants do not follow the same ratio, the lines are parallel and the pair is inconsistent. When a₁/a₂ = b₁/b₂ = c₁/c₂, the two equations represent the same line and the pair is dependent. Students often forget that all three ratios must be compared after writing each equation in the form ax + by + c = 0, especially the sign of c.
The algebraic methods also find the solution exactly. In substitution, one equation is solved for one variable and the result is put into the other, leaving a linear equation in one variable. In elimination, the equations are multiplied by suitable numbers so that the coefficients of one variable become equal in magnitude; adding or subtracting then cancels that variable. Both methods are useful: substitution often works well when a variable already has coefficient 1, and elimination works well when the coefficients can be matched easily.
Most board problems are word problems. The main skill is translating the situation into two correct equations: name the unknowns, express each sentence as an equation, then solve by substitution or elimination. Clearly state the units and the final answer. A quick check in the original statement is essential, because an answer can satisfy the equations but violate the wording if a sign or a ratio was set wrong.
Key terms
- Pair of linear equations in two variables
- Two simultaneous equations, usually written as a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, in which x and y appear only to the first power. In each equation the coefficients a and b are real numbers and are not both zero.
- Solution of the pair
- An ordered pair (x, y) that makes both equations true at the same time. Graphically, it is a point common to the two lines; for coincident lines every point is a solution.
- Consistent pair
- A pair that has at least one solution. The equations are consistent whether the lines intersect at one point (unique solution) or coincide (infinitely many solutions).
- Inconsistent pair
- A pair that has no common solution. This happens when the two lines are parallel and distinct.
- Dependent pair
- A pair in which the two equations represent the same line, so every solution of one equation is also a solution of the other. It has infinitely many solutions and satisfies a₁/a₂ = b₁/b₂ = c₁/c₂.
- Intersecting lines
- Two lines that cross at exactly one point. For such a pair, a₁/a₂ ≠ b₁/b₂, and that one point gives the unique solution.
- Substitution method
- From one equation, express one variable in terms of the other; replace that expression in the second equation. This gives a linear equation in one variable, which is solved and the value substituted back.
- Elimination method
- Multiply the equations by suitable constants so that the coefficients of x or y are equal in magnitude. Add or subtract the equations to cancel that variable, solve for the other, and then find the first variable.
Formula sheet
| What | Formula | Notes |
|---|---|---|
| Unique solution | a₁/a₂ ≠ b₁/b₂ | For the equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, when this holds the corresponding lines intersect at one point, so the pair is consistent and has one solution. |
| No solution (inconsistent) | a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | For the same pair written in that form, the lines are parallel and distinct. There is no ordered pair that satisfies both equations. |
| Infinitely many solutions (dependent) | a₁/a₂ = b₁/b₂ = c₁/c₂ | For the same pair, the lines coincide, so every point on the common line satisfies both equations. |
Practice questions with answers
1. Solve by substitution: x + 2y = 7 and 2x − y = −1.
From the second equation, y = 2x + 1. Substituting into x + 2y = 7 gives x + 2(2x + 1) = 7, so 5x + 2 = 7 and x = 1. Then y = 2(1) + 1 = 3. Check: 1 + 6 = 7 and 2 − 3 = −1, so the solution is (1, 3).
2. Solve by elimination: 3x + 4y = 10 and 2x − 2y = 2.
Multiply the second equation by 2 to get 4x − 4y = 4. Adding it to the first equation cancels y: 3x + 4y + 4x − 4y = 10 + 4, so 7x = 14 and x = 2. Substituting x = 2 in 3x + 4y = 10 gives 6 + 4y = 10, so y = 1. The solution is (2, 1).
3. Without solving, determine whether 2x + 3y − 8 = 0 and 4x + 6y − 7 = 0 are consistent or inconsistent.
Here a₁/a₂ = 2/4 = 1/2 and b₁/b₂ = 3/6 = 1/2, while c₁/c₂ = (−8)/(−7) = 8/7. Since a₁/a₂ = b₁/b₂ but c₁/c₂ is different, the lines are parallel and distinct. The pair has no common solution, so it is inconsistent.
4. For what value of k will the equations x + 2y = 3 and 2x + ky = 8 have no solution?
Write the equations as x + 2y − 3 = 0 and 2x + ky − 8 = 0. For no solution, a₁/a₂ = b₁/b₂, so 1/2 = 2/k, which gives k = 4. Also c₁/c₂ = (−3)/(−8) = 3/8, which is not 1/2, so the lines are parallel and distinct for k = 4.
5. The sum of two numbers is 26 and their difference is 4. Find the numbers.
Let the larger number be x and the smaller be y. Then x + y = 26 and x − y = 4. Adding the equations gives 2x = 30, so x = 15. Substituting into x + y = 26 gives y = 11. Check 15 − 11 = 4, so the numbers are 15 and 11.
6. A father is three times as old as his son. In 10 years, he will be twice as old as his son. Find their present ages.
Let the son's present age be x years and the father's be y years. The conditions give y = 3x and y + 10 = 2(x + 10). Substituting 3x for y gives 3x + 10 = 2x + 20, so x = 10. Then y = 30. In 10 years the son will be 20 and the father 40, which is twice the son's age, so the present ages are 30 and 10.
7. Two kg of apples and one kg of grapes cost ₹160. Four kg of apples and three kg of grapes cost ₹360. Find the cost per kg of apples and grapes.
Let x be the price per kg of apples and y be the price per kg of grapes. Then 2x + y = 160 and 4x + 3y = 360. Multiplying the first equation by 2 gives 4x + 2y = 320; subtracting it from the second gives y = 40. Substituting y = 40 in 2x + y = 160 gives 2x = 120, so x = 60. Thus apples cost ₹60 per kg and grapes cost ₹40 per kg.
8. The denominator of a fraction is 4 more than its numerator. If 1 is added to both the numerator and the denominator, the fraction becomes 2/3. Find the original fraction.
Let the numerator be x and the denominator be y. Then y = x + 4 and (x + 1)/(y + 1) = 2/3. Cross-multiplying gives 3(x + 1) = 2(y + 1). Substitute y = x + 4: 3x + 3 = 2x + 10, so x = 7 and y = 11. The original fraction is 7/11, since adding 1 gives 8/12 = 2/3.
9. State the condition under which a pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 is dependent.
A pair is dependent when the two equations represent the same line. This happens when a₁/a₂ = b₁/b₂ = c₁/c₂. Then every point on the common line satisfies both equations, so there are infinitely many solutions.
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