Chapter 3: Pair of Linear Equations in Two Variables

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, where a₁, b₁, c₁, a₂, b₂, c₂ are real numbers.

A pair that has a solution is consistent. A pair with no solution is inconsistent. A pair with infinitely many solutions is dependent (and consistent).

(i) The lines intersect at a point: unique solution (consistent). (ii) The lines are parallel: no solution (inconsistent). (iii) The lines coincide: infinitely many solutions (dependent).

Intersecting: a₁/a₂ ≠ b₁/b₂. Coincident: a₁/a₂ = b₁/b₂ = c₁/c₂. Parallel: a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

a₁/a₂ = 5/7 and b₁/b₂ = −4/6. They are not equal, so the lines intersect at a point.

9/18 = 3/6 = 12/24 = 1/2. All three ratios are equal, so the lines are coincident.

6/2 = 3 and −3/−1 = 3 are equal, but 10/9 is different, so the lines are parallel.

Write them as 2x − 3y − 8 = 0 and 4x − 6y − 9 = 0. 2/4 = −3/−6 = 1/2, but the ratio of c-values 8/9 is different, so the lines are parallel and the pair is inconsistent.

5/−10 = −3/6 = 11/−22 = −1/2. All the ratios are equal, so the lines coincide and the pair is dependent (consistent).

Let the width be x and the length x + 4. Then x + (x + 4) = 36, so x = 16. Width is 16 m and length is 20 m.

Let a pencil cost x and a pen y: 5x + 7y = 50 and 7x + 5y = 46. Solving gives x = 3 and y = 5. A pencil costs ₹3 and a pen costs ₹5.

Let girls be x and boys y: x + y = 10 and x − y = 4. Solving gives x = 7, y = 3. So there are 7 girls and 3 boys.

Steps: (1) write one variable in terms of the other from one equation, (2) put it in the other equation and solve the one-variable equation, (3) put the value back to get the other variable. Here x = 3 − 2y, then 7(3 − 2y) − 15y = 2 gives y = 19/29 and x = 49/29.

Putting x = 14 − y gives 14 − y − y = 4, so y = 5 and x = 9.

Let the larger be x and the smaller y: x − y = 26 and x = 3y. Then 3y − y = 26, so y = 13 and x = 39.

Let a bat cost x and a ball y: 7x + 6y = 3800 and 3x + 5y = 1750. Solving gives x = 500, y = 50. A bat costs ₹500 and a ball ₹50.

Let the fixed charge be x and the rate per km y: x + 10y = 105 and x + 15y = 155. Subtracting gives 5y = 50, so y = 10 and x = 5. Fare for 25 km = 5 + 25 × 10 = ₹255.

Steps: (1) multiply the equations so that the coefficients of one variable become equal, (2) add or subtract to remove that variable, (3) find the remaining variable and then the other. Here multiply the first equation by 3 to get 3x + 3y = 15. Adding gives 5x = 19, so x = 19/5 and y = 6/5.

Let Nuri be x and Sonu y: x − 5 = 3(y − 5) and x + 10 = 2(y + 10). Simplifying: x − 3y = −10 and x − 2y = 10. Solving gives y = 20, x = 50. Nuri is 50 years old and Sonu is 20.

Let there be x notes of ₹50 and y notes of ₹100: x + y = 25 and 50x + 100y = 2000. Solving gives x = 10, y = 15. She got 10 notes of ₹50 and 15 notes of ₹100.