Pair of Linear Equations in Two Variables — Karnataka (SSLC) Class 10 Mathematics Solutions (Free)

Linear Equation in Two Variables: A linear equation in two variables is an equation of the form ax + by + c = 0, where a, b, and c are real numbers and at least one of a or b is non-zero. General form: ax + by + c = 0 or ax + by = c where a = coefficient of x b = coefficient of y c = constant term Examples: - 2x + 3y - 6 = 0 - x - 2y = 5 - 3x + 4y = 12 Properties: - The graph of a linear equation in two variables is a straight line - The equation has infinitely many solutions (ordered pairs) - Each solution corresponds to a point on the line

Consistency of a Pair of Linear Equations: Consider the pair: a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 The pair of equations is CONSISTENT if it has at least one solution. Two cases of consistency: 1) Consistent and Independent: Condition: a₁/a₂ ≠ b₁/b₂ Number of solutions: Exactly one (unique) Geometrically: The lines intersect at one point 2) Consistent and Dependent: Condition: a₁/a₂ = b₁/b₂ = c₁/c₂ Number of solutions: Infinitely many Geometrically: The lines coincide (are the same line) Inconsistent: Condition: a₁/a₂ = b₁/b₂ ≠ c₁/c₂ Number of solutions: None Geometrically: The lines are parallel and never intersect

Given equations: 2x + y = 7 ... (1) x - y = 2 ... (2) Step 1: Add equations (1) and (2) to eliminate y (2x + y) + (x - y) = 7 + 2 3x = 9 x = 3 Step 2: Substitute x = 3 into equation (2) 3 - y = 2 y = 1 Step 3: Verification Substitute x = 3, y = 1 into equation (1): 2(3) + 1 = 6 + 1 = 7 ✓ Substitute x = 3, y = 1 into equation (2): 3 - 1 = 2 ✓ Therefore, the solution is x = 3, y = 1

Given equations: 3x + 4y = 10 ... (1) 2x - 2y = 2 ... (2) Step 1: Simplify equation (2) by dividing by 2 x - y = 1 x = y + 1 ... (3) Step 2: Substitute x = y + 1 into equation (1) 3(y + 1) + 4y = 10 3y + 3 + 4y = 10 7y + 3 = 10 7y = 7 y = 1 Step 3: Find x using equation (3) x = 1 + 1 = 2 Step 4: Verification Substitute x = 2, y = 1 into equation (1): 3(2) + 4(1) = 6 + 4 = 10 ✓ Substitute x = 2, y = 1 into equation (2): 2(2) - 2(1) = 4 - 2 = 2 ✓ Therefore, the solution is x = 2, y = 1

Given equations: x + 2y = 5 ... (1) 2x + 3y = 8 ... (2) Rearranging in the form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0: x + 2y - 5 = 0 2x + 3y - 8 = 0 Here: a₁ = 1, b₁ = 2, c₁ = -5 a₂ = 2, b₂ = 3, c₂ = -8 Using cross-multiplication method: x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁) Calculate each denominator: b₁c₂ - b₂c₁ = 2(-8) - 3(-5) = -16 + 15 = -1 c₁a₂ - c₂a₁ = (-5)(2) - (-8)(1) = -10 + 8 = -2 a₁b₂ - a₂b₁ = 1(3) - 2(2) = 3 - 4 = -1 So: x/(-1) = y/(-2) = 1/(-1) Therefore: x = -1/(-1) = 1 y = -2/(-1) = 2 Verification: 1 + 2(2) = 1 + 4 = 5 ✓ 2(1) + 3(2) = 2 + 6 = 8 ✓ Therefore, the solution is x = 1, y = 2

For a system of linear equations to have infinitely many solutions, the equations must be consistent and dependent. This means: a₁/a₂ = b₁/b₂ = c₁/c₂ For the given equations: 2x - 3y = 7 → a₁ = 2, b₁ = -3, c₁ = 7 4x - ky = 14 → a₂ = 4, b₂ = -k, c₂ = 14 Applying the condition: a₁/a₂ = b₁/b₂ = c₁/c₂ First, check c₁/c₂: 7/14 = 1/2 So we need a₁/a₂ = 1/2: 2/4 = 1/2 ✓ (This is satisfied) Now, we need b₁/b₂ = 1/2: (-3)/(-k) = 1/2 3/k = 1/2 k = 6 Verification: With k = 6: The second equation becomes 4x - 6y = 14, which simplifies to 2x - 3y = 7 (dividing by 2). The equations are identical, so there are infinitely many solutions. Therefore, k = 6

For the system: 3x + 2y = 5 ... (1) 6x + 4y = 8 ... (2) Identify coefficients: a₁ = 3, b₁ = 2, c₁ = 5 a₂ = 6, b₂ = 4, c₂ = 8 Check the ratios: a₁/a₂ = 3/6 = 1/2 b₁/b₂ = 2/4 = 1/2 c₁/c₂ = 5/8 Observation: a₁/a₂ = b₁/b₂ = 1/2, but c₁/c₂ = 5/8 ≠ 1/2 Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the system is INCONSISTENT. Geometrically: The lines represented by these equations are parallel and do not intersect. To verify, simplify equation (2): 6x + 4y = 8 Dividing by 2: 3x + 2y = 4 Now we have: 3x + 2y = 5 3x + 2y = 4 These are two different parallel lines with the same slope but different y-intercepts. They never intersect, so the system has NO SOLUTION. Therefore, the system is inconsistent

Let the two numbers be x and y. Step 1: Form equations from the given conditions Sum = 15: x + y = 15 ... (1) Difference = 3: x - y = 3 ... (2) Step 2: Solve using elimination method Add equations (1) and (2): (x + y) + (x - y) = 15 + 3 2x = 18 x = 9 Step 3: Find y Substitute x = 9 into equation (1): 9 + y = 15 y = 6 Step 4: Verification Sum: 9 + 6 = 15 ✓ Difference: 9 - 6 = 3 ✓ Therefore, the two numbers are 9 and 6

Let the original speed of the train be x km/h. Step 1: Form equations using distance = speed × time For original journey: Time = 300/x hours For increased speed journey: Time = 300/(x + 5) hours Given that the faster journey takes 2 hours less: 300/x - 300/(x + 5) = 2 Step 2: Solve the equation 300(x + 5) - 300x = 2x(x + 5) 300x + 1500 - 300x = 2x² + 10x 1500 = 2x² + 10x 2x² + 10x - 1500 = 0 Divide by 2: x² + 5x - 750 = 0 Step 3: Factorise Find two numbers that multiply to -750 and add to 5. Factors: 30 and -25 (since 30 × (-25) = -750 and 30 + (-25) = 5) x² + 5x - 750 = (x + 30)(x - 25) = 0 Step 4: Find the roots Either x + 30 = 0 or x - 25 = 0 Therefore: x = -30 or x = 25 Since speed cannot be negative, x = 25 km/h. Step 5: Verification Original time: 300/25 = 12 hours Time at increased speed (30 km/h): 300/30 = 10 hours Difference: 12 - 10 = 2 hours ✓ Therefore, the original speed of the train is 25 km/h