Pair of Linear Equations in Two Variables — Andhra Pradesh (SSC) Class 10 Mathematics Solutions (Free)
Free step-by-step Andhra Pradesh (SSC) Class 10 Mathematics solutions for "Pair of Linear Equations in Two Variables" — important questions with detailed answers, download PDF for board exam preparation.
TL;DR: Free step-by-step Andhra Pradesh (SSC) Class 10 Mathematics solutions for "Pair of Linear Equations in Two Variables" — important questions with detai…
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Q1: Define a linear equation in two variables and state the general form.
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 (or…
Q2: A pair of linear equations is consistent if which condition is satisfied?
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…
Q3: Solve the system of equations 2x + y = 7 and x - y = 2 using the elimination method.
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
Q4: Solve the system 3x + 4y = 10 and 2x - 2y = 2 using substitution method.
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
Q5: Solve using the cross-multiplication method: x + 2y = 5 and 2x + 3y = 8
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)
Ther…
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