Linear Equations in Two Variables — Karnataka (SSLC) Class 9 Mathematics Solutions (Free)
Free step-by-step Karnataka (SSLC) Class 9 Mathematics solutions for "Linear Equations in Two Variables" — important questions with detailed answers, download PDF for board exam preparation.
TL;DR: Free step-by-step Karnataka (SSLC) Class 9 Mathematics solutions for "Linear Equations in Two Variables" — important questions with detailed answers,…
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Q1: Find four solutions of the linear equation 2x + 3y = 12.
Equation: 2x + 3y = 12
Method: Choose values for x and solve for y (or vice versa).
Solution 1: Let x = 0
2(0) + 3y = 12
3y = 12
y = 4
Solution: (0, 4)
Solution 2: Let x = 3
2(3) + 3y = 12
6 + 3y = 12
3y = 6
y = 2
Solution: (3, 2)
Solution 3: Let x = 6
2(6) + 3y = 12
12 + 3y = 12
3y = 0
y = 0
Solution: (6, 0)
Solution 4: Let x = -3
2(-3) + 3y = 12
-6 + 3y = 12
3y = 18
y = 6
Solution: (-3, 6)
Answer: Four solutions are (0, 4), (3, 2), (6, 0), and (-3, 6)
Q2: Solve the system of equations: 2x + y = 7 and x - y = 2
Equations:
2x + y = 7 ... (1)
x - y = 2 ... (2)
Method: Elimination
Step 1: Add equations (1) and (2) to eliminate y
(2x + y) + (x - y) = 7 + 2
3x = 9
Step 2: Solve for x
x = 9/3 = 3
Step 3: Substitute x = 3 in equation (2)
3 - y = 2
y = 3 - 2 = 1
Step 4: Verify solution in both equations
Equation (1): 2(3) + 1 = 6 + 1 = 7 ✓
Equation (2): 3 - 1 = 2 ✓
Answer: x = 3, y = 1
Q3: Express 2x + 3y - 6 = 0 in the form y = mx + c and identify the slope and y-intercept.
Equation: 2x + 3y - 6 = 0
Step 1: Solve for y
3y = -2x + 6
y = -2x/3 + 6/3
y = (-2/3)x + 2
Step 2: Compare with y = mx + c
This is in the form y = mx + c where:
m = -2/3 (slope)
c = 2 (y-intercept)
Step 3: Interpret the slope
Slope m = -2/3 < 0, so the line is falling (decreasing from left to right).
For every 3 units moved right, the line drops 2 units.
Step 4: Interpret the y-intercept
The line crosses the y-axis at (0, 2).
Answer:
y = (-2/3)x + 2
Slope (m) = -2/3
y-intercept (c) = 2
Q4: Determine if (2, 1) is a solution of the system: x + 2y = 4 and 2x - y = 3
Point: (2, 1) means x = 2, y = 1
Equations:
x + 2y = 4 ... (1)
2x - y = 3 ... (2)
Step 1: Check equation (1)
Substitute x = 2, y = 1:
2 + 2(1) = 2 + 2 = 4 ✓
The point satisfies equation (1).
Step 2: Check equation (2)
Substitute x = 2, y = 1:
2(2) - 1 = 4 - 1 = 3 ✓
The point satisfies equation (2).
Step 3: Conclusion
Since (2, 1) satisfies both equations, it is the solution of the system.
Answer: Yes, (2, 1) is a solution of the given system.
Q5: Solve graphically: x + y = 5 and x - y = 1
Equations:
x + y = 5 ... (1)
x - y = 1 ... (2)
Step 1: Find solutions for equation (1): x + y = 5
Let x = 0: y = 5 → (0, 5)
Let x = 5: y = 0 → (5, 0)
Let x = 2: y = 3 → (2, 3)
Step 2: Find solutions for equation (2): x - y = 1
Let x = 0: y = -1 → (0, -1)
Let x = 1: y = 0 → (1, 0)
Let x = 3: y = 2 → (3, 2)
Step 3: Plot both lines
Line 1 (x + y = 5): passes through (0, 5) and (5, 0)
Line 2 (x - y = 1): passes through (0, -1) and (1, 0)
Step 4: Find intersection point
Both lines intersect at a …
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