Linear Equations in Two Variables
Punjab Board · Class 9 · Mathematics
Flashcards for Linear Equations in Two Variables — Punjab Board Class 9 Mathematics. Quick Q&A cards covering key concepts, definitions, and formulas.
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Convert 2x + 3y = 4.37 to the form ax + by + c = 0 and identify a, b, and c.
Answer
Step 1: Subtract 4.37 from both sides → 2x + 3y - 4.37 = 0 Step 2: Identify coefficients → a = 2, b = 3, c = -4.37 Answer: The equation in standard form is 2x + 3y - 4.37 = 0 with a = 2, b = 3, c = -4…
Express the equation x - 4 = √3y in the form ax + by + c = 0.
Answer
Step 1: Move all terms to left side → x - √3y - 4 = 0 Step 2: Rearrange to match ax + by + c = 0 → 1·x + (-√3)·y + (-4) = 0 Step 3: Identify: a = 1, b = -√3, c = -4 Answer: x - √3y - 4 = 0…
Check whether (3, 2) is a solution of 2x + 3y = 12.
Answer
Step 1: Substitute x = 3 and y = 2 into 2x + 3y = 12 Step 2: Calculate left side → 2(3) + 3(2) = 6 + 6 = 12 Step 3: Compare → 12 = 12 ✓ Answer: Yes, (3, 2) is a solution because both sides are equal.
Is (1, 4) a solution of 2x + 3y = 12? Show your work.
Answer
Step 1: Substitute x = 1 and y = 4 into 2x + 3y = 12 Step 2: Calculate left side → 2(1) + 3(4) = 2 + 12 = 14 Step 3: Compare → 14 ≠ 12 ✗ Answer: No, (1, 4) is NOT a solution because 14 ≠ 12.
Find one solution of the equation x + 2y = 6 by choosing x = 0.
Answer
Step 1: Substitute x = 0 into x + 2y = 6 Step 2: Equation becomes → 0 + 2y = 6 Step 3: Solve for y → 2y = 6 → y = 3 Answer: (0, 3) is a solution of x + 2y = 6…
Find one solution of the equation x + 2y = 6 by choosing y = 0.
Answer
Step 1: Substitute y = 0 into x + 2y = 6 Step 2: Equation becomes → x + 2(0) = 6 Step 3: Solve for x → x + 0 = 6 → x = 6 Answer: (6, 0) is a solution of x + 2y = 6…
Find two solutions of 4x + 3y = 12.
Answer
Solution 1: Let x = 0 → 4(0) + 3y = 12 → 3y = 12 → y = 4. So (0, 4) is a solution. Solution 2: Let y = 0 → 4x + 3(0) = 12 → 4x = 12 → x = 3. So (3, 0) is a solution. Answer: Two solutions are (0, 4) a…
Find two solutions of 2x + 5y = 0.
Answer
Solution 1: Let x = 0 → 2(0) + 5y = 0 → 5y = 0 → y = 0. So (0, 0) is a solution. Solution 2: Let x = 1 → 2(1) + 5y = 0 → 2 + 5y = 0 → 5y = -2 → y = -2/5. So (1, -2/5) is a solution. Answer: Two soluti…
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