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✧ CLASS -9 -MATHS- CHAPTER 4- LINEAR EQUATIONS IN TWO VARIABLES
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Answer:
3 . check which of the following are solutions of the equation 2x + 3y = k
Step-by-step explanation:
i) ( 0,-2)
Substituting x=0 and y = -2 in x-2y = 4 , we get
LHS = x-2y = 0 - 2x -2 = 0 + 4 = 4 = RHS
so, x = 0, y = -2 or ( 0, -2 ) is solution of x-2y = 4.
ii) (2,0)
Substituting x = 2 and y = 0 in x - 2y = 4, we get
LHS = 2 x 1 - 2 x 0 = 2 ≠ RHS
so, (2,0) is not a solution of the equation x - 2y=4.
iii) (4,0)
substituting x = 4 and y = 0 in x - 2y = 4, we get
LHS = 4 x 1 - 2 x 0 = 4 = RHS
so, (4,0) is a solution of the given equation.
iv) ( √2, 4√2 )
substituting x = √2 and y = 4√2 in x - 2y = 4, we get
LHS = 4 x √2 - 2 x 4√2 = -4√2 ≠ RHS
so, ( √2, 4√2 = -4√2 ≠ RHS.
so, ( √2, 4√2 ) is not a solution of the given equation.
v) ( 1, 1 )
substituting x = 1 and y = 1 in x - 2y = 4 , we get
LHS = 1 - 2 x 1 = -1 ≠ RHS
so, ( 1, 1 ) is not the solution of the given equation.
2) Write four solutions for each of the following equation.
ii) πx + y = 9
ans. we have , πx + y = 9
substituting x = 0 in this equation, we get
π x 0 + y = 9 → 0 + y = 9 → y = 9
so, ( 0, 9 ) is solution of the given equation, we get
substituting x = 1 , in the given equation , we get
π x 1 + y = 9 → π + y = 9 → y = 9 - π
so, ( 1, 9 -π ) is solution of the given equation.
similarly, by substituting x = - 1 and x = 2 respectively, we obtain ( -1, 9 + π ) and ( 2, 9 - 2π ) as solutions of the given equation.
hope so this helps you.
I only know this much.
Question 1st Which one of the following options is true, and why? y = 3x + 5 has
- A unique solution.
- Only two solutions.
- Infinitely many solutions.
Answer 1st y = 3x + 5 has infinitely many solutions. Infinitely many solutions that is option three is true. Because every value for x have a corresponding value for y and vice-versa. For example,
~ Put x as 0 in equation
→ y = 3x + 5
→ y = 3(0) + 5
→ y = 0 + 5
→ y = 5
→ Therefore, (0,5) is one solution
~ Now put y as 0 in equation
→ y = 3x + 5
→ 0 = 3x + 5
→ 0 - 5 = 3x
→ -5 = 3x
→ -5/3 = x
→ Therefore, (-5/3,0) is 2nd solution
~ Similarly if we are going to put x as 1 then we get new solution etc. Therefore, the option third is absolutely correct!
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Question 2nd Write four solutions for each of the following equations:
a) 2x + y = 7
Firstly put x as 0
→ 2(0) + y = 7
→ 0 + y = 7
→ y = 7
- → Therefore, (0,7) is 1st solution
Now put y as 0
→ 2x + 0 = 7
→ 2x = 7
→ x = 7/2
→ x = 3.5
- → Therefore, (3.5,0) is 2nd solution
Now put x as 1
→ 2(1) + y = 7
→ 2 + y = 7
→ y = 7-2
→ y = 5
- Therefore, (0,5) is 3rd solution
Now put y as 1
→ 2x + 1 = 7
→ 2x = 7-1
→ 2x = 6
→ x = 6/2
→ x = 3
- Therefore, (3,0) is 4th solution
b) πx + y = 9
Firstly put x as 0
→ π(0) + y = 9
→ 0 + y = 9
→ y = 9
- Therefore, (0,9) is 1st solution
Now put y as 0
→ πx + 0 = 9
→ πx = 9
→ x = 9/π
- Therefore, (9/π,0) is 2nd solution
Now put x as 1
→ π(1) + y = 9
→ π1 + y = 9
→ y = 9-π1
→ y = 9-π
- Therefore, (1,9-π) is 3rd solution
Now put y as 1
→ πx + 1 = 9
→ πx = 9 - 1
→ πx = 8
→ x = 8/π
- Therefore, (8/π,0) is 4th solution
c) x = 4y
Firstly put x as 0
→ 0 = 4y
→ 0/4 = y
- Therefore, (0,0/4) is 1st solution
Now put y as 0
→ x = 4(0)
→ x = 0
- Therefore, (0,0) is 2nd solution
Now put x as 1
→ 1 = 4y
→ 1/4 = y
- Therefore, (0,1/4) is 3rd solution
Now put y as 1
→ x = 4(1)
→ x = 4
- Therefore, (4,0) is 4th solution
~ You can put any digit at the place of x or y, vice versa!
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Question 3rd Check whether of the following are solutions of the equation x - 2y = 4 and which are not:
a) (0,2)
Put x as 0 and y as 2 in equation
→ 0 - 2(2) = 4
→ 0 - 4 = 4
→ -4 ≠ 4
b) (2,0)
Put x as 2 and y as 0 in equation
→ 2 - 2(0) = 4
→ 2 - 0 = 4
→ 2 ≠ 4
c) (4,0)
Put x as 4 and y as 0 in equation
→ 4 - 2(0) = 4
→ 4 - 0 = 4
→ 4 = 4
d) (√2 , 4√2)
Put x as √2 and y as 4√2 in equation
→ √2 - 2(4√2) = 4
→ √2 - 8√2 = 4
→ -7√2 ≠ 4
e) (1,1)
Put x as 1 and y as 1 in equation
→ 1 - 2(1) = 4
→ 1 - 2 = 4
→ -2 ≠ 4
~ Therefore, (4,0) is the solution of the given equation and others are not!
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Question 4th Find the value of k, if x = 2, y = 1 is the solution of 2x + 3y = k
Putting x as 2 and y as 1 in equation
→ 2(2) + 3(1) = k
→ 4 + 3 = k
→ 7 = k
→ k = 7
Therefore, the value ok k = 7