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