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✧ CLASS -9 -MATHS- CHAPTER 4- LINEAR EQUATIONS IN TWO VARIABLES

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Answered by TrustedAnswerer19
78

Answer:

2)

i) (0,7) (1,5) (-1,9) ( 2,3)

ii) (0,9) (1, 9-π) (2,9-2π) (3, 9-3π)

iii) (0,0) (4,1) (8,2) (12,3)

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Answered by tennetiraj86
19

Step-by-step explanation:

Solutions :-

1.

Option (iii)

Reason :-

Given equation is y = 3x+5

It is a linear equation in two variables

So it has infinitely number of many solutions.

2.

(i)

Given equation is 2x+y = 7

Put x = 0 then

2(0)+y = 7

=> 0+y = 7

=> y = 7

The first solution = (0,7)

Put y = 0 then

2x+0 = 7

=> 2x = 7

=> x = 7/2

The second solution = (7/2,0)

Put x = 1 then

2(1)+y = 7

=> 2+y = 7

=> y = 7-2

=> y = 5

The third solution = (1,5)

Put y= 1 then

2x+1 = 7

=> 2x = 7-1

=> 2x = 6

=> x = 6/2

=> x = 3

The fourth solution = (3,1)

ii)

Given equation is πx+y = 9

Put x = 0 then

π(0)+y = 9

=> 0+y = 9

=> y = 9

The first solution = (0,9)

Put y = 0 then

πx+0 = 9

=> πx = 9

=> x = 9/π

The second solution = (9/π,0)

Put x = 1 then

π(1)+y = 9

=> π+y = 9

=> y = 9-π

The third solution = (1, 9-π)

Put y= 1 then

πx+1 = 9

=> πx = 9-1

=> πx = 8

=> x = 8/π

The fourth solution = (8/π,1)

iii)

Given equation is x = 4y

Put x = 0 then

0=4y

=> y = 0/4

=> y = 0

The first solution = (0,0)

Put y = 1 then

x = 4(1)

=> x = 4

The second solution = (4,1)

Put x = 1 then

1 = 4y

=> y = 1/4

The third solution = (1,1/4)

Put y= -1 then

x= 4(-1)

=> x = -4

The fourth solution = (-1,-4)

3)

Given equation is x-2y =4

I.(0,2):-

Put x = 0 and y = 2 then LHS

=> 0-2(2)

=> 0-4

=> -4

LHS ≠ RHS.

(0,2) is not a solution.

ii)(2,0):-

=>Put x = 2 and y = 0 then LHS

=> 2-2(0)

=> 2-0

=> 2

LHS ≠ RHS.

(2,0) is not a solution.

iii)(4,0)

Put x = 4 and y = 0 then LHS

=> 4-2(0)

=> 4-0

=> 4

LHS = RHS.

(4,0) is a solution.

iv)(2,42):-

Put x = √2 and y = 4√2 then LHS

=> √2-2(4√2)

=> √2-8√2

=> (1-8)√2

=> -7√2

LHS ≠ RHS.

(√2,4√2) is not solution.

v)(1,1):-

Put x =1 and y = 1 then LHS

=> 1-2(1)

=> 1-2

=> -2

LHS ≠ RHS.

(1,1) is not solution.

4)

Given equation is 2x+3y = k

If x = 2 and y = 1 is a solution then it satisfies the given equation.

=> 2(2)+3(1) = k

=> 4+3 = k

=> 7=k

Therefore,k = 7

The value of k = 7

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