Problem based on Linear equation in two variable

1. y = 4 x + 5 have how many solutions find it ?

Sol. Y = 4 x + 5 have infinite many solution because for all the value of x, a value of y we will find which satisfied the equation.

maths ncert solutions

2. Find the four solutions for given equations :-

a). 3 x + 2 y = 5

b). y = 3 x + 7

c). 2 x + 5 y = 6

d). 5 x + 2 y = 3

e). ∏ X + 2 Y = 2

f). 3 x - 2 y = 5

g). y = 3 x - 7

h). 2 x - 5 y = 6

i). 5 x - 2 y = 3

j). ∏ X - 2 Y = 2

a). 3 x + 2 y = 5

For x = 0

3 (0) + 2 y = 5 => y = 5/2

So, (0 , 5/2 ) is the solution of equation.

3 x + 2 y = 5

For x = 1

3 (1) + 2 y = 5 => 3 + 2 y = 5 => 2 y = 2 => y = 1

So, (1 , 1) is the solution of equation.

3 x + 2 y = 5

For x = -1

3 (-1) + 2 y = 5 => -3 + 2 y = 5 => 2 y = 8 => y = 4

So, (-1, 4) is the solution of equation.

3 x + 2 y = 5

For x = 2

3 (2) + 2 y = 5 => 6 + 2 y = 5 => 2 y = -1 => y = -1/2

So, (2 , -1/2) is the solution of equation.

b). y = 3 x + 7

For x = 0

y = 3 (0) + 7 => y = 7

So, (0 , 7) is the solution of equation.

y = 3 x + 7

For x = 1

y = 3 (1) + 7 => y = 3 + 7 => y = 10

So, (1 , 10) is the solution of equation.

y = 3 x + 7

For x = -1

y = 3 (-1) + 7 => y = -3 + 7 => y = 4

So, (-1 , 4) is the solution of equation.

y = 3 x + 7

For x = -2

y = 3 (-2) + 7 => y = -6 + 7 => y = 1

So, (-2 , 1) is the solution of equation.

c). 2 x + 5 y = 6

For x = 0

2 (0) + 5 y = 6 => y = 6/5

So, (0 , 6/5) is the solution of equation.

2 x + 5 y = 6

For x = 1

2 (1) + 5 y = 6 => 2 + 5 y = 6 => 5 y = 4 => y = 4/5

So, (1 , 4/5) is the solution of equation.

2 x + 5 y = 6

For x = -1

2 (-1) + 5 y = 6 => -2 + 5 y = 6 => y = 8/5

So, (-1, 8/5) is the solution of equation.

2 x + 5 y = 6

For x = 2

2 (2) + 5 y = 6 => 4 + 5 y = 6 => 5 y = 2 => y = 2/5

So, (2 , 2/5) is the solution of equation.

d). 5 x + 2 y = 3

For x = 0

5 (0) + 2 y = 3 => y = 3/2

So, (0 , 3/2) is the solution of equation.

5 x + 2 y = 3

For x = 1

5 (1) + 2 y = 3 => 5 + 2 y = 3 => 2 y = -2 => y = -1

So, (1 , -1) is the solution of equation.

5 x + 2 y = 3

For x = -1

5 (-1) + 2 y = 3 => -5 + 2 y = 3 => 2 y = 8 => y = 4

So, (-1, 4) is the solution of equation.

5 x + 2 y = 3

For x = 2

5 (2) + 2 y = 3 => 10 + 2 y = 3 => 2 y = -7 => y = -7/2

So, (2 , -7/2) is the solution of equation.

e). ∏ X + 2 Y = 2

For x = 0

∏ (0) + 2 y = 2 => y = 1

So, (0 , 1) is the solution of equation.

∏ x + 2 y = 2

For x = 1

∏ (1) + 2 y = 2 => ∏ + 2 y = 2 => 2 y = 2 - ∏ => y = 1 - ∏/2

So, {1 ,( 1 - ∏/2)} is the solution of equation.

∏ x + 2 y = 2

For x = -1

∏ (-1) + 2 y = 2 => -∏ + 2 y = 2 => 2 y = 2 + ∏ => y = 1 + ∏/2

So, {-1 ,( 1 + ∏/2)} is the solution of equation.

∏ x + 2 y = 2

For x = 2

∏ (2) + 2 y = 2 => 2 ∏ + 2 y = 2 => 2 y = 2 - 2 ∏ => y = 1 - ∏

So, {1 ,( 1 - ∏)} is the solution of equation.

f). 3 x - 2 y = 5

For x = 0

3 (0) - 2 y = 5 => y = -5/2

So, (0 , -5/2 ) is the solution of equation.

3 x - 2 y = 5

For x = 1

3 (1) - 2 y = 5 => 3 - 2 y = 5 => -2 y = 2 => y = -1

So, (1 , -1) is the solution of equation.

3 x - 2 y = 5

For x = -1

3 (-1) - 2 y = 5 => -3 - 2 y = 5 => -2 y = 8 => y = -4

So, (-1, -4) is the solution of equation.

3 x - 2 y = 5

For x = 2

3 (2) - 2 y = 5 => 6 - 2 y = 5 => -2 y = -1 => y = 1/2

So, (2 , 1/2) is the solution of equation.

g). y = 3 x - 7

For x = 0

y = 3 (0) - 7 => y = -7

So, (0 , -7) is the solution of equation.

y = 3 x - 7

For x = 1

y = 3 (1) - 7 => y = 3 - 7 => y = -4

So, (1 , -4) is the solution of equation.

y = 3 x - 7

For x = -1

y = 3 (-1) - 7 => y = -3 - 7 => y = -10

So, (-1 , -10) is the solution of equation.

y = 3 x - 7

For x = -2

y = 3 (-2) - 7 => y = -6 - 7 => y = -13

So, (-2 , -13) is the solution of equation.

h). 2 x - 5 y = 6

For x = 0

2 (0) - 5 y = 6 => y = -6/5

So, (0 , -6/5) is the solution of equation.

2 x - 5 y = 6

For x = 1

2 (1) - 5 y = 6 => 2 - 5 y = 6 => -5 y = 4 => y = -4/5

So, (1 , -4/5) is the solution of equation.

2 x - 5 y = 6

For x = -1

2 (-1) - 5 y = 6 => -2 - 5 y = 6 => y = -8/5

So, (-1, -8/5) is the solution of equation.

2 x - 5 y = 6

For x = 2

2 (2) - 5 y = 6 => 4 - 5 y = 6 => -5 y = 2 => y = -2/5

So, (2 , -2/5) is the solution of equation.

i). 5 x - 2 y = 3

For x = 0

5 (0) - 2 y = 3 => y = -3/2

So, (0 , -3/2) is the solution of equation.

5 x - 2 y = 3

For x = 1

5 (1) - 2 y = 3 => 5 - 2 y = 3 => -2 y = -2 => y = 1

So, (1 , 1) is the solution of equation.

5 x - 2 y = 3

For x = -1

5 (-1) - 2 y = 3 => -5 - 2 y = 3 => -2 y = 8 => y = -4

So, (-1, -4) is the solution of equation.

5 x - 2 y = 3

For x = 2

5 (2) - 2 y = 3 => 10 - 2 y = 3 => -2 y = -7 => y = 7/2

So, (2 , 7/2) is the solution of equation.

j). ∏ X - 2 Y = 2

For x = 0

∏ (0) - 2 y = 2 => y = -1

So, (0 , -1) is the solution of equation.

∏ x - 2 y = 2

For x = 1

∏ (1) - 2 y = 2 => ∏ - 2 y = 2 => -2 y = 2 - ∏ => y = ∏/2 - 1

So, {1 ,(∏/2 - 1)} is the solution of equation.

∏ x - 2 y = 2

For x = -1

∏ (-1) - 2 y = 2 => -∏ - 2 y = 2 => -2 y = 2 + ∏ => y = -1 - ∏/2

So, {-1 ,( -1 - ∏/2)} is the solution of equation.

∏ x - 2 y = 2

For x = 2

∏ (2) - 2 y = 2 => 2 ∏ - 2 y = 2 => 2 y = 2 ∏ - 2 => y = ∏ - 1

So, {1 ,(∏ - 1)} is the solution of equation.

Milan Tomic

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