Linear Equations in Two Variables – Maharashtra Board Class 9 Solutions for Algebra

Linear Equations in Two Variables – Maharashtra Board Class 9 Solutions for Algebra

AlgebraGeometryScience and TechnologyHindi

Exercise – 4.1

Solution 1:

1. 2x + 5y = 7; x – 10y = 11; 4x + 9y = 13
2. No. The given equation is not of the type ax + by = c, hence it is not a linear equation.
3. P is a real number. P > 0.
4. (1, 17), (2, 16), etc. There are infinite solutions.
5. Substituting x = 1 and y = a in the equation x + 3y = 10, we get,1 + 3a = 10
   ∴3a = 10 – 1
   ∴3a = 9
   ∴ a = 3

Solution 2(i):

Solution 2(ii):

Solution 2(iii):

Solution 2(iv):

Solution 2(v):

Exercise – 4.2

Solution 1(i):

2x + 3y = -4 …(i)
x – 5y = 11 …(ii)
From equation (ii), we can express x in terms of y,
∴ x = 5y + 11 …(iii)
Substitute this value of x in equation (i)
∴ 2(5y + 11) + 3y = -4
∴ 10y + 22 + 3y = -4
∴ 13y +22 = -4
∴ 13y = -4 – 22
∴ 13y = -26
∴ y = -2
Substituting y = -2 in equation (iii),
x = 5(-2) + 11
∴ x = -10 + 11
∴ x = 1
∴ x = 1 and y = -2

Solution 1(ii):

x + 2y = 0 …(i)
10x + 15y = 105 …(ii)
Dividing both sides of equation (ii) by 5, we get
2x + 3y = 21 …(iii)
Expressing x in terms of y in equation (i)
x = 2y …(iv)
Substitute x = 2y in equation (iii)
∴ 2(2y) + 3y = 21
∴ 4y + 3y = 21
∴ 7y = 21
∴ y = 3
Substituting y = 3 in equation (iii),
x = 2(3)
∴ x = 6
∴ x = 6 and y = 3

Solution 1(iii):

Solution 1(iv):

Solution 1(v):

2x – y – 3 = 0 …(i)
4x – y – 5 = 0 …(ii)
∴ 2x – y – 3 = 0
∴ y = 2x – 3 …(iii)
Substitute y = 2x – 3 in equation (ii),
∴ 4x – (2x – 3) – 5 = 0
∴ 4x – 2x + 3 – 5 = 0
∴ 2x = 5 – 3
∴ 2x = 2
∴ x = 1
Put x = 1 in equation (iii),
y = 2(1) – 3
∴ y = 2 – 3
∴ y = -1
∴ x = 1 and y = -1

Solution 1(vi):

Exercise – 4.3

Solution 1(i):

Solution 1(ii):

Solution 1(iii):

Solution 1(iv):

Solution 2(i):

Solution 2(ii):

Solution 2(iii):

Solution 2(iv):

Solution 2(v):

Exercise – 4.3

Solution 1(i):

Let the two numbers be x and y. x > y
According to the first condition,
x + y = 125 …(i)
According to the second condition,
x – y = 25 …(ii)

Solution 1(ii):

Let the complementary angles be x and y. x > y
So, x + y = 90° (∵Sum of complementary angles is 90°)
According to the given condition,
x – y = 6°

Solution 1(iii):

Let the length of the rectangle be x cm and the breadth be y cm.
According to the first condition,
x = y + 4
x – y = 4 …(i)
According to the second condition,
2(x + y) = 40
x + y = 20 …(ii)

Solution 1(iv):

Let Sonali’s age be x years and Monali’s age of y years.
According to the first condition,
x + y = 29 …(i)
According to the second condition,
y = x – 3
x – y = 3 …(ii)

Solution 1(v):

Let the father’s age be x years and the son’s age be y years.
According to the first condition,
x = 4y
x – 4y = 0 …(i)
According to the second condition,
x – y = 30 …(ii)

Solution 2(i):

Solution 2(ii):

Solution 2(iii):

Solution 2(iv):

Solution 2(v):

Solution 2(vi):