Pair of Linear Equations in Two Variables Questions

Pair of linear equations in two Variables questions with answers are provided here. These questions are very helpful for students of Class 10 to practise for exams. The pair of linear equations in two variables problems are prepared as per the NCERT guidelines and the latest CBSE syllabus (2022-2023).

Also check: Pair Of Linear Equation In Two Variables notes.

Definition: When an equation is written in the form of ax+by+c = 0, such equations are called linear equations in two variables. Here, a, b and c are constants (real numbers). Also, the coefficients of x and y i.e., a and b are never equal to zero.

Now, if we are given pairs of linear equations in two variables, we have to solve such equations to find the value of unknown variables. To find the solution we can use:

• Cross multiplication method
• Elimination method
• Substitution method

Rules for Number of Solution: If m₁ and m₂ are the slopes of pairs of linear equations in two variables, then;

• For m₁ ≠ m₂, we have a unique solution
• For m₁ = m₂, we have no solution

Pair of Linear Equations in Two Variables Questions and Solutions

Q.1: Solve the following pair of linear equations:

y -4x= 1

6x- 5y= 9

Solution: Given,

y -4x= 1

6x- 5y= 9

We can write;

-4x + y = 1 ………(i)

6x – 5y = 9 ………(ii)

Multiplying equation (i) by 5 and adding with equation (ii), we get;

-20x + 5y = 5

+6x – 5y = 9

————————-

-14 x = 14

———————–

x = 14/-14 = -1

Now putting the value of x = -1 in equation (i)

-4 (-1) + y = 1

y = 1 – 4

y = -3

Q.2: Solve the following using the cross multiplication method.

x + y = 1

2x – 3y = 11

Solution: Given,

x+y=1

2x – 3y = 11

By cross multiplication we have;

x/32 = y/3 = -1/-5

x/32 = ⅕

x = 32/5

y/3 = ⅕

y = 3/5

Q.3: How many solutions does the pair of equations y = 0 and y = -10 have?

Solution: Given,

y = 0 and y = -10

They are parallel lines,

Therefore, no solution.

Q.4: Solve for x and y.

7(y+3) – 2 (x+2) = 14

4 (y-2) + 3(x-3) = 2

Solution: Given,

7(y+3) – 2 (x+2) = 14

4 (y-2) + 3(x-3) = 2

We can write the above equations as:

7(y+3) 2 (x+2) = 14

7y + 21- 2x – 4 = 14

2x – 7y – 3 = 0 ……….(i)

4 (y-2) + 3(x-3) = 2

4y – 8 + 3x – 9 = 2

3x + 4y -19 = 0 ……….(ii)

Solving equations (i) and (ii) we get;

x = 5 and y = 1

Q.4: The sum of numerator and denominator of a fraction is 3 less than twice the denominator. If each of the numerator and denominator is decreased by 1, the fraction becomes ½. Find the fraction.

Solution: Let the numerator be x and denominator be y.

Let the fraction be x/y

x+y+3=2y

x−y=−3

x=y−3 ………(i)

Also,

(x-1)/(y-1) = ½

(x-1) = (y-1)/2

2x=y−1+2=y+1

2x−y=1 …………..(ii)

Substituting (i) in equation (ii), we get;

2(y−3)−y=1

⇒2y−6−y=1

⇒y=7

Now, put y=7 in (i) we get;

x=7−3=4

Thus, x=4,y=7

Therefore, the required fraction is 4/7.

Q.5: Find whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident.

2x – 3y + 6 = 0,4x – 5y + 2 = 0

Solution: Given,

2x – 3y + 6 = 0

4x – 5y + 2 = 0

a1/a2 = 2/4 = ½

b1/b2 = -3/-5 = ⅗

Since, a1/a2 ≠ b1/b2

Thus, the given linear equations have a unique solution and they intersect at a point.

Q.6: Find the solution of the given pair of linear equations.

x + 2y – 3= 0

3x – 2y + 7 = 0

Solution: Given,

x + 2y – 3= 0

3x – 2y + 7 = 0

We can write the given equations as:

x+2y = 3

3x – 2y = -7

On adding equations (i) and (ii) we get;

4x = -4

x = -1

Putting x = -1 in eq.(i), we get;

-1 + 2y = 3

2y = 4

y = 2

Therefore, the solutions are x = -1 and y = 2.

Q.7: Solve the following:

6(ax + by) = 3a + 2b

6(bx – ay) = 3b -2a

Solution:

Given,

6(ax + by) = 3a + 2b

6ax + 6by = 3a + 2b ……………(i)

6(bx – ay) = 3b -2a

6bx – 6ay = 3b – 2a …………..(ii)

On multiplying eq.(i) by a and eq.(ii) by b and adding them we get;

6a²x + 6aby = 3a² + 2ab

+6b²x – 6aby = -2ab + 3b²

—————————————————

6(a² + b²)x = 3(a² + b²)

x = ½

Now putting the value of x = ½ in equation (i), we have;

6a x ½ + 6by = 3a + 2b

6by = 2b

y = ⅓

Therefore, the solutions of the given equations are x = ½ and y = ⅓.

Q.8: If the system of equations has a unique solution, find the value of k.

6x + 2y = 3 and kx + y = 2

Solution: Given,

6x + 2y = 3 and kx + y = 2 have unique solutions

Thus, a1/a2 ≠ b1/b2

6/k ≠ 2/1

k ≠ 3

Therefore, k will have any real value apart from 3.

Q.9: Solve by elimination method.

3x = y + 5

5x – y = 11

Solution: Given,

3x = y + 5

3x – y = 5 ………..(i)

5x – y = 11 ………(ii)

Subtracting equation (i) and (ii) we get;

3x – y = 5

5x – y = 11

–     +      –

———————————

-2x = -6

——————————–

x = 3

Now, the value of x in (i), we get

3x – y = 5

⇒ 3(3) – y = 5

9 – 5 = y

⇒ y = 4

Therefore, the required solutions are x = 3 and y = 4.

Q.10: A two-digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.

Solution:

Let unit’s place digit be x and ten’s place digit be y.

Then original number = x + 10y

and reversed number = 10x + y

According to the given condition;

x + 10y = 7(x + y)

x + 10y = 7x + 7y

⇒ 10y – 7y = 7x – x

⇒ 3y = 6x ⇒ y = 2x ……….…(i)

Again, given;

(x + 10y) – (10x + y) = 18

x + 10y – 10x – y = 18

⇒ 9y – 9x = 180

⇒ y – x = 2

⇒ 2x – x = 2

Therefore, x = 2

Putting the value of ‘x’ in (i), we get :

y = 2(2) = 4

Hence,

Required number = x + 10y

= 2 + 10(4)

= 42

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