RBSE Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise 4.3

Chapter 4 Linear Equations in Two Variables Ex 4.3

Question 1. In each of the following system of linear equation determine whether the system of equations has a unique solution or no solution or infinitely many solutions. In case there is a unique solution, find it.
2x + y = 35
3x + 4y = 65
Answer: The given system of equations is:
\( 2x + y = 35 \) ... (i)
\( 3x + 4y = 65 \) ... (ii)
We compare the coefficients:
\( \frac{a_1}{a_2} = \frac{2}{3} \)
\( \frac{b_1}{b_2} = \frac{1}{4} \)
\( \frac{c_1}{c_2} = \frac{35}{65} = \frac{7}{13} \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the system of equations is consistent and has a unique solution. This means the lines intersect at one point.
To find the unique solution, we can use the elimination method:
Multiply equation (i) by 4:
\( 4(2x + y) = 4(35) \)
\( 8x + 4y = 140 \) ... (iii)
Now, subtract equation (ii) from equation (iii):
\( (8x + 4y) - (3x + 4y) = 140 - 65 \)
\( 8x + 4y - 3x - 4y = 75 \)
\( 5x = 75 \)
\( \implies x = \frac{75}{5} \)
\( \implies x = 15 \)
Substitute the value of \( x = 15 \) into equation (i):
\( 2(15) + y = 35 \)
\( 30 + y = 35 \)
\( \implies y = 35 - 30 \)
\( \implies y = 5 \)
Thus, the required unique solution is \( x = 15 \) and \( y = 5 \).
In simple words: First, we check if the lines made by these equations meet at just one point. Since they do, we then use a method called elimination to find the exact values for x and y that solve both equations.

🎯 Exam Tip: Remember to always check the ratios of coefficients (\( \frac{a_1}{a_2} \), \( \frac{b_1}{b_2} \), \( \frac{c_1}{c_2} \)) first to determine the nature of the solution (unique, no, or infinite).

Question 2. In each of the following system of linear equation determine whether the system of equations has a unique solution or no solution or infinitely many solutions. In case there is a unique solution, find it.
2x - y = 6
x - y = 2
Answer: The given system of equations is:
\( 2x - y = 6 \) ... (i)
\( x - y = 2 \) ... (ii)
We compare the coefficients:
\( \frac{a_1}{a_2} = \frac{2}{1} = 2 \)
\( \frac{b_1}{b_2} = \frac{-1}{-1} = 1 \)
\( \frac{c_1}{c_2} = \frac{6}{2} = 3 \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) (because \( 2 \neq 1 \)), the system of equations is consistent and has a unique solution. This means the two lines intersect at a single point.
To find the unique solution, we can use the elimination method:
Subtract equation (ii) from equation (i):
\( (2x - y) - (x - y) = 6 - 2 \)
\( 2x - y - x + y = 4 \)
\( x = 4 \)
Substitute the value of \( x = 4 \) into equation (ii):
\( 4 - y = 2 \)
\( \implies -y = 2 - 4 \)
\( \implies -y = -2 \)
\( \implies y = 2 \)
Thus, the required unique solution is \( x = 4 \) and \( y = 2 \).
In simple words: First, we confirm that these two equations will have only one answer. Then, by subtracting one equation from the other, we easily find the values for x and y.

🎯 Exam Tip: When the coefficients of one variable are the same (like 'y' in this case), subtraction is often the quickest way to eliminate that variable.

Question 3. In each of the following system of linear equation determine whether the system of equations has a unique solution or no solution or infinitely many solutions. In case there is a unique solution, find it.
3x + 2y + 25 = 0
2x + y + 10 = 0
Answer: The given system of equations is:
\( 3x + 2y + 25 = 0 \) ... (i)
\( 2x + y + 10 = 0 \) ... (ii)
We compare the coefficients:
\( \frac{a_1}{a_2} = \frac{3}{2} \)
\( \frac{b_1}{b_2} = \frac{2}{1} = 2 \)
\( \frac{c_1}{c_2} = \frac{25}{10} = \frac{5}{2} \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) (because \( \frac{3}{2} \neq 2 \)), the system of equations is consistent and has a unique solution. This means the lines cross at only one point.
To find the unique solution, we can use the elimination method:
Multiply equation (ii) by 2:
\( 2(2x + y + 10) = 2(0) \)
\( 4x + 2y + 20 = 0 \) ... (iii)
Now, subtract equation (iii) from equation (i):
\( (3x + 2y + 25) - (4x + 2y + 20) = 0 - 0 \)
\( 3x + 2y + 25 - 4x - 2y - 20 = 0 \)
\( -x + 5 = 0 \)
\( \implies -x = -5 \)
\( \implies x = 5 \)
Substitute the value of \( x = 5 \) into equation (ii):
\( 2(5) + y + 10 = 0 \)
\( 10 + y + 10 = 0 \)
\( 20 + y = 0 \)
\( \implies y = -20 \)
Thus, the required unique solution is \( x = 5 \) and \( y = -20 \).
In simple words: After checking that there's a unique answer, we multiply the second equation by 2 to make the 'y' terms match. Then, we subtract the equations to find 'x', and finally use 'x' to find 'y'.

🎯 Exam Tip: When using the elimination method, aim to make the coefficients of one variable equal (or opposite) in both equations so that adding or subtracting them cancels out that variable.

Question 4. In each of the following system of linear equation determine whether the system of equations has a unique solution or no solution or infinitely many solutions. In case there is a unique solution, find it.
x + 2y + 1 = 0
2x - 3y - 12 = 0
Answer: The given system of equations is:
\( x + 2y + 1 = 0 \) ... (i)
\( 2x - 3y - 12 = 0 \) ... (ii)
We compare the coefficients:
\( \frac{a_1}{a_2} = \frac{1}{2} \)
\( \frac{b_1}{b_2} = \frac{2}{-3} = -\frac{2}{3} \)
\( \frac{c_1}{c_2} = \frac{1}{-12} = -\frac{1}{12} \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) (because \( \frac{1}{2} \neq -\frac{2}{3} \)), the system of equations is consistent and has a unique solution. This means the two lines intersect at just one point.
To find the unique solution, we use the elimination method:
Multiply equation (i) by 2:
\( 2(x + 2y + 1) = 2(0) \)
\( 2x + 4y + 2 = 0 \) ... (iii)
Subtract equation (ii) from equation (iii):
\( (2x + 4y + 2) - (2x - 3y - 12) = 0 - 0 \)
\( 2x + 4y + 2 - 2x + 3y + 12 = 0 \)
\( 7y + 14 = 0 \)
\( \implies 7y = -14 \)
\( \implies y = \frac{-14}{7} \)
\( \implies y = -2 \)
Substitute the value of \( y = -2 \) into equation (i):
\( x + 2(-2) + 1 = 0 \)
\( x - 4 + 1 = 0 \)
\( x - 3 = 0 \)
\( \implies x = 3 \)
Thus, the required unique solution is \( x = 3 \) and \( y = -2 \).
In simple words: We first check the coefficient ratios to see that there is one unique solution. Then, we multiply the first equation by 2, and subtract the second equation from it. This helps us find 'y', which we then use to find 'x'.

🎯 Exam Tip: Be careful with signs when subtracting equations, especially when terms like \( -(-3y) \) become \( +3y \).

Question 5. Find the value of 'k' for which the system
(i) 2x + ky = 1, 3x – 5y = 7
(ii) kx + 2y = 5, 3x + y = 1 has no solution.
Answer: For a system of linear equations to have no solution, the condition is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). This means the lines are parallel and distinct.

(i) The given equations are:
\( 2x + ky - 1 = 0 \)
\( 3x - 5y - 7 = 0 \)
Here, \( a_1 = 2, b_1 = k, c_1 = -1 \)
\( a_2 = 3, b_2 = -5, c_2 = -7 \)
Applying the condition for no solution:
\( \frac{2}{3} = \frac{k}{-5} \)
\( \implies 2(-5) = 3k \)
\( \implies -10 = 3k \)
\( \implies k = -\frac{10}{3} \)
Now, we must also check the second part of the condition: \( \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \frac{k}{-5} \neq \frac{-1}{-7} \)
\( \frac{k}{-5} \neq \frac{1}{7} \)
Substitute \( k = -\frac{10}{3} \):
\( \frac{-10/3}{-5} \neq \frac{1}{7} \)
\( \frac{2}{3} \neq \frac{1}{7} \), which is true.
So, for the system (i) to have no solution, \( k = -\frac{10}{3} \).

(ii) The given equations are:
\( kx + 2y - 5 = 0 \)
\( 3x + y - 1 = 0 \)
Here, \( a_1 = k, b_1 = 2, c_1 = -5 \)
\( a_2 = 3, b_2 = 1, c_2 = -1 \)
Applying the condition for no solution:
\( \frac{k}{3} = \frac{2}{1} \)
\( \implies k = 3 \times 2 \)
\( \implies k = 6 \)
Now, we must also check the second part of the condition: \( \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \frac{2}{1} \neq \frac{-5}{-1} \)
\( 2 \neq 5 \), which is true.
So, for the system (ii) to have no solution, \( k = 6 \).
In simple words: For equations to have no solution, their lines must be parallel and never cross. This happens when the ratios of their 'x' and 'y' coefficients are equal, but this common ratio is not equal to the ratio of their constant terms. We use this rule to find the specific 'k' value for each part.

🎯 Exam Tip: Always remember the three conditions for linear equations: unique solution (\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)), no solution (\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)), and infinitely many solutions (\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)).

Question 6. Solve the following system of equations
mx - ny = m² + n²
x + y = 2m
Answer: The given equations are:
\( mx - ny = m^2 + n^2 \) ... (i)
\( x + y = 2m \) ... (ii)
We can solve this system using the elimination method. To eliminate 'y', multiply equation (ii) by 'n':
\( n(x + y) = n(2m) \)
\( nx + ny = 2mn \) ... (iii)
Now, add equation (i) and equation (iii):
\( (mx - ny) + (nx + ny) = (m^2 + n^2) + (2mn) \)
\( mx + nx = m^2 + 2mn + n^2 \)
Factor out 'x' on the left side and recognize the perfect square trinomial on the right side:
\( x(m + n) = (m + n)^2 \)
\( \implies x = \frac{(m + n)^2}{(m + n)} \)
\( \implies x = m + n \) (assuming \( m+n \neq 0 \))
Now, substitute the value of \( x = m + n \) into equation (ii):
\( (m + n) + y = 2m \)
\( \implies y = 2m - (m + n) \)
\( \implies y = 2m - m - n \)
\( \implies y = m - n \)
Thus, the required solution is \( x = m + n \) and \( y = m - n \). This method is effective for systems involving literal coefficients.
In simple words: We have two equations with 'x' and 'y' and also 'm' and 'n'. We multiply the second equation by 'n' and add it to the first equation to get rid of 'y'. This helps us find 'x', and then we use that 'x' to find 'y'.

🎯 Exam Tip: When solving equations with literal coefficients, treat the letters (like 'm' and 'n') as constants and use the same algebraic methods (substitution or elimination) you would for numbers.

Question 7. Find the value of 'λ' for which the system
3x + λy + 1 = 0
2x + y - 9 = 0
Answer: The given equations are:
\( 3x + \lambda y + 1 = 0 \) ... (i)
\( 2x + y - 9 = 0 \) ... (ii)
Here, \( a_1 = 3, b_1 = \lambda, c_1 = 1 \)
\( a_2 = 2, b_2 = 1, c_2 = -9 \)

(i) Condition for a unique solution:
For a unique solution, the condition is \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
\( \frac{3}{2} \neq \frac{\lambda}{1} \)
\( \implies \lambda \neq \frac{3}{2} \)
So, for the system to have a unique solution, \( \lambda \) can be any real number except \( \frac{3}{2} \). This means the lines will intersect at a single point if their slopes are different.

(ii) Condition for no solution:
For no solution, the condition is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
First, apply \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \):
\( \frac{3}{2} = \frac{\lambda}{1} \)
\( \implies \lambda = \frac{3}{2} \)
Next, check the second part of the condition: \( \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \frac{\lambda}{1} \neq \frac{1}{-9} \)
Substitute \( \lambda = \frac{3}{2} \):
\( \frac{3}{2} \neq -\frac{1}{9} \), which is true.
So, for the system to have no solution, \( \lambda = \frac{3}{2} \). In this case, the lines will be parallel and distinct.
In simple words: We find values of lambda that determine if the two lines will cross (unique solution) or stay parallel without ever meeting (no solution). For a unique solution, the ratio of x coefficients must not equal the ratio of y coefficients. For no solution, these ratios must be equal, but not equal to the ratio of the constant terms.

🎯 Exam Tip: Clearly state and apply the correct condition (\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) for unique, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) for no solution, and \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) for infinite solutions) for each part of the question to avoid errors.