RBSE Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables Important Questions

Multiple Choice Questions

Question 1. The lines representing the linear equations \( 2x - y = 3 \) and \( 4x - y = 5 \)
(a) intersect at a point
(b) are parallel
(c) are coincident
(d) intersect at exactly two points
Answer: (a) intersect at a point
For the given equations:
First equation: \( 2x - y = 3 \implies a_1 = 2, b_1 = -1, c_1 = -3 \)
Second equation: \( 4x - y = 5 \implies a_2 = 4, b_2 = -1, c_2 = -5 \)
Now, we compare the ratios of the coefficients:
\( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \)
\( \frac{b_1}{b_2} = \frac{-1}{-1} = 1 \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the lines will intersect at a single point. This means there is a unique solution where the two lines cross each other.
In simple words: When you compare the numbers next to 'x' and 'y' in both equations, if their ratios are not equal, the lines will cross at one single spot. This is how we know they have one unique solution.

🎯 Exam Tip: Remember to write the equations in the standard form \( ax + by + c = 0 \) before comparing the coefficients to avoid sign errors, even if the source places 'c' on the other side.

Question 2. If the pair of linear equations in two variables has infinite number of solutions, then the lines represented by these equations are:
(a) intersecting lines
(b) coincident lines
(c) parallel lines
Answer: (b) coincident lines
If a pair of linear equations has an infinite number of solutions, it means that every point on one line is also a point on the other line. This happens when the two lines are exactly the same and lie directly on top of each other. These lines are called coincident lines because they occupy the same space.
In simple words: When two lines are exactly the same, they touch at every point, meaning they have endless solutions. Such lines are called coincident.

🎯 Exam Tip: Coincident lines occur when the ratios of all three coefficients (a, b, and c) are equal: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).

Question 3. The pair of linear equations \( 5x + 4y = 20 \) and \( 10x + 8y = 16 \) have
(a) no solution
(b) many solutions
(c) two solutions
(d) one solution
Answer: (a) no solution
For the first equation: \( 5x + 4y = 20 \implies a_1 = 5, b_1 = 4, c_1 = -20 \)
For the second equation: \( 10x + 8y = 16 \implies a_2 = 10, b_2 = 8, c_2 = -16 \)
Let's compare the ratios of the coefficients:
\( \frac{a_1}{a_2} = \frac{5}{10} = \frac{1}{2} \)
\( \frac{b_1}{b_2} = \frac{4}{8} = \frac{1}{2} \)
\( \frac{c_1}{c_2} = \frac{-20}{-16} = \frac{5}{4} \)
Here we see that \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). This condition means that the lines are parallel and will never intersect. When lines are parallel, they do not share any common points, so there is no solution.
In simple words: If the ratios of the first two numbers (from x and y) are the same, but the ratio of the last numbers (constants) is different, the lines are parallel. Parallel lines never meet, so there is no answer that works for both.

🎯 Exam Tip: The condition \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) is a key indicator for parallel lines, resulting in no solution. Memorize this relation for quick problem-solving.

Question 4. The value of k for which the pair of linear equations \( 4x + 6y - 1 = 0 \) and \( 2x + ky - 7 = 0 \) represent parallel lines is:
(a) k = 3
(b) k = 2
(c) k = 4
(d) k = - 2
Answer: (a) k = 3
For the lines to be parallel, the condition is:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Here, \( a_1 = 4, b_1 = 6, c_1 = -1 \)
And \( a_2 = 2, b_2 = k, c_2 = -7 \)
Substituting these values into the condition:
\( \frac{4}{2} = \frac{6}{k} \neq \frac{-1}{-7} \)
First, let's use the equality part: \( \frac{4}{2} = \frac{6}{k} \)

\( \implies 2 = \frac{6}{k} \)

\( \implies 2k = 6 \)

\( \implies k = 3 \)
Now, let's check the inequality part with \( k=3 \):
\( \frac{6}{3} \neq \frac{1}{7} \)
\( 2 \neq \frac{1}{7} \)
This is true, so \( k=3 \) satisfies the condition for parallel lines. Thus, when \( k=3 \), the lines are indeed parallel.
In simple words: For lines to be parallel, the ratio of the 'x' numbers must be the same as the ratio of the 'y' numbers, but different from the ratio of the constant numbers. By setting the first two ratios equal, we can find the value of 'k'.

🎯 Exam Tip: Always verify both parts of the parallel lines condition (\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) AND \( \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)) with your calculated 'k' value to ensure it's a valid solution.

Question 5. The value of k for which the pair of linear equations \( 6x + ky + C_1 = 0 \) and \( x + y + C_2 = 0 \) has a unique solution is:
(a) k = 6
(b) k ≠ 6
(c) k = 0
(d) k has any value
Answer: (b) k ≠ 6
For a pair of linear equations to have a unique solution, the lines must intersect at exactly one point. The condition for a unique solution is:
\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Here, from the assumed equations for \( a_1/a_2 = 6/1 \) and \( b_1/b_2 = k/1 \), we substitute the coefficients:
\( \frac{6}{1} \neq \frac{k}{1} \)

\( \implies 6 \neq k \)
So, for the given equations to have a unique solution, the value of k must not be equal to 6. This ensures that the slopes of the lines are different.
In simple words: For lines to cross at only one spot, the numbers in front of 'x' and 'y' must not have the same ratio. This means 'k' cannot be 6, otherwise the lines would be parallel or the same.

🎯 Exam Tip: For unique solutions, the most important thing is that the ratio of 'a' coefficients is not equal to the ratio of 'b' coefficients, ensuring the lines are not parallel.

Question 6. The value of \( \lambda \) for which \( x + 2y + 7 = 0 \) and \( 2x + \lambda y + 14 = 0 \) represent coincident lines is
(a) 3
(b) 4
(c) - 4
(d) - 3
Answer: (b) 4
For the lines to be coincident (meaning they are the same line and have infinite solutions), the ratios of all corresponding coefficients must be equal:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Here, \( a_1 = 1, b_1 = 2, c_1 = 7 \)
And \( a_2 = 2, b_2 = \lambda, c_2 = 14 \)
Substituting these values:
\( \frac{1}{2} = \frac{2}{\lambda} = \frac{7}{14} \)
From the first part of the equality:
\( \frac{1}{2} = \frac{2}{\lambda} \)

\( \implies \lambda \times 1 = 2 \times 2 \)

\( \implies \lambda = 4 \)
We can also check with the second part: \( \frac{7}{14} = \frac{1}{2} \). Since \( \frac{2}{\lambda} = \frac{1}{2} \), this also gives \( \lambda = 4 \). When \( \lambda = 4 \), all ratios are equal, making the lines coincident.
In simple words: If two lines are exactly the same, all the numbers in their equations (a, b, and c) must be in the same proportion. By comparing these proportions, we can find the missing number, \( \lambda \).

🎯 Exam Tip: Coincident lines are essentially multiples of each other. If you multiply the first equation by a constant and get the second, the lines are coincident.

Question 7. A pair of linear equations \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) is inconsistent if
Answer: A pair of linear equations is inconsistent if it has no solution. This occurs when the lines represented by the equations are parallel. The condition for inconsistent equations (or parallel lines) is:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
This means the lines have the same slope but different y-intercepts, so they never meet.
In simple words: Equations are "inconsistent" if they have no solution. This happens when the lines they draw are parallel and never cross, meaning their 'x' and 'y' ratios are the same, but the constant ratio is different.

🎯 Exam Tip: The term "inconsistent" is synonymous with "no solution" and "parallel lines" in the context of linear equations.

Question 8. If the system of equations \( 2x + 3y = 7 \) and \( 2ax + (a + b)y = 28 \) represent coincident lines, then the required condition is
(a) b = 2a
(b) a = 2b
(c) 2a + b = 0
(d) a + 2b = 0
Answer: (a) b = 2a
For the given equations to represent coincident lines, the ratios of their corresponding coefficients must be equal:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Here, the equations are: \( 2x + 3y - 7 = 0 \) and \( 2ax + (a + b)y - 28 = 0 \)
So, \( a_1 = 2, b_1 = 3, c_1 = -7 \)
And \( a_2 = 2a, b_2 = (a+b), c_2 = -28 \)
Substitute these into the condition:
\( \frac{2}{2a} = \frac{3}{a+b} = \frac{-7}{-28} \)
First, simplify the last ratio: \( \frac{-7}{-28} = \frac{1}{4} \)
Now, equate the first ratio to \( \frac{1}{4} \):
\( \frac{2}{2a} = \frac{1}{4} \)
\( \frac{1}{a} = \frac{1}{4} \)

\( \implies a = 4 \)
Next, equate the second ratio to \( \frac{1}{4} \):
\( \frac{3}{a+b} = \frac{1}{4} \)

\( \implies 4 \times 3 = a+b \)

\( \implies 12 = a+b \)
Substitute the value of \( a=4 \) into this equation:
\( 12 = 4 + b \)

\( \implies b = 12 - 4 \)

\( \implies b = 8 \)
Now, let's check which of the given options is true for \( a=4 \) and \( b=8 \):
(a) \( b = 2a \implies 8 = 2(4) \implies 8 = 8 \) (This is true)
(b) \( a = 2b \implies 4 = 2(8) \implies 4 = 16 \) (This is false)
(c) \( 2a + b = 0 \implies 2(4) + 8 = 0 \implies 8 + 8 = 0 \implies 16 = 0 \) (This is false)
(d) \( a + 2b = 0 \implies 4 + 2(8) = 0 \implies 4 + 16 = 0 \implies 20 = 0 \) (This is false)
So, the required condition is \( b=2a \).
In simple words: For two lines to be exactly the same (coincident), the numbers in their equations must all match up perfectly in proportion. We use this rule to find the specific connection between 'a' and 'b' that makes this happen.

🎯 Exam Tip: When dealing with coincident lines, equate all three ratios \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) and solve the resulting system of equations for the unknowns.

Question 9. If the system of equations \( 2x + 3y = 7 \) and \( (a + b)x + (2a - b)y = 21 \) has infinitely many solutions, then
(a) a = 1, b = 5
(b) a = 5, b = 1
(c) a = - 1, b = + 5
Answer: (b) a = 5, b = 1
For a system of linear equations to have infinitely many solutions, the lines must be coincident. This means all the ratios of their coefficients must be equal:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
The given equations are: \( 2x + 3y - 7 = 0 \) and \( (a + b)x + (2a - b)y - 21 = 0 \)
So, \( a_1 = 2, b_1 = 3, c_1 = -7 \)
And \( a_2 = (a+b), b_2 = (2a-b), c_2 = -21 \)
Substitute these into the condition:
\( \frac{2}{a+b} = \frac{3}{2a-b} = \frac{-7}{-21} \)
First, simplify the last ratio: \( \frac{-7}{-21} = \frac{1}{3} \)
Now, we have two equalities:
1. \( \frac{2}{a+b} = \frac{1}{3} \)

\( \implies 3 \times 2 = a+b \)

\( \implies a+b = 6 \) ... (i)
2. \( \frac{3}{2a-b} = \frac{1}{3} \)

\( \implies 3 \times 3 = 2a-b \)

\( \implies 2a-b = 9 \) ... (ii)
Now we solve the system of two linear equations (i) and (ii) for 'a' and 'b':
Add equation (i) and equation (ii):
\( (a+b) + (2a-b) = 6 + 9 \)
\( a+2a+b-b = 15 \)
\( 3a = 15 \)

\( \implies a = \frac{15}{3} \)

\( \implies a = 5 \)
Substitute the value of \( a=5 \) into equation (i):
\( 5 + b = 6 \)

\( \implies b = 6 - 5 \)

\( \implies b = 1 \)
So, the values are \( a=5 \) and \( b=1 \).
In simple words: When equations have endless solutions, it means they are the same line. We use the rule that all their parts must be proportional to create two smaller equations. Solving these gives us the exact numbers for 'a' and 'b'.

🎯 Exam Tip: Problems with infinitely many solutions often reduce to solving a system of two linear equations in terms of the unknown constants (like 'a' and 'b' here).

Question 10. The line \( 2x - 3y - 6 = 0 \) meets x-axis at
(a) (0, 3)
(b) (3, 0)
(c) (2, 3)
(d) (6, 2)
Answer: (b) (3, 0)
When a line meets the x-axis, the y-coordinate of that point is always zero. We can find where the line intersects the x-axis by setting \( y=0 \) in the equation of the line.
Given equation: \( 2x - 3y - 6 = 0 \)
Substitute \( y=0 \):
\( 2x - 3(0) - 6 = 0 \)
\( 2x - 0 - 6 = 0 \)
\( 2x - 6 = 0 \)
Now, solve for x:
\( 2x = 6 \)

\( \implies x = \frac{6}{2} \)

\( \implies x = 3 \)
So, the point where the line meets the x-axis is \( (3, 0) \).
In simple words: To find where a line crosses the 'x' axis, always put zero for 'y' in the equation. Then, solve to find the 'x' value. The point will be (x-value, 0).

🎯 Exam Tip: Remember that any point on the x-axis has a y-coordinate of 0, and any point on the y-axis has an x-coordinate of 0. This is crucial for finding intercepts.