RBSE Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise 4.1

Question 1. Solve the following pair of equations graphically: x + 3y = 6; 2x – 3y = 12
Answer:
First, consider the equation \( x + 3y = 6 \). We can rewrite this to find \( x \) in terms of \( y \):
\( x = 6 - 3y \) ... (i)

Now, we find some points for this equation:
When \( y = 1 \), \( x = 6 - 3(1) = 3 \). So, the point is (3, 1).
When \( y = 2 \), \( x = 6 - 3(2) = 0 \). So, the point is (0, 2).
When \( y = 0 \), \( x = 6 - 3(0) = 6 \). So, the point is (6, 0).
When \( y = -1 \), \( x = 6 - 3(-1) = 9 \). So, the point is (9, -1).

Table for equation (i):

Next, consider the second equation: \( 2x - 3y = 12 \). We can rewrite this to find \( x \) in terms of \( y \):
\( 2x = 3y + 12 \)
\( \implies x = \frac{3y + 12}{2} \) ... (ii)

Now, we find some points for this equation:
When \( y = 1 \), \( x = \frac{3(1) + 12}{2} = \frac{15}{2} = 7.5 \). So, the point is (7.5, 1).
When \( y = 0 \), \( x = \frac{3(0) + 12}{2} = \frac{12}{2} = 6 \). So, the point is (6, 0).
When \( y = -1 \), \( x = \frac{3(-1) + 12}{2} = \frac{9}{2} = 4.5 \). So, the point is (4.5, -1).

Table for equation (ii):

By plotting these points and drawing lines on a graph paper, we observe that the two lines intersect at the point (6, 0). This intersection point represents the common solution to both equations. Plotting graphs helps to visually find where equations meet.
Therefore, the solution to the given pair of equations is \( x = 6 \) and \( y = 0 \).
In simple words: First, find several points for each equation. Then, draw both lines on a graph. The spot where the lines cross is the answer. Here, they cross at (6, 0).

🎯 Exam Tip: Always use a ruler and sharp pencil for plotting points and drawing lines on graph paper to ensure accuracy, as even small errors can lead to incorrect intersection points.

Question 2. Solve the following pair of equations graphically: 2x + y = 6; 2x - y + 2 = 0
Answer:
Let's start with the first equation: \( 2x + y = 6 \). We can express \( y \) in terms of \( x \):
\( y = 6 - 2x \) ... (i)

Now, we find some points for this equation:
When \( x = 1 \), \( y = 6 - 2(1) = 4 \). So, the point is (1, 4).
When \( x = 2 \), \( y = 6 - 2(2) = 2 \). So, the point is (2, 2).
When \( x = 3 \), \( y = 6 - 2(3) = 0 \). So, the point is (3, 0).
When \( x = 0 \), \( y = 6 - 2(0) = 6 \). So, the point is (0, 6).

Table for equation (i):

Next, consider the second equation: \( 2x - y + 2 = 0 \). We can express \( y \) in terms of \( x \):
\( y = 2x + 2 \) ... (ii)

Now, we find some points for this equation:
When \( x = 0 \), \( y = 2(0) + 2 = 2 \). So, the point is (0, 2).
When \( x = 1 \), \( y = 2(1) + 2 = 4 \). So, the point is (1, 4).
When \( x = -1 \), \( y = 2(-1) + 2 = 0 \). So, the point is (-1, 0).
When \( x = 2 \), \( y = 2(2) + 2 = 6 \). So, the point is (2, 6).

Table for equation (ii):

After plotting these points and drawing the lines on a graph paper, we see that the lines intersect at the point (1, 4). This intersection gives the unique solution for the system. Graphing linear equations helps visualize the nature of their solutions.
Thus, the solution to this pair of equations is \( x = 1 \) and \( y = 4 \).
In simple words: Find points for each equation, then draw the lines on a graph. Where the lines cross, that is your answer. Here, they cross at (1, 4).

🎯 Exam Tip: When choosing points for a table, pick a mix of positive, negative, and zero values for x (or y) to ensure the line is accurately plotted across different quadrants.

Question 3. Solve the following pair of equations graphically: x - 2y = 6; 3x - 6y = 0
Answer:
Let's begin with the first equation: \( x - 2y = 6 \). We can express \( x \) in terms of \( y \):
\( x = 2y + 6 \) ... (i)

Now, we find some points for this equation:
When \( y = 0 \), \( x = 2(0) + 6 = 6 \). So, the point is (6, 0).
When \( y = 1 \), \( x = 2(1) + 6 = 8 \). So, the point is (8, 1).
When \( y = -1 \), \( x = 2(-1) + 6 = 4 \). So, the point is (4, -1).
When \( y = 2 \), \( x = 2(2) + 6 = 10 \). So, the point is (10, 2).

Table for equation (i):

Next, consider the second equation: \( 3x - 6y = 0 \). We can express \( x \) in terms of \( y \):
\( 3x = 6y \)
\( \implies x = 2y \) ... (ii)

Now, we find some points for this equation:
When \( y = 0 \), \( x = 2(0) = 0 \). So, the point is (0, 0).
When \( y = 1 \), \( x = 2(1) = 2 \). So, the point is (2, 1).
When \( y = -1 \), \( x = 2(-1) = -2 \). So, the point is (-2, -1).
When \( y = 2 \), \( x = 2(2) = 4 \). So, the point is (4, 2).

Table for equation (ii):

After plotting these points and drawing the lines on a graph paper, we find that the lines are parallel to each other. Parallel lines never intersect, which means there is no common solution. This indicates that the system of equations is inconsistent. If you simplify the second equation by dividing by 3, you get \( x - 2y = 0 \), which clearly shows it's parallel to \( x - 2y = 6 \).
Therefore, this system of equations has no solution.
In simple words: Find points for each line and draw them. When the lines run side-by-side and never cross, it means there is no answer that works for both equations.

🎯 Exam Tip: If the coefficients of x and y in two linear equations are proportional but the constant terms are not, the lines will be parallel and have no solution. For example, \( Ax + By = C \) and \( kAx + kBy = D \) with \( C \neq D/k \).

Question 4. Solve the following pair of equations graphically: x + y = 4; 2x – 3y = 3
Answer:
Let's consider the first equation: \( x + y = 4 \). We can write \( y \) in terms of \( x \):
\( y = 4 - x \) ... (i)

Now, we find some points for this equation:
When \( x = 0 \), \( y = 4 - 0 = 4 \). So, the point is (0, 4).
When \( x = 1 \), \( y = 4 - 1 = 3 \). So, the point is (1, 3).
When \( x = 2 \), \( y = 4 - 2 = 2 \). So, the point is (2, 2).
When \( x = 3 \), \( y = 4 - 3 = 1 \). So, the point is (3, 1).
When \( x = 4 \), \( y = 4 - 4 = 0 \). So, the point is (4, 0).

Table for equation (i):

Next, consider the second equation: \( 2x - 3y = 3 \). We can express \( x \) in terms of \( y \):
\( 2x = 3y + 3 \)
\( \implies x = \frac{3y + 3}{2} \) ... (ii)

Now, we find some points for this equation:
When \( y = 1 \), \( x = \frac{3(1) + 3}{2} = \frac{6}{2} = 3 \). So, the point is (3, 1).
When \( y = 2 \), \( x = \frac{3(2) + 3}{2} = \frac{9}{2} = 4.5 \). So, the point is (4.5, 2).
When \( y = 0 \), \( x = \frac{3(0) + 3}{2} = \frac{3}{2} = 1.5 \). So, the point is (1.5, 0).

Table for equation (ii):

After plotting these points and drawing the lines on a graph paper, we observe that both lines intersect at the point (3, 1). This common point is the solution to the system of equations. The coordinate system helps pinpoint the exact solution.
Therefore, the solution to this pair of equations is \( x = 3 \) and \( y = 1 \).
In simple words: Get a few points for each line and draw them. The place where the lines cross over each other is the answer, which is (3, 1).

🎯 Exam Tip: Always verify your graphical solution by substituting the intersection point's coordinates back into both original equations to ensure they are satisfied. This confirms the accuracy of your plot.

Question 5. Solve the following pair of equations graphically: 2x – 3y + 13 = 0; 3x - 2y + 12 = 0
Answer:
Let's start with the first equation: \( 2x - 3y + 13 = 0 \). We can express \( x \) in terms of \( y \):
\( 2x = 3y - 13 \)
\( \implies x = \frac{3y - 13}{2} \) ... (i)

Now, we find some points for this equation:
When \( y = 1 \), \( x = \frac{3(1) - 13}{2} = \frac{-10}{2} = -5 \). So, the point is (-5, 1).
When \( y = 0 \), \( x = \frac{3(0) - 13}{2} = \frac{-13}{2} = -6.5 \). So, the point is (-6.5, 0).
When \( y = -1 \), \( x = \frac{3(-1) - 13}{2} = \frac{-16}{2} = -8 \). So, the point is (-8, -1).
When \( y = 2 \), \( x = \frac{3(2) - 13}{2} = \frac{-7}{2} = -3.5 \). So, the point is (-3.5, 2).

Table for equation (i):

Next, consider the second equation: \( 3x - 2y + 12 = 0 \). We can express \( y \) in terms of \( x \):
\( 2y = 3x + 12 \)
\( \implies y = \frac{3x + 12}{2} \) ... (ii)

Now, we find some points for this equation:
When \( x = 0 \), \( y = \frac{3(0) + 12}{2} = \frac{12}{2} = 6 \). So, the point is (0, 6).
When \( x = -1 \), \( y = \frac{3(-1) + 12}{2} = \frac{9}{2} = 4.5 \). So, the point is (-1, 4.5).
When \( x = -2 \), \( y = \frac{3(-2) + 12}{2} = \frac{6}{2} = 3 \). So, the point is (-2, 3).

Table for equation (ii):

After plotting these points and drawing the lines on a graph paper, we find that the lines intersect at the point (-2, 3). This point is the unique solution to this system of equations. Using a consistent scale on both axes is crucial for accurate results.
Therefore, the required solution is \( x = -2 \) and \( y = 3 \).
In simple words: Find some points for each equation and plot them. The point where the two lines meet, which is (-2, 3), is the answer.

🎯 Exam Tip: When dealing with negative coordinates, pay extra attention to plotting them correctly in the respective quadrants. Double-check your calculations for negative values.

Question 6. Solve the following pair of equations graphically: 3x – 4y = 1; -2x + \( \frac{8}{3} \)y = 5
Answer:
Let's consider the first equation: \( 3x - 4y = 1 \). We can express \( y \) in terms of \( x \):
\( 4y = 3x - 1 \)
\( \implies y = \frac{3x - 1}{4} \) ... (i)

Now, we find some points for this equation:
When \( x = 0 \), \( y = \frac{3(0) - 1}{4} = \frac{-1}{4} = -0.25 \). So, the point is (0, -0.25).
When \( x = -1 \), \( y = \frac{3(-1) - 1}{4} = \frac{-4}{4} = -1 \). So, the point is (-1, -1).
When \( x = 3 \), \( y = \frac{3(3) - 1}{4} = \frac{8}{4} = 2 \). So, the point is (3, 2).

Table for equation (i):

Next, consider the second equation: \( -2x + \frac{8}{3}y = 5 \). To simplify, we can multiply the entire equation by 3:
\( -6x + 8y = 15 \)
Now, we express \( y \) in terms of \( x \):
\( 8y = 15 + 6x \)
\( \implies y = \frac{15 + 6x}{8} \) ... (ii)

Now, we find some points for this equation:
When \( x = 0 \), \( y = \frac{15 + 6(0)}{8} = \frac{15}{8} = 1.875 \). So, the point is (0, 1.875).
When \( y = 0 \), \( 0 = \frac{15 + 6x}{8} \)
\( \implies 15 + 6x = 0 \)
\( \implies 6x = -15 \)
\( \implies x = \frac{-15}{6} = -2.5 \). So, the point is (-2.5, 0).

Table for equation (ii):

After plotting these points and drawing the lines on a graph paper, we observe that the two lines are parallel to each other. This means they will never intersect, and thus the system of equations has no solution. Such systems are sometimes called inconsistent systems, as no common solution satisfies both conditions. One way to check this is to see that the slopes of the lines are the same but their y-intercepts are different.
Therefore, the system of equations has no solution.
In simple words: Get a few points for each line and draw them. If the lines end up being parallel (never touching), it means there's no answer that works for both equations.

🎯 Exam Tip: Fractional coefficients can be tricky; always simplify the equation by multiplying by the LCM of the denominators to work with whole numbers, reducing chances of calculation errors.

Question 7. Solve the following pair of equations graphically: 4x + y - 10 = 0; \( \frac{x}{2} \) + y = -4
Answer:
Let's take the first equation: \( 4x + y - 10 = 0 \). We can express \( y \) in terms of \( x \):
\( y = -4x + 10 \)
\( \implies y = 10 - 4x \) ... (i)

Now, we find some points for this equation:
When \( x = 1 \), \( y = 10 - 4(1) = 6 \). So, the point is (1, 6).
When \( x = 2 \), \( y = 10 - 4(2) = 2 \). So, the point is (2, 2).
When \( x = 3 \), \( y = 10 - 4(3) = -2 \). So, the point is (3, -2).
When \( x = 4 \), \( y = 10 - 4(4) = -6 \). So, the point is (4, -6).

Table for equation (i):

Next, consider the second equation: \( \frac{x}{2} + y = -4 \). To remove the fraction, multiply by 2:
\( x + 2y = -8 \)
Now, we express \( y \) in terms of \( x \):
\( 2y = -8 - x \)
\( \implies y = \frac{-8 - x}{2} \) ... (ii)

Now, we find some points for this equation:
When \( x = -2 \), \( y = \frac{-8 - (-2)}{2} = \frac{-6}{2} = -3 \). So, the point is (-2, -3).
When \( x = -4 \), \( y = \frac{-8 - (-4)}{2} = \frac{-4}{2} = -2 \). So, the point is (-4, -2).
When \( x = -6 \), \( y = \frac{-8 - (-6)}{2} = \frac{-2}{2} = -1 \). So, the point is (-6, -1).
When \( x = 4 \), \( y = \frac{-8 - 4}{2} = \frac{-12}{2} = -6 \). So, the point is (4, -6).

Table for equation (ii):

After plotting all these points and drawing the lines on a graph paper, we find that the point of intersection of both lines is (4, -6). This means the values \( x=4 \) and \( y=-6 \) satisfy both equations simultaneously. Graphical solutions provide a clear visual of where two linear equations meet.
Therefore, the required solution is \( x = 4 \) and \( y = -6 \).
In simple words: Find points for both equations and draw them as lines. The point where they cross is the answer. Here, the lines cross at (4, -6).

🎯 Exam Tip: When an equation has fractions, multiply all terms by the least common multiple of the denominators to eliminate fractions and make calculations simpler.

Question 8. Solve the following pair of equations graphically: 0.3x + 0.4y = 3.2; 0.6x + 0.8y = 2.4
Answer:
Let's convert the equations with decimals into equations with whole numbers by multiplying each by 10.
The first equation: \( 0.3x + 0.4y = 3.2 \)
\( \implies 3x + 4y = 32 \) ... (i)

From equation (i), we express \( y \) in terms of \( x \):
\( 4y = 32 - 3x \)
\( \implies y = \frac{32 - 3x}{4} \)

Now, we find some points for this equation:
When \( x = 0 \), \( y = \frac{32 - 3(0)}{4} = \frac{32}{4} = 8 \). So, the point is (0, 8).
When \( x = 2 \), \( y = \frac{32 - 3(2)}{4} = \frac{32 - 6}{4} = \frac{26}{4} = 6.5 \). So, the point is (2, 6.5).
When \( x = 4 \), \( y = \frac{32 - 3(4)}{4} = \frac{32 - 12}{4} = \frac{20}{4} = 5 \). So, the point is (4, 5).

Table for equation (i):

Now, for the second equation: \( 0.6x + 0.8y = 2.4 \)
\( \implies 6x + 8y = 24 \) ... (ii)

From equation (ii), we express \( y \) in terms of \( x \):
\( 8y = 24 - 6x \)
\( \implies y = \frac{24 - 6x}{8} \)

Now, we find some points for this equation:
When \( x = 0 \), \( y = \frac{24 - 6(0)}{8} = \frac{24}{8} = 3 \). So, the point is (0, 3).
When \( x = 2 \), \( y = \frac{24 - 6(2)}{8} = \frac{24 - 12}{8} = \frac{12}{8} = 1.5 \). So, the point is (2, 1.5).
When \( x = 4 \), \( y = \frac{24 - 6(4)}{8} = \frac{24 - 24}{8} = \frac{0}{8} = 0 \). So, the point is (4, 0).

Table for equation (ii):

After plotting these points and drawing the lines on a graph paper, we find that the two lines are parallel to each other. This means they do not intersect at any point, so the system of equations has no solution. Multiplying by 10 makes the calculation easier and avoids decimal errors. These equations represent parallel lines with different y-intercepts.
Therefore, the system of equations has no solution.
In simple words: First, remove decimals by multiplying. Then find points for each line and draw them. If the lines are parallel and never meet, there is no answer that works for both.

🎯 Exam Tip: When solving equations with decimals, always convert them to integers first by multiplying by an appropriate power of 10. This minimizes calculation errors and simplifies plotting.

Question 9. Solve the following pair of equations graphically: 2x + 3y = 8; 4x - \( \frac{3}{2} \)y = 1
Answer:
Let's take the first equation: \( 2x + 3y = 8 \). We can express \( y \) in terms of \( x \):
\( 3y = 8 - 2x \)
\( \implies y = \frac{8 - 2x}{3} \) ... (i)

Now, we find some points for this equation:
When \( x = 1 \), \( y = \frac{8 - 2(1)}{3} = \frac{6}{3} = 2 \). So, the point is (1, 2).
When \( x = -2 \), \( y = \frac{8 - 2(-2)}{3} = \frac{8 + 4}{3} = \frac{12}{3} = 4 \). So, the point is (-2, 4).
When \( x = -5 \), \( y = \frac{8 - 2(-5)}{3} = \frac{8 + 10}{3} = \frac{18}{3} = 6 \). So, the point is (-5, 6).

Table for equation (i):

Next, consider the second equation: \( 4x - \frac{3}{2}y = 1 \). To remove the fraction, multiply the entire equation by 2:
\( 8x - 3y = 2 \)
Now, we express \( y \) in terms of \( x \):
\( 3y = 8x - 2 \)
\( \implies y = \frac{8x - 2}{3} \) ... (ii)

Now, we find some points for this equation:
When \( x = 1 \), \( y = \frac{8(1) - 2}{3} = \frac{6}{3} = 2 \). So, the point is (1, 2).
When \( x = -2 \), \( y = \frac{8(-2) - 2}{3} = \frac{-16 - 2}{3} = \frac{-18}{3} = -6 \). So, the point is (-2, -6).

Table for equation (ii):

After plotting the points from both tables and drawing the lines on a graph paper, we find that the point of intersection of both lines is (1, 2). This means that \( x=1 \) and \( y=2 \) is the solution that satisfies both original equations. Always ensure your graph paper has accurate markings and consistent scaling for the best results.
Therefore, the required solution is \( x = 1 \) and \( y = 2 \).
In simple words: Find points for each equation and draw them. The place where the two lines cross, which is (1, 2), is the answer.

🎯 Exam Tip: Pay close attention to fractions in equations. It's often helpful to clear fractions by multiplying by the least common multiple (LCM) before finding points, as this reduces calculation complexity.

Question 9. \( 2x + 3y = 8 \); \( 4x - \frac{3}{2}y = 1 \)
Answer: First, let's rearrange the given equations to make them easier to plot. For the first equation, \( 2x + 3y = 8 \):
\( 3y = 8 - 2x \)
\( y = \frac{8 - 2x}{3} \)
Let's find some points for this line:
When \( x = 1 \), \( y = \frac{8 - 2(1)}{3} = \frac{6}{3} = 2 \). So, point (1, 2).
When \( x = -2 \), \( y = \frac{8 - 2(-2)}{3} = \frac{8 + 4}{3} = \frac{12}{3} = 4 \). So, point (-2, 4).
When \( x = -5 \), \( y = \frac{8 - 2(-5)}{3} = \frac{8 + 10}{3} = \frac{18}{3} = 6 \). So, point (-5, 6).

For the second equation, \( 4x - \frac{3}{2}y = 1 \):
Multiply by 2 to clear the fraction:
\( 8x - 3y = 2 \)
\( 3y = 8x - 2 \)
\( y = \frac{8x - 2}{3} \)
Let's find some points for this line:
When \( x = 1 \), \( y = \frac{8(1) - 2}{3} = \frac{6}{3} = 2 \). So, point (1, 2).
When \( x = -2 \), \( y = \frac{8(-2) - 2}{3} = \frac{-16 - 2}{3} = \frac{-18}{3} = -6 \). So, point (-2, -6).
When \( x = 4 \), \( y = \frac{8(4) - 2}{3} = \frac{32 - 2}{3} = \frac{30}{3} = 10 \). So, point (4, 10).

When we plot these points and draw the lines on a graph paper, we observe that both lines cross each other at the point (1, 2). This means that (1, 2) is the solution where both equations are true at the same time. Graphing helps us visualize where the lines meet, which gives us the solution.
In simple words: We found points for each equation and plotted them. The two lines crossed at the point (1, 2), which is the answer that works for both equations.

🎯 Exam Tip: Always check your intersection point by plugging the x and y values back into both original equations to ensure they satisfy both equations.

Question 10. \( 3x - y = 2 \); \( 6x - 2y = 4 \)
Answer: Let's consider the first equation, \( 3x - y = 2 \). To make plotting easier, we can express \( y \) in terms of \( x \):
\( y = 3x - 2 \)
Now, let's find some points for this line:
When \( x = 0 \), \( y = 3(0) - 2 = -2 \). So, point (0, -2).
When \( x = 1 \), \( y = 3(1) - 2 = 1 \). So, point (1, 1).
When \( x = 2 \), \( y = 3(2) - 2 = 4 \). So, point (2, 4).

Next, let's consider the second equation, \( 6x - 2y = 4 \). We can also express \( y \) in terms of \( x \):
\( 2y = 6x - 4 \)
\( y = \frac{6x - 4}{2} \)
\( y = 3x - 2 \)
Notice that this is the exact same equation as the first one. This means that both equations represent the same line. When plotted, the lines will completely overlap, showing that every point on the line is a solution.

Plotting these points and drawing the lines shows that they lie directly on top of each other. This means they are the same line. When two lines are exactly the same, they have endlessly many points in common. So, there are infinitely many solutions to this system of equations.
In simple words: Both equations are actually for the same line. When you graph them, they sit right on top of each other, meaning there are countless solutions.

🎯 Exam Tip: If one equation is a simple multiple of the other (like \( 6x - 2y = 4 \) is \( 2 \times (3x - y = 2) \)), the lines will be identical, resulting in infinitely many solutions.

Question 11. \( 3x + 2y = 0 \); \( 2x + y = -1 \)
Answer: Let's take the first equation, \( 3x + 2y = 0 \). We can express \( y \) in terms of \( x \):
\( 2y = -3x \)
\( y = \frac{-3x}{2} \)
Let's find some points for this line:
When \( x = 2 \), \( y = \frac{-3(2)}{2} = -3 \). So, point (2, -3).
When \( x = 4 \), \( y = \frac{-3(4)}{2} = -6 \). So, point (4, -6).
When \( x = -2 \), \( y = \frac{-3(-2)}{2} = 3 \). So, point (-2, 3).

Now, let's take the second equation, \( 2x + y = -1 \). We can express \( y \) in terms of \( x \):
\( y = -1 - 2x \)
Let's find some points for this line:
When \( x = 1 \), \( y = -1 - 2(1) = -3 \). So, point (1, -3).
When \( x = -1 \), \( y = -1 - 2(-1) = -1 + 2 = 1 \). So, point (-1, 1).
When \( x = -2 \), \( y = -1 - 2(-2) = -1 + 4 = 3 \). So, point (-2, 3).

After plotting these points and drawing the lines for both equations on a graph paper, we can see where they intersect. The point where the two lines cross is (-2, 3). This point is the solution that satisfies both equations at the same time. This method is effective for visually determining the common solution.
In simple words: We found points for each equation and drew their lines. The spot where the two lines met on the graph was (-2, 3), which is our final answer.

🎯 Exam Tip: When graphing, use a ruler and sharp pencil to draw straight lines accurately. Even a small error in drawing can lead to an incorrect intersection point.