Get the most accurate MSBSHSE Solutions for Class 10 Maths Chapter 1 Linear Equations in Two Variables Set 1.2 here. Updated for the 2026-27 academic session, these solutions are based on the latest MSBSHSE textbooks for Class 10 Maths. Our expert-created answers for Class 10 Maths are available for free download in PDF format.
Detailed Chapter 1 Linear Equations in Two Variables Set 1.2 MSBSHSE Solutions for Class 10 Maths
For Class 10 students, solving MSBSHSE textbook questions is the most effective way to build a strong conceptual foundation. Our Class 10 Maths solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 1 Linear Equations in Two Variables Set 1.2 solutions will improve your exam performance.
Class 10 Maths Chapter 1 Linear Equations in Two Variables Set 1.2 MSBSHSE Solutions PDF
Question 1. Complete the following table to draw graph of the equations.
(i) \( x + y = 3 \)
| \( x \) | 3 | ||
|---|---|---|---|
| \( y \) | 5 | 3 | |
| \( (x, y) \) | (3, 0) | (0, 3) |
(ii) \( x - y = 4 \)
| \( x \) | -1 | ||
|---|---|---|---|
| \( y \) | 0 | -4 | |
| \( (x, y) \) | (0, -4) |
Answer:
(i) \( x + y = 3 \)
For \( x = 3 \):
\( 3 + y = 3 \)
\( \implies y = 0 \)
For \( y = 5 \):
\( x + 5 = 3 \)
\( \implies x = -2 \)
For \( y = 3 \):
\( x + 3 = 3 \)
\( \implies x = 0 \)
The completed table is:
| \( x \) | 3 | -2 | 0 |
|---|---|---|---|
| \( y \) | 0 | 5 | 3 |
| \( (x, y) \) | (3, 0) | (-2, 5) | (0, 3) |
(ii) \( x - y = 4 \)
For \( y = 0 \):
\( x - 0 = 4 \)
\( \implies x = 4 \)
For \( x = -1 \):
\( -1 - y = 4 \)
\( \implies -y = 5 \)
\( \implies y = -5 \)
For \( y = -4 \):
\( x - (-4) = 4 \)
\( \implies x + 4 = 4 \)
\( \implies x = 0 \)
The completed table is:
| \( x \) | 4 | -1 | 0 |
|---|---|---|---|
| \( y \) | 0 | -5 | -4 |
| \( (x, y) \) | (4, 0) | (-1, -5) | (0, -4) |
In simple words: To complete the tables, we substitute the given values of \( x \) or \( y \) into each equation to find the missing coordinate, which gives us the points needed to draw the graphs.
🎯 Exam Tip: Double-check your calculations by substituting both coordinates back into the original equation to ensure they satisfy it before plotting them on the graph.
Question 1. Complete the following activity to draw the graphs of the equations.
(i) \( x + y = 3 \)
(ii) \( x - y = 4 \)
Answer:
(i) \( x + y = 3 \)
| \( x \) | 3 | -2 | 0 |
|---|---|---|---|
| \( y \) | 0 | 5 | 3 |
| \( (x, y) \) | (3, 0) | (-2, 5) | (0, 3) |
(ii) \( x - y = 4 \)
| \( x \) | 4 | -1 | 0 |
|---|---|---|---|
| \( y \) | 0 | -5 | -4 |
| \( (x, y) \) | (4, 0) | (-1, -5) | (0, -4) |
In simple words: To complete the table, substitute the given value of \( x \) or \( y \) into the equation to find the missing coordinate, then write them together as an ordered pair \( (x, y) \).
🎯 Exam Tip: Double-check your calculations by substituting both coordinates of the ordered pair back into the original equation to ensure they satisfy it.
Question 2. Solve the following simultaneous equations graphically.
(i) \( x + y = 6 \); \( x - y = 4 \)
(ii) \( x + y = 5 \); \( x - y = 3 \)
(iii) \( x + y = 0 \); \( 2x - y = 9 \)
(iv) \( 3x - y = 2 \); \( 2x - y = 3 \)
(v) \( 3x - 4y = -7 \); \( 5x - 2y = 0 \)
(vi) \( 2x - 3y = 4 \); \( 3y - x = 4 \)
Answer:
(i) \( x + y = 6 \); \( x - y = 4 \)
For \( x + y = 6 \)
\( \implies y = 6 - x \):
| \( x \) | 0 | 1 | 2 | 6 |
|---|---|---|---|---|
| \( y \) | 6 | 5 | 4 | 0 |
| \( (x, y) \) | (0, 6) | (1, 5) | (2, 4) | (6, 0) |
For \( x - y = 4 \)
\( \implies y = x - 4 \):
| \( x \) | 0 | 1 | 4 | 5 |
|---|---|---|---|---|
| \( y \) | -4 | -3 | 0 | 1 |
| \( (x, y) \) | (0, -4) | (1, -3) | (4, 0) | (5, 1) |
Plot the points and draw the lines. The two lines intersect at the point \( (5, 1) \).
\( \implies \) The solution of the given simultaneous equations is \( x = 5, y = 1 \).
(ii) \( x + y = 5 \); \( x - y = 3 \)
For \( x + y = 5 \)
\( \implies y = 5 - x \):
| \( x \) | 0 | 1 | 4 | 5 |
|---|---|---|---|---|
| \( y \) | 5 | 4 | 1 | 0 |
| \( (x, y) \) | (0, 5) | (1, 4) | (4, 1) | (5, 0) |
For \( x - y = 3 \)
\( \implies y = x - 3 \):
| \( x \) | 0 | 3 | 4 | 5 |
|---|---|---|---|---|
| \( y \) | -3 | 0 | 1 | 2 |
| \( (x, y) \) | (0, -3) | (3, 0) | (4, 1) | (5, 2) |
The two lines intersect at the point \( (4, 1) \).
\( \implies \) The solution of the given simultaneous equations is \( x = 4, y = 1 \).
(iii) \( x + y = 0 \); \( 2x - y = 9 \)
For \( x + y = 0 \)
\( \implies y = -x \):
| \( x \) | 0 | 1 | 3 | -3 |
|---|---|---|---|---|
| \( y \) | 0 | -1 | -3 | 3 |
| \( (x, y) \) | (0, 0) | (1, -1) | (3, -3) | (-3, 3) |
For \( 2x - y = 9 \)
\( \implies y = 2x - 9 \):
| \( x \) | 3 | 4 | 5 | 2 |
|---|---|---|---|---|
| \( y \) | -3 | -1 | 1 | -5 |
| \( (x, y) \) | (3, -3) | (4, -1) | (5, 1) | (2, -5) |
The two lines intersect at the point \( (3, -3) \).
\( \implies \) The solution of the given simultaneous equations is \( x = 3, y = -3 \).
(iv) \( 3x - y = 2 \); \( 2x - y = 3 \)
For \( 3x - y = 2 \)
\( \implies y = 3x - 2 \):
| \( x \) | 0 | 1 | 2 | -1 |
|---|---|---|---|---|
| \( y \) | -2 | 1 | 4 | -5 |
| \( (x, y) \) | (0, -2) | (1, 1) | (2, 4) | (-1, -5) |
For \( 2x - y = 3 \)
\( \implies y = 2x - 3 \):
| \( x \) | 0 | 1 | 2 | -1 |
|---|---|---|---|---|
| \( y \) | -3 | -1 | 1 | -5 |
| \( (x, y) \) | (0, -3) | (1, -1) | (2, 1) | (-1, -5) |
The two lines intersect at the point \( (-1, -5) \).
\( \implies \) The solution of the given simultaneous equations is \( x = -1, y = -5 \).
(v) \( 3x - 4y = -7 \); \( 5x - 2y = 0 \)
For \( 3x - 4y = -7 \)
\( \implies y = \frac{3x + 7}{4} \):
| \( x \) | -1 | 3 | -5 |
Question 2. Solve the following simultaneous equations graphically.
For \( x - y = 4 \):
Plot the points and draw the lines on a graph paper. The two lines intersect at point (5, 1). \( \therefore x = 5 \) and \( y = 1 \) is the solution of the simultaneous equations \( x + y = 6 \) and \( x - y = 4 \). (ii) \( x + y = 5 \) and \( x - y = 3 \) The given simultaneous equations are: \( x + y = 5 \) \( \therefore y = 5 - x \)
And, \( x - y = 3 \) \( \dots y = x - 3 \)
Plot these points on the same coordinate plane. The two lines intersect at point (4, 1). \( \therefore x = 4 \) and \( y = 1 \) is the solution of the simultaneous equations \( x + y = 5 \) and \( x - y = 3 \). In simple words: To solve equations using a graph, we find a few points for each line, draw them on graph paper, and find where they cross. The crossing point is the answer that works for both equations. 🎯 Exam Tip: Always write the scale on the top right corner of your graph paper (e.g., Scale: On both axes 1 cm = 1 unit) and clearly label the intersection point to secure full marks. Question 2(ii). Solve the following simultaneous equations graphically: \( x+y=5 \) and \( x-y=3 \). 🎯 Exam Tip: Always write the scale on the top right corner of the graph paper and clearly label the coordinates of the intersection point to secure full marks.
Question 2(iii). Solve the following simultaneous equations graphically: \( x+y=0 \) and \( 2x-y=9 \).
\( 2x-y = 9 \) \( \therefore y = 2x-9 \)
In simple words: To solve these equations graphically, we first find four coordinate points for each equation by choosing values for x and calculating y. Then, we plot these points on a graph to draw the lines. 🎯 Exam Tip: Choose simple integer values for x (like 0, 1, 2) to make calculating y easier and to avoid plotting fractional coordinates on the graph.
Question 1. Solve the simultaneous equations \(x + y = 0\) and \(2x - y = 9\) graphically. 🎯 Exam Tip: Always write the scale of the graph (e.g., Scale: On both axes 1 cm = 1 unit) in the top right corner to avoid losing marks.
Question 2. (iv) Solve the following simultaneous equations graphically: \(3x - y = 2\) and \(2x - y = 3\).
\(2x - y = 3\) \(\dots y = 2x - 3\)
🎯 Exam Tip: Choose integer values for \(x\) that result in small, whole-number values for \(y\) so that they are easy to plot accurately on your graph paper. Question 2. (iv) Solve the following simultaneous equations graphically: \(3x - y = 2\) and \(2x - y = 3\) 🎯 Exam Tip: Always write the coordinates of the intersection point clearly on the graph and state the final solution explicitly with variables \(x\) and \(y\).
Question 2. (v) Solve the following simultaneous equations graphically: \(3x - 4y = -7\) and \(5x - 2y = 0\)
Table of values for \(5x - 2y = 0\):
In simple words: We rearrange each equation to express \(y\) in terms of \(x\). Then, we choose different values for \(x\) to find the corresponding values of \(y\), which gives us points to plot on the graph. 🎯 Exam Tip: Choose integer values for \(x\) that result in whole numbers or simple decimals (like .5) for \(y\) to make plotting on the graph paper much easier and more accurate.
Question 2. (v) Solve the following simultaneous equations graphically: \(3x - 4y = -7\), \(5x - 2y = 0\) 🎯 Exam Tip: Always write the coordinates of the intersection point clearly and state the final values of x and y to secure full marks.
Question 2. (vi) Solve the following simultaneous equations graphically: \(2x - 3y = 4\), \(3y - x = 4\)
\(3y - x = 4\) \(\therefore 3y = x + 4\) \(\therefore y = \frac{x + 4}{3}\)
In simple words: We find four points for each equation by choosing values for x and calculating y. Then we plot these points on a graph to draw two straight lines. 🎯 Exam Tip: Choose integer values for x that result in integer values for y to make plotting on the graph paper much easier and more accurate.
Question 1. Solve the following simultaneous equations by graphical method. Complete the following tables to get ordered pairs.
For equation \(5x - 3y = 1\):
i. and ii. Plotting the points and drawing the graphs: Plot the points \((0, -1)\), \((1, 0)\), \((3, 2)\), and \((-2, -3)\) on the graph paper and draw a straight line passing through them. This line represents the equation \(x - y = 1\). Next, plot the points \((2, 3)\), \((5, 8)\), \((-1, -2)\), and \((-4, -7)\) on the same coordinate plane and draw a straight line passing through them. This line represents the equation \(5x - 3y = 1\). iii. Finding the point of intersection: The two lines intersect each other at the point \((-1, -2)\). This intersection point represents the unique set of values that satisfies both equations simultaneously. Therefore, the solution of the simultaneous equations is \(x = -1, y = -2\), which can also be written as \((x, y) = (-1, -2)\). In simple words: To solve these equations using a graph, we find a few points for each equation and draw them as straight lines. The place where these two lines cross each other is the final answer, which is \(x = -1\) and \(y = -2\). 🎯 Exam Tip: Always write the scale on the top right corner of your graph paper (e.g., Scale: On both axes 1 cm = 1 unit) and clearly label the coordinates of the intersection point to secure full marks. Question. Solve the following simultaneous equations graphically: \( x - y = 1 \) and \( 5x - 3y = 1 \)
Table for \( 5x - 3y = 1 \):
Plot the points on a graph paper and draw a line passing through them for each equation. The two lines intersect at point \( (-1, -2) \). \( \therefore (x, y) = (-1, -2) \) is the solution of the given simultaneous equations. In simple words: To solve equations using a graph, we find points for both lines, draw them, and find where they cross. The crossing point is the answer. 🎯 Exam Tip: Always choose simple integer values for x to find integer values of y, making it much easier to plot accurately on the graph paper.
Question 1. Solve the above equations by method of elimination. Check your solution with the graphical solution. 🎯 Exam Tip: When using the elimination method, always label your equations clearly as (I), (II), etc., to make your steps easy for the examiner to follow.
Question 1. Solve the simultaneous equations \( x - y = 1 \) and \( 5x - 3y = 1 \) by the elimination method. 🎯 Exam Tip: Always write down the final solution as an ordered pair \( (x, y) \) at the end of your answer to secure full marks.
Question 2. The following table contains the values of x and y co-ordinates for ordered pairs to draw the graph of \( 5x - 3y = 1 \).
(i) Is it easy to plot these points? (ii) Which precaution is to be taken to find ordered pairs so that plotting of points becomes easy? Answer: (i) No, it is not easy to plot these points because most of the coordinates are fractional values which are difficult to mark precisely on standard graph paper. (ii) To make plotting easy, we should choose integer values for both x and y. We can achieve this by selecting values of x that make the numerator a multiple of the coefficient of y, resulting in clean integer coordinates. In simple words: Fractions are hard to locate accurately on a graph. We should choose whole numbers for our coordinates so that drawing the line is simple and precise. 🎯 Exam Tip: When creating your own table of values, try to find at least three integer coordinate pairs to ensure your plotted line is perfectly straight and easy to draw. Here, \( -\frac{1}{3} = -0.33... \), \( \frac{4}{3} = 1.33... \), \( -\frac{11}{3} = -3.66... \)
Question 3. To solve simultaneous equations \( x + 2y = 4 \); \( 3x + 6y = 12 \) graphically, following are the ordered pairs.
For \( 3x + 6y = 12 \):
Plotting the above ordered pairs, graph is drawn. Observe it and find answers of the following questions. In simple words: To solve these equations graphically, we find three sets of coordinates for each equation, plot them on a graph, and draw lines through them to see where they intersect. 🎯 Exam Tip: Always choose integer values for coordinates when plotting graphs to ensure high accuracy and avoid plotting fractional points which can lead to errors.
Question 1. Observe the graph and answer the following questions: 🎯 Exam Tip: To quickly check if two equations represent the same line, simplify them to their simplest form. If they are identical, they will have infinitely many solutions and their coefficient ratios will be equal.
Question 4. Draw graphs of \( x - 2y = 4 \), \( 2x - 4y = 12 \) on the same co-ordinate plane. Observe it. Think of the relation between the coefficients of x, coefficients of y and the constant terms and draw the inference. (Textbook pg. no. 10)
For equation \( 2x - 4y = 12 \): \( 2x - 4y = 12 \) \( \implies x - 2y = 6 \) ...[Dividing both sides by 2] \( \implies 2y = x - 6 \) \( \implies y = \frac{x - 6}{2} \)
Graph Plotting Details: • The line for \( x - 2y = 4 \) passes through the points \( (0, -2) \), \( (2, -1) \), \( (-2, -3) \), and \( (4, 0) \). • The line for \( 2x - 4y = 12 \) passes through the points \( (0, -3) \), \( (-2, -4) \), \( (2, -2) \), and \( (4, -1) \). • Both plotted lines run parallel to each other on the coordinate plane and do not intersect at any point. ii. Let us compare the ratios of the coefficients and constant terms: Ratio of coefficients of x: \( \frac{a_1}{a_2} = \frac{1}{2} \) Ratio of coefficients of y: \( \frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2} \) Ratio of constant terms: \( \frac{c_1}{c_2} = \frac{4}{12} = \frac{1}{3} \) Inference: Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \), the two lines are parallel to each other and do not intersect. Therefore, this system of simultaneous equations has no solution. In simple words: When we plot these two equations, they form two parallel lines that never cross each other. This happens because the ratios of their x and y coefficients are equal, but different from the ratio of their constant numbers, meaning they have no common solution. 🎯 Exam Tip: When drawing parallel lines, ensure your plotted points are highly accurate and clearly labeled on the graph. Always write the final inference showing the ratio comparison \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) to secure full marks. p>Ratio of coefficients of y = \( \frac{-2}{-4} = \frac{1}{2} \)
Ratio of constant terms = \( \frac{4}{12} = \frac{1}{3} \) \( \therefore \) Ratio of coefficients of x = ratio of coefficients of y \( \neq \) ratio of constant terms iii. If ratio of coefficients of x = ratio of coefficients of y \( \neq \) ratio of constant terms, then the graphs of the two equations will be parallel to each other. Condition of Consistency in Equations
🎯 Exam Tip: Memorize this consistency table thoroughly, as questions on identifying the number of solutions or the nature of graphical lines based on ratio comparisons are frequently asked in exams. |
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MSBSHSE Solutions Class 10 Maths Chapter 1 Linear Equations in Two Variables Set 1.2
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Detailed Explanations for Chapter 1 Linear Equations in Two Variables Set 1.2
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