All these RS Aggarwal class 10 solutions Chapter 3 Linear equations in two variables are solved by Byju's top ranked professors as per CBSE guidelines.

**Question 1:** **On a graph paper, draw a horizontal line X???OX and a vertical line YOY??? as the x-axis and the y-axis respectively.**

**2x + 3y = 2,??****x – 2y = 8**

**Solution:**

Given equations are 2x + 3y = 2 and x – 2y = 8

Graph of 2x + 3y = 2:

\( y = \frac{2(1 – x)}{3} \)Putting x = 1, we get y = 0

Putting x = -2, we get y = 2

Putting x = 4, we get y = -2

Hence, the table is:

x |
1 | -2 | 4 |

y |
0 | 2 | -2 |

Plot the points A (1, 0), B (-2, 2) and C (4, -2) on the graph paper. Join AB and AC to get the graph line BC. Extend it both ways.

Thus, line BC is the graph of 2x + 3y = 2.

Graph of x – 2y = 8:

\( y = \frac{x – 8}{2} \)Putting x = 2, we get y = -3

Putting x = 4, we get y = -2

Putting x = 0, we get y = -4

Hence, the table is:

x |
2 | 4 | 0 |

y |
-3 | -2 | -4 |

Now, on the same graph paper as above plot the points P (0, -4) and Q (2, -3). The point C (4, -2) has already been plotted. Join QC and extend it. Thus, the line PC is the graph of x – 2y = 8.

The two graph lines intersect at C (4, -2). Therefore, x = 4, y = -2 is the solution of given system of equations.

**Question 2:** **On a graph paper, draw a horizontal line X???OX and a vertical line YoY??? as the x-axis and the y-axis respectively.**

**3x + 2y = 4** **and 2x – 3y = 7**

**Solution:**

Given equations are 3x + 2y = 4 and 2x – 3y = 7

Graph of 3x + 2y = 4:

\( y = \frac{4 – 3x}{2} \)Putting x = 0, we get y = 2

Putting x = 2, we get y = -1

Putting x = -2, we get y = 5

Hence, the table is:

x |
0 | 2 | -2 |

y |
2 | -1 | 5 |

Plot the points A (0, 2), B (2, -1) and C (-2, 5) on the graph paper. Join AB and AC to get the graph line BC. Extend it both ways.

Thus, line BC is the graph of 3x + 2y = 4.

Graph of 2x – 3y = 7:

\( y = \frac{x – 8}{2} \)Putting x = 2, we get y = -1

Putting x = -1, we get y = -3

Putting x = 5, we get y = 1

Hence, the table is:

x |
2 | -1 | 5 |

y |
-1 | -3 | 1 |

Now, on the same graph paper as above plot the points P (-1, -3) and Q (5, 1). The point B (2, -1) has already been plotted. Join QB and extend it. Thus, the line PQ is the graph of 2x – 3y = 7.

The two graph lines intersect at B (2,-1). Therefore, x = 2, y = -1 is the solution of given system of equations.

**Question 3:** **On a graph paper, draw a horizontal line X???OX and a vertical line YoY??? as the x-axis and the y-axis respectively.**

**x – y + 1 = 0,**

**3x + 2y – 12 = 0**

**Solution:**

Given equations are x – y + 1 = 0 and 3x + 2y – 12 = 0

Graph of x – y + 1 = 0:

x – y + 1 = 0 \( \Rightarrow \) y = x + 1 ??????..(1)

Putting x = 0, we get y = 1

Putting x = -1, we get y = 0

Putting x = 2, we get y = 3

Hence, the table is:

x |
0 | -1 | 2 |

y |
1 | 0 | 3 |

Plot the points A (0, 1), B (-1, 0) and C (2, 3) on the graph paper. Join AB and AC to get the graph line BC. Extend it both ways.

Thus, line BC is the graph of x – y + 1 = 0.

Graph of 3x + 2y – 12 = 0:

\( y = \frac{12 – 3x}{2} \)Putting x = 0, we get y = 6

Putting x = 2, we get y = 3

Putting x = 4, we get y = 0

Hence, the table is:

x |
0 | 2 | 4 |

y |
6 | 3 | 0 |

Now, on the same graph paper as above plot the points P (0, 6) and Q (4, 0). The point C (2, 3) has already been plotted. Join QB and extend it. Thus, the line PQ is the graph of 3x + 2y – 12 = 0.

The two graph lines intersect at (2, 3). Therefore, x = 2, y = 3 is the solution of given system of equations.

**Question 4:** **On a graph paper, draw a horizontal line X???OX and a vertical line YoY??? as the x-axis and the y-axis respectively.**

**2x + 3y = 4,??****3x – y = -5**

**Solution:**

Given equations are 2x + 3y = 4 and 3x – y = -5

Graph of 2x + 3y = 4:

2x + 3y = 4 \( \Rightarrow y = \frac{4 – 2x}{3} \)

Putting x = -1, we get y = 2

Putting x = 2, we get y = 0

Putting x = 5, we get y = -2

Hence, the table is:

x |
-1 | 2 | 5 |

y |
2 | 0 | -2 |

Plot the points A (-1, 2), B (2, 0) and C (5, -2) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line BC is the graph of 2x + 3y = 4.

Graph of 3x – y = -5:

y = 3x + 5

Putting x = -1, we get y = -2

Putting x = 0, we get y = 5

Putting x = -2, we get y = -1

Hence, the table is:

x |
-1 | 0 | -2 |

y |
2 | 5 | -1 |

Now, on the same graph paper as above plot the points P(0, 5) and Q(-2, -1). The point A (-1, 2) has already been plotted. Join PA and amd QA to get line PQ. Thus, the line PQ is the graph of 3x – y = -5.

The two graph lines intersect at A (-1, 2). Therefore, x = -1, y = 2 is the solution of given system of equations.

**Question 5:**

**2x – 3y = 1,??****3x – 4y = 1 **

**Solution:**

Given equations are 2x – 3y = 1 and 3x – 4y = 1

Graph of 2x – 3y = 1:

2x – 3y = 1 \( \Rightarrow y = \frac{2x – 1}{3} \)

Putting x = -1, we get y = -1

Putting x = 2, we get y = 1

Putting x = 5, we get y = 3

Hence, the table is:

x |
-1 | 2 | 5 |

y |
-1 | 1 | 3 |

Plot the points A (-1, -1), B (2, 1) and C (5, 3) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 2x – 3y = 1.

Graph of 3x – 4y = 1:

3x – 4y = 1 \( \Rightarrow y = \frac{3x – 1}{4} \)

Putting x = -1, we get y = -1

Putting x = 3, we get y = 2

Putting x = 2, we get y = -4

Hence, the table is:

x |
-1 | 3 | -5 |

y |
-1 | 2 | -4 |

Now, on the same graph paper as above plot the points P (3, 2) and Q(-5, -4). The point A (-1, -1) has already been plotted. Join PA and QA. Thus, the line PQ is the graph of 3x – 4y = 1.

The two graph lines intersect at A (-1, -1). Therefore, x = -1, y = -1 is the solution of given system of equations.

**Question 6:**

**4x + 3y = 5,??****2y – x = 7 **

**Solution:**

Given equations are 4x + 3y = 5 and 2y – x = 7

Graph of 4x + 3y = 5:

4x + 3y = 5 \( \Rightarrow y = \frac{5 – 4x}{3} \)

Putting x = -1, we get y = 3

Putting x = 2, we get y = -1

Putting x = 5, we get y = -5

Hence, the table is:

x |
-1 | 2 | 5 |

y |
3 | -1 | -5 |

Plot the points A (-1, 3), B (2, -1) and C (5, -5) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 4x + 3y = 5.

Graph of 2y – x = 7:

2y – x = 7\( \Rightarrow y = \frac{7 + x}{2} \)

Putting x = -1, we get y = 3

Putting x = 3, we get y = 5

Putting x = -3, we get y = 2

Hence, the table is:

x |
-1 | 3 | -3 |

y |
3 | 5 | 2 |

Now, on the same graph paper as above plot the points P (3, 5) and Q (-3, 2). The point A (-1, 3) has already been plotted. Join PA and PQ. Thus, the line PQ is the graph of 2y – x = 7.

The two graph lines intersect at A(-1, -3). Therefore, x = -1, y = 3 is the solution of given system of equations.

**Question 7:**

**x + 2y + 2 = 0,??****3x + 2y + 2 = 0**

**Solution:**

Given equations are x + 2y + 2 = 0 and 3x + 2y + 2 = 0

Graph of x + 2y + 2 = 0:

x + 2y + 2 = 0 \( \Rightarrow y = \frac{-x -2}{2} \)

Putting x = -2, we get y = 0

Putting x = 0, we get y = -1

Putting x = 2, we get y = -2

Hence, the table is:

x |
-2 | 0 | 2 |

y |
0 | -1 | -2 |

Plot the points A (-2, 0), B (0, -1) and C (2, -2) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of x + 2y + 2 = 0.

Graph of 3x + 2y + 2 = 0:

3x + 2y + 2 = 0 \( \Rightarrow y = \frac{-3x + 2}{2} \)

Putting x = 0, we get y = 1

Putting x = 2, we get y = -2

Putting x = 4, we get y = -5

Hence, the table is:

x |
0 | 2 | 4 |

y |
1 | -2 | -5 |

Now, on the same graph paper as above plot the points P (0, 1) and Q (4, -5). The point C (2, -2) has already been plotted. Join PC and QC. Thus, the line PQ is the graph of 3x + 2y + 2 = 0.

The two graph lines intersect at C(2, -2). Therefore, x = 2, y = -2 is the solution of given system of equations.

**Question 8:**

**2x + 3y = 8,??****x – 2y + 3 = 0 **

**Solution:**

Given equations are 2x + 3y = 8 and x – 2y + 3 = 0

Graph of 2x + 3y = 8:

2x + 3y = 8 \( \Rightarrow y = \frac{8 – 2x}{3} \)

Putting x = 1, we get y = 2

Putting x = -5, we get y = 6

Putting x = 7, we get y = -2

Hence, the table is:

x | 1 | -5 | 7 |

y | 2 | 6 | -2 |

Plot the points A (1, 2), B (-5, 6) and C (7, -2) on the graph paper. Join AB and BC to get the graph line BC. Extend it both ways.

Thus, line AC is the graph of 2x + 3y = 8.

Graph of x – 2y + 3 = 0:

x – 2y + 3 = 0 \( \Rightarrow y = \frac{x + 3}{2} \)

Putting x = 1, we get y = 2

Putting x = 3, we get y = 3

Putting x = -3, we get y = 0

Hence, the table is:

x |
1 | 3 | -3 |

y |
2 | 3 | 0 |

Now, on the same graph paper as above plot the points P (3, 3) and Q (-3, 0). The point A (1, 2) has already been plotted. Join PA and QA. Thus, the line PQ is the graph of x – 2y + 3 = 0.

The two graph lines intersect at A (1, 2). Therefore, x = 1, y = 2 is the solution of given system of equations.

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