The linear equation in two variables contains two different variables in the equation which can be solved for and the generalized form can be written as ax+by=c, where x, y are variables and a,b,c are constants. The two variables must be in two separate equations, to obtain the solution of another variable.

An example of a linear equation in two variable are: 3x+5y= 6

Methods to solve linear equations in two variables are:

- Graphing Method
- Substitution Method
- Elimination Method

Check out the RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables available below:

## RS Aggarwal Class 9 Solutions Chapter 8

**Q.1:** **Draw the graph of each of the following equations:**

**(i) x = 5**

**(ii) y = -2**

**(iii) x + 6 = 0**

**(iv) x + 7 = 0**

**(v) y = 0**

**(vi) x = 0**

**Sol:**

**(i)** **The given equation is x = 5**

For the given equation, many solutions are possible. Out of them, we take, (5, 1) and (5, -1)

Thus, forming the table:

x |
5 | 5 |

y |
1 | -1 |

On plotting the points, on the graph sheet, the following graph is obtained, in which the line AB is the required graph:

**(ii) The given equation is y = -2**

For the given equation, many solutions are possible. Of them, we take, (1, -2) and (3, -2)

Thus, forming the table:

x |
1 | 3 |

y |
-2 | -2 |

On plotting the points, on the graph sheet, the following graph is obtained, in which the line CD is the required graph:

**(iii) The given equation is x + 6 = 0**

On solving, x = -6

For the given equation, many solutions are possible. Of them, we take, (-6, 1) and (-6, -5)

Thus, forming the table:

x |
-6 | -6 |

y |
1 | -5 |

On plotting the points, on the graph sheet, the following graph is obtained, in which the line EF is the required graph:

**(iv)** **The given equation is x + 7 = 0**

On solving, x = -7

For the given equation, many solutions are possible. Of them, we take, (-7, 6) and (-7, -1)

Thus, forming the table, we get:

x |
-7 | -7 |

y |
6 | -1 |

On plotting the points, on the graph sheet, the following graph is obtained, in which the line GH is the required graph:

**(v)** The given equation is x = 0, which on plotting using the above method represents the x-axis itself.

**(vi)** The given equation is y = 0, which on plotting using the above method represents the y-axis itself.

**Q. 2:** **Draw the graph of the equation y = 3x. From your graph, find the value of y when x = -2.**

**Sol:**

The given equation is y = 3x.

Assuming the values for any one variable, we can get the value for the other.

Thus, assuming the values of x as 0, 1 and 2 we get,

x = 0 â‡’ y = 3(0) = 0

x = 1 â‡’ y = 3(1) = 3

x = 2 â‡’ y = 3(2) = 6

**Forming the table:**

x |
0 | 1 | 2 |

y |
0 | 3 | 6 |

Plotting the points on the graph sheet, we obtain the required graph:

(NOTE: To draw a line, only two points is required. Here, three points are taken for better understanding)

Take a point A on the left of y-axis such that the distance of point A from the y-axis is 2 units.

Draw AB parallel to y-axis cutting the line y = 3x at B. Draw BC parallel to x-axis meeting y-axis at C.

Hence, y = OC = -6

**Q.3:** **Draw the graph of the equation x + 2y â€“ 3 = 0. Find your graph, find the value of y when x = 5.**

**Sol:**

The given equation is x + 2y â€“ 3 = 0.

Assuming the values for any one variable, we can get the value for the other.

Thus, assuming the values of y as -2 and 2 we get:

y = -2â‡’ x = 3 â€“ 2(-2) = 7

y = 2 â‡’ x = 3 â€“ 2(2) = -1

**Forming the table:**

x |
7 | -1 |

y |
-2 | 2 |

Plotting the points on the graph sheet, we obtain the required graph:

Take a point P on x-axis such that OP = 5

Draw PQ parallel to y-axis cutting the line x = 3 â€“ 2y at Q. Through Q, draw PN parallel to the x-axis at N. Hence, y = ON = -1

**Q. 4:** **Draw the graph of each of the following equations:**

**(i) y = x**

**(ii) y = -x**

**(iii) y + 3x = 0**

**(iv) 2x + 3y = 0**

**(v) 3x â€“ 2y = 0**

**(vi) 2x + y = 0**

**Sol:**

**(i)** The given equation is y = x

Thus, when x = 1, y = 1 and when x = -1, y = -1

Thus, the table looks as:

x |
1 | -1 |

y |
1 | -1 |

Plotting the points, the graph is obtained as follows:

**(ii) The given equation is y = -x.**

Thus, when x = 1, y = -1 and when x = -1, y = 1

Thus, the table looks as:

x |
1 | -1 |

y |
-1 | 1 |

Plotting the points, the graph is obtained as follows:

**(iii) The given equation is y + 3x = 0.**

Solving, y = -3x

Thus, when x = 1, y = -3 and when x = -1, y = 3

Thus, the table looks as:

x |
1 | -1 |

y |
-3 | 3 |

Plotting the points, the graph is obtained as follows:

**(iv) The given equation is 2x + 3y = 0.**

On solving, y=âˆ’23Ã—x

Thus, when x = 3, y = -2 and when x = -3, y = 2

Thus, the table looks as:

x |
3 | -3 |

y |
-2 | 2 |

Plotting the points, the graph is obtained as follows:

**(v) The given equation is 3x â€“ 2y = 0.**

On solving, y=32Ã—x

Thus, when x = 2, y = 3 and when x = -2, y = -3

Thus, the table looks as:

x |
2 | -2 |

y |
3 | -3 |

Plotting the points, the graph is obtained as follows:

**(vi) The given equation is 2x + y = 0.**

On solving, Â y = -2x.

Thus, when x = 1, y = -2 and when x = -1, y = 2

Thus, the table looks as:

x | 1 | -1 |

y | -2 | 2 |

Plotting the points, the graph is obtained as follows:

**Q.5:** **Draw the graph of the equation 2x â€“ 3y = 5. From the graph, find (i) the value of y when x = 4, and (ii) the value of x when y = 3**

**Sol:**

The given equation is 2x â€“ 3y = 5.

On solving, y=2xâ€“53

Thus, when x = 4, y = 1 and when x = -2, y = -3

Thus, the table looks as:

x |
4 | -2 |

y |
1 | -3 |

Plotting the points, the graph is obtained as follows:

**Q.6:** **Draw the graph of the equation 2x + y = 6. Find the coordinates of the point, where the graph cuts the x-axis.**

**Sol:**

The given equation is 2x + y = 6.

On solving, y = 6 â€“ 2x

Thus, when x = 1, y = 4 and when x = -1, y = 8

Thus, the table looks as:

x |
1 | -1 |

y |
4 | 8 |

Plotting the points, the graph is obtained as follows:

We find that the line cuts the x-axis at a point P which is at a distance of 3 units to the right of the y-axis.

Therefore, the coordinates of P are (3, 0).

**Q.7:** **Draw the graph of the equation 3x + 2y = 6. Find the coordinates of the point, where the graph cuts the y-axis.**

**Sol:**

The given equation is 3x + 2y = 6.

On solving, y=6âˆ’3x2

Thus, when x = 0, y = 0 and when x = -2, y = 6

Thus, the table looks as,

x |
0 | -2 |

y |
0 | 6 |

Plotting the points, the graph is obtained as follows:

We find that the given line cuts the y-axis at a point Z which is 3 units above the x-axis.

Therefore, the coordinates of Z are (0,3).

**CCE Questions:**

**Question 1:**

(b) the y-axis.

X = 0 is the equation of the y-axis.

Therefore, the graph of x = 0 is YOYâ€™

**Question 2:**

**(a)** the x-axis.

Y = 0 is the equation of x-axis.

Therefore, the graph of y = 0 is Xâ€™OX.

**Question 3:**

**(b)** parallel to the y-axis and passing through (-3, 0)

Given equation: x + 3 = 0.

Or, x = -3.

The equation of the line is parallel to the y-axis and passes through (-3, 0).

**Question 4:**

**(b)** parallel to the x-axis and passing through (0, 4)

Given equation: y â€“ 4 = 0.

Or, y = -4.

The equation of the line parallel to the x-axis and passes through (0, 4).

**Question 5:**

**(c)** the line y = x

Given, a point of the form (a, a), where aâ‰ 0.

When a = 1, the point is (1, 1)

When a = 2, the point is (2, 2)â€¦. And so on.

When these points i.e., (1, 1), (2, 2)â€¦. And so on are plotted, and the line obtained is extended on both the sides, the equation of the line y = x is obtained.

**Question 6:**

**(d)** the line x + y = 0.

Given, a point of the form (a, -a), where aâ‰ 0.

When a = 1, the point is (1, -1)

When a = 2, the point is (2, -2)â€¦. And so on.

When these points i.e., (1, 1), (2, 2)â€¦. And so on are plotted, and the line obtained is extended on both the sides, the equation of line x + y = 0 is obtained.

**Question 7:**

**(c)** infinitely many solutions.

Given linear equation: 3x â€“ 5y = 15

Or, x=5y+153

When y = 0, x = 5

When y = 3, x = 10

When y = -3, x = 0

Thus, we have the following table:

x |
5 | 10 | 0 |

y |
0 | 3 | -3 |

Plotting the points (5, 0), (10, 3) and (0, -3) and joining them and extending both the ends, we get infinite points that satisfy the given equation.

Hence, the linear equation has infinitely many solutions.

**Question 8:**

**(c)** (0, 3)

Given linear equation: 3x + 2y = 6

Or, y=6âˆ’3x2

Putting, x = 0, we get y = 3

Thus, the graph cuts the y-axis at (0, 3)

**Question 9:**

**(d)** (3, 0)

Given equation: 4x + 3y = 12

Or, x=12âˆ’3y4

When, y =0, x = 3

Thus, the graph cuts the x-axis at (3, 0)