Jharkhand Board JAC Class 9 Maths Important Questions Chapter 4 Linear Equations in Two Variables Important Questions and Answers.

## JAC Board Class 9th Maths Important Questions Chapter 4 Linear Equations in Two Variables

Question 1.

Prove that x = 3, y = 2 is a solution of 3x – 2y = 5.

Solution :

x = 3, y = 2 is a solution of 3x – 2y = 5, because L.H.S. = 3x – 2y = 3 × 3 – 2 × 2 = 9 – 4 = 5 = R.H.S.

i.e. x = 3, y = 2 satisfies the equation 3x – 2y = 5.

∴ It is a solution of the given equation.

Question 2.

Prove that x = 1, y = 1 as well as x = 2, y = 5 is a solution of 4x – y – 3 = 0.

Solution :

Given eq, is 4x – y – 3 = 0 …(i)

First we put x = 1, y = 1 in L.H.S. of eq (i)

Here L.H.S. = 4x – y – 3 = 4 × 1 – 1 – 3 = 4 – 4 = 0 = R.H.S.

Now we put x = 2, y = 5 in eq. (i)

L.H.S. = 4x – y – 3 = 4 × 2 – 5 – 3 = 8 – 8 = 0 = R.H.S.

Since, x = 1, y = 1 and x = 2, y = 5, both pairs satisfied the given equation, therefore they are the solutions of given equation.

Question 3.

Determine whether x = 2, y = – 1 is a solution of equation 3x + 5y – 2 = 0.

Solution :

Given eq. is 3x + 5y – 2 = 0 ….(i)

Taking L.H.S. = 3x + 5y – 2 = 3 × 2 + 5 × (-1) – 2 = 6 – 5 – 2 = – 1 + 0 = R.H.S.

Here LH.S. ≠ R.H.S. therefore x = 2, y = – 1 is not a solution of given equation.

Question 4.

Draw the graph of

(i) 2x + 5 = 0

(ii) 3y – 15 = 0

Solution :

(i) Graph of 2x + 5 = 0

On simplifying it we get 2x = – 5

x = – \(\frac {5}{2}\)

First we plot point A_{1} (-\(\frac {5}{2}\), 0) and then we plot any other point A_{2} (-\(\frac {5}{2}\), 2) on the graph paper, then we join these two points we get required line l as shown in adjoining figure

(ii) Graph of 3y – 15 = 0

On simplifying it we get 3y = 15 y = 5.

⇒ y = \(\frac {15}{3}\) = 5

First we plot the point B_{1}(0, 5) and then we plot any other point B_{2}(3, 5) on the graph paper, then we join these two points we get required line m as shown in figure.

Note: A point which lies on the line is a solution of that equation. A point not lying on the line is not a solution of the equation.

Question 5.

Draw the graph of the line x – 2y = 3, from the graph find the coordinates of the point when

(i) x = – 5

(ii) y = 0

Solution :

Here given equation is x – 2y = 3.

Solving it for y we get 2y = x – 3

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

Let x = 0, then y = \(\frac {1}{2}\)(0) – \(\frac {3}{2}\) = – \(\frac {3}{2}\)

x = 3, then y = \(\frac {1}{2}\)(3) – \(\frac {3}{2}\) = 0

x = – 2, then y Hence, we get y = \(\frac {1}{2}\)(-2) – \(\frac {3}{2}\) = – \(\frac {5}{2}\)

Hence, we get

x | 0 | 3 | – 2 |

Y | –\(\frac {3}{2}\) | 0 | –\(\frac {5}{2}\) |

Clearly from the graph, when x = -5 then y = -4 so corresponding coordinates are (-5, -4) and when y = 0 then x = 3, so corresponding coordinates are (3, 0).

Question 6.

Draw the graphs of the lines represented by the equations x + y = 4 and 2x – y = 2 in the same graph. Also find the coordinates of the point where the two lines intersect.

Solution :

Given equations are x + y = 4 ………….(i)

and 2x – y = 2 ….. (ii)

(i) We have, y = 4 – x

x | 0 | 2 | 4 |

Y | 4 | 2 | 0 |

(ii) We have, y = 2x – 2

x | 1 | 0 | 3 |

Y | 0 | – 2 | 4 |

By drawing the lines on a graph paper, clearly we can say that P is the point of intersection where coordinates are x = 2, y = 2, i.e., P(2, 2).

Question 7.

Solve : \(\frac {x}{2}\) = 3 + \(\frac {x}{3}\)

Solution :

Given \(\frac {x}{2}\) = 3 + \(\frac {x}{3}\)

⇒ \(\frac {x}{2}\) – \(\frac {x}{3}\) = 3

⇒ \(\frac{3 x-2 x}{6}\) = 3

⇒ \(\frac {x}{6}\) = 3

⇒ x = 18

Question 8.

Solve the following system of equations:

2x – 3y = 5

3x + 2y = 1

Solution :

Given eq. are 2x – 3y = 5 ………(i)

and 3x + 2y = 1 ……………(ii)

Multiplying eq. (i) by 3 and eq. (ii) by 2. we get 6x – 9y = 15 and 6x + 4y = 2 respectively.

⇒ -13y = 13

⇒ y = – 1

Putting the value of y in eq. (i) we get

2x – (3) × (-1) = 5

2x + 3 = 5

⇒ 2x = 5 – 3

⇒ 2x = 2

⇒ x = 1

∴ x = 1, y = – 1 is the solution of given system of linear equations.

Question 9.

Solve the following system of equations:

x + 4y = 14

7x – 3y = 5

Solution :

Let x + 4y = 14 ……..(i)

and 7x – 3y = 5 ……….(ii)

From equation (i)

x = 14 – 4y

Substitute the value of x in equation (ii)

⇒ 7(14 – 4y) – 3y = 5

⇒ 98 – 28y – 3y = 5

⇒ 98 – 31y = 5

⇒ 93 = 3ly

⇒ y = \(\frac {93}{31}\)

⇒ y = 3

x = 14 – 4y = 14 – 4 × 3 = 14 – 12 = 2

So, solution is x = 2 and y = 3.

Multiple Choice Questions

Question 1.

Which of the following equations is not a linear equation?

(a) 2x + 3 = 7x – 2

(b) \(\frac {2}{3}\)x + 5 = 3x – 4

(c) x^{2} + 3 = 5x – 3

(d) (x – 2)^{2} = x^{2} + 8

Solution :

(c) x^{2} + 3 = 5x – 3

Question 2.

Solution of equation \(\sqrt{3}\)x – 2 = 2\(\sqrt{3}\) + 4 is

(a) 2(\(\sqrt{3}\) – 1)

(b) 2(1 – \(\sqrt{3}\))

(c) 1 + \(\sqrt{3}\)

(d) 2(1 + \(\sqrt{3}\))

Solution :

(d) 2(1 + \(\sqrt{3}\))

Question 3.

The value of x which satisfies \(\frac{6 x+5}{4 x+7}=\frac{3 x+5}{2 x+6}\) is :

(a) -1

(b) 1

(c) 2

(d) -2

Solution :

(b) 1

Question 4.

Solution of \(\frac{x-a}{b+c}+\frac{x-b}{c+a}+\frac{x-c}{a+b}\) = 3 is

(a) a + b – c

(b) a – b + c

(c) – a + b + c

(d) a + b + c

Solution :

(d) a + b + c

Question 5.

One-fourth of one-third of one-half of a number is 12, then number is

(a) 284

(b) 286

(c) 288

(d) 290

Solution :

(c) 288

Question 6.

A linear equation in two variables has maximum

(a) one solution

(b) two solutions

(c) infinitely many solutions

(d) None of these

Solution :

(c) infinitely many solutions

Question 7.

Solutions of the equation x – 2y = 2 is/are

(a) x = 4, y = 1

(b) x = 2, y = 0

(c) x = 6, y = 2

(d) All of these

Solution :

(d) All of these

Question 8.

The graph of line 5x + 3y = 4 cuts y-axis at the point

(a) (0, \(\frac {4}{3}\))

(b) (0, \(\frac {3}{4}\))

(c) (\(\frac {4}{3}\), 0)

(d) (\(\frac {4}{3}\), 0)

Solution :

(a) (0, \(\frac {4}{3}\))

Question 9.

If x = 1, y = 1 is a solution of equation 9ax + 12ay = 63, then the value of a is

(a) – 3

(b) 3

(c) 7

(d) 5

Solution :

(b) 3