WBBSE Class 9 Maths Algebra Chapter 3 Linear Simultaneous Equations Multiple Choice Questions

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WBBSE Class 9 Maths Algebra Chapter 3 Linear Simultaneous Equations Multiple Choice Questions

Example 1. The two equations 4x + 3y = 7 and 7x – 3y = 4 have

  1. Only one solution
  2. Infinite number of solutions
  3. No solution
  4. None of these

Solution: The correct answer is 1. Only one solution

Here \(\frac{4}{7} \neq \frac{3}{-3}\)

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Example 2. The two equations 3x + 6y = 15 and 6x + 12y = 30 have

  1. Only one solution
  2. Infinite number of solutions
  3. No solution
  4. None of these

Solution: The correct answer is 2. Infinite number of solutions

Here \(\frac{3}{6}=\frac{6}{12}=\frac{15}{30}\)

Example 3. The two equations 4x + 4y = 20 and 5x + 5y = 30 have solution (1,1)

  1. Only one solution
  2. Infinite no of solutions
  3. No solution
  4. None of these.

Solution: The correct answer is 3. No solution

Here \(\frac{4}{5}=\frac{4}{5} \neq \frac{20}{30}\)

Example 4. Which of the following equations have a solution (1, 1)

  1. 2x + 3y = 9
  2. 6x + 2y = 9
  3. 3x + 2y = 5
  4. 4x + 6y = 8

Solution: The correct answer is 3. 3x + 2y = 5

3.1 +2.1 = 5

WBBSE Class 9 Maths Algebra Chapter 3 Linear Simultaneous Equations Multiple Choice Questions

 

 

 

 

 

 

 

 

 

 

Example 5. The two equations 4x + 3y = 25 and 5x – 2y = 14 have the solutions

  1. x = 4, y = 3
  2. x = 3, y = 4
  3. x = 3, y = 3
  4. x = 4, y = -3

Solution: The correct answer is 1. x = 4, y = 3

4x + 3y 25….. (1)

5x-2y= 14……..(2)

Multiplying equation (1) by 2 and equation (2) by 3

⇒ \(\begin{array}{r}
8 x+6 y=50 \\
15 x-6 y=42 \\
\hline 23 x=92
\end{array}\)

x = 4

∴ \(y=\frac{25-4 \times 4}{3}=3\)

Example 6. The solution of these equation x + y = 7 are

  1. (1, 6), (3, 4)
  2. (1, 6), (4, 3)
  3. (1, 6) (4, 3)
  4. (1, 6) (-4, 3)

Solution: The correct answer is 3. (1, 6) (4, 3)

1 + 6 = 7, 4 + 3 = 7

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Example 7. The equations \(\frac{x}{4}\) + \(\frac{x}{2}\) + 6 = 0 and 8x + \(\frac{y}{k}\) + 2 = 0 can have no solution if the value of k is one of the following.

  1. 2
  2. ±\(\frac{1}{2}\)
  3. ±2
  4. 4

Solution: The correct answer is 2. ±\(\frac{1}{2}\)

\(\frac{\frac{1}{k}}{8}=\frac{\frac{1}{2}}{\frac{1}{k}} \neq \frac{6}{2}\)

or, \(\frac{1}{k^2}=4\)

k= +\(\frac{1}{2}\)

The value of k is +\(\frac{1}{2}\)

Example 8. The equations (1+k)x + \(\frac{y}{b}\)= 1 and 3x = \(\frac{y}{k}\) = 1 cannot be solves if k is

  1. 1
  2. 2
  3. – 2
  4. – 3

Solution: The correct answer is 3. -2

\(\frac{1+k}{3}=\frac{\frac{1}{6}}{\frac{1}{k}} \neq \frac{1}{1}\)

 

or, \(\frac{1+k}{3}=\frac{k}{6}\)

or, 6+ 6k 3k, 3k = -6,

k = -2

The value of k is -2

Example 9. 2x + 3t = 1, y = \(\frac{t}{3}\) + 1 can be solved

  1. 2x + 9y = 10
  2. 2x – 9y = 10
  3. 2x – 9y = 0
  4. 9x + 2y = 0

Solution: The correct answer is 1. 2x + 9y = 10

2x + 3 (3y – 3) = 1 or, 2x + 9y = 10.

 

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