NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.1 Question 4(iii)

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.1 Question 4(iii)

Question:

Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

\( 2x + y - 6 = 0 \),
\( 4x - 2y - 4 = 0 \)

Given:

  • Equation 1: \( 2x + y - 6 = 0 \)
  • Equation 2: \( 4x - 2y - 4 = 0 \)

To Find:

  • Whether the given pair of equations is consistent or inconsistent.
  • If consistent, find the solution graphically.

Formula:

For two linear equations in the form \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \):

  • If \( \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \), the system is consistent and has a unique solution.
  • If \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \), the system is inconsistent (no solution).
  • If \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \), the system has infinitely many solutions (consistent).

Solution:

Step 1: Write the equations in standard form.

$$ \begin{align*} \text{Equation 1:} & \quad 2x + y - 6 = 0 \\ \text{Equation 2:} & \quad 4x - 2y - 4 = 0 \end{align*} $$

Step 2: Identify the coefficients.

$$ \begin{align*} a_1 = 2,\quad b_1 = 1,\quad c_1 = -6 \\ a_2 = 4,\quad b_2 = -2,\quad c_2 = -4 \end{align*} $$

Step 3: Check consistency using the formula.

$$ \dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2} \\ \dfrac{b_1}{b_2} = \dfrac{1}{-2} = -\dfrac{1}{2} $$ Since \( \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \), the pair of equations is consistent and has a unique solution.

Step 4: Express each equation in the form \( y = mx + c \).

$$ \begin{align*} \text{Equation 1:} & \quad 2x + y - 6 = 0 \implies y = 6 - 2x \\ \text{Equation 2:} & \quad 4x - 2y - 4 = 0 \implies 4x - 2y = 4 \implies 2x - y = 2 \implies y = 2x - 2 \end{align*} $$

Step 5: Prepare a table of values for each equation.

For Equation 1: \( y = 6 - 2x \)

xy
06
14
22
30

For Equation 2: \( y = 2x - 2 \)

xy
0-2
10
22
34

Step 6: Find the point of intersection algebraically.

$$ \begin{align*} 2x + y &= 6 \quad \cdots (1) \\ 2x - y &= 2 \quad \cdots (2) \\ \text{Add (1) and (2):} \\ (2x + y) + (2x - y) &= 6 + 2 \\ 4x &= 8 \\ x &= 2 \\ \text{Substitute } x = 2 \text{ into (1):} \\ 2(2) + y &= 6 \\ 4 + y &= 6 \\ y &= 2 \end{align*} $$ The solution is \( (x, y) = (2, 2) \).

Step 7: Graphical representation.

Plot the points from the tables above for each equation. The two lines will intersect at the point \( (2, 2) \), which is the solution to the system.

Result:

  • The given pair of equations is consistent (they have a unique solution).
  • The solution is \((x, y) = (2, 2)\).
  • Graphically, the two lines intersect at the point \((2, 2)\).
© Kaliyuga Ekalavya. All rights reserved.

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