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Q4(iii):
Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

Solution :

Given: A pair of linear equations in two variables:
Equation 1: $2x + y - 6 = 0$
Equation 2: $4x - 2y - 4 = 0$

To Find: Determine if the system is consistent or inconsistent. If consistent, find the solution graphically.

Step 1: Comparing coefficients to check for consistency.
For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, the consistency is determined by the ratios of coefficients:
$a_1 = 2, b_1 = 1, c_1 = -6$
$a_2 = 4, b_2 = -2, c_2 = -4$

Calculate the ratios:
$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{1}{-2} = -\frac{1}{2}$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a unique point. Therefore, the system is consistent.

Step 2: Finding coordinates for graphical representation.
For Equation 1 ($y = 6 - 2x$):
If $x = 0, y = 6 - 2(0) = 6 \implies (0, 6)$
If $x = 3, y = 6 - 2(3) = 0 \implies (3, 0)$

For Equation 2 ($2y = 4x - 4 \implies y = 2x - 2$):
If $x = 0, y = 2(0) - 2 = -2 \implies (0, -2)$
If $x = 1, y = 2(1) - 2 = 0 \implies (1, 0)$

Step 3: Visual Representation
x y

Step 4: Solving the equations algebraically to verify the intersection point.
Multiply Equation 1 by 2:
$2(2x + y - 6) = 0 \implies 4x + 2y - 12 = 0$ (Equation 3)
Add Equation 3 and Equation 2:
$(4x + 2y - 12) + (4x - 2y - 4) = 0$
$8x - 16 = 0$
$8x = 16 \implies x = 2$
Substitute $x = 2$ into Equation 1:
$2(2) + y - 6 = 0$
$4 + y - 6 = 0$
$y - 2 = 0 \implies y = 2$

Final Answer: The system is consistent, and the solution is $x = 2, y = 2$.


More Questions from Class 10 Mathematics Pair of linear equations in two variable EXERCISE 3.1


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