Question

x+2y-4=0 3x+6y-12=0​

Answer 1

Step-by-step explanation:

x+2y-4=0--(1)

3x+6y-12=0--(2)

3×eq(1) 3x+6y-12=0

1×eq(2) 3x+6y-12=0

solving both equations

0

Answer 2

Given:

x+2y-4=0

3x+6y-12=0​

To find:

Solution of the given equation.

Solution:

The given equation is the linear equation in two variables.

• a1x+b1y+c1=0

• a2x+b2y+c2=0

If,

• $\frac{a1}{a2} =\frac{b1}{b2} =\frac{c1}{c2}$  [Concurrent lines]

• $\frac{a1}{a2} =\frac{b1}{b2} \neq \frac{c1}{c2}$[parallel lines]

• $\frac{a1}{a2} \neq \frac{b1}{b2}$ [intersecting lines]

as per the equations given in the question then the ratio of the its coefficients

• $\frac{1}{3} =\frac{2}{6} =\frac{-4}{-12}$

• $\frac{1}{3}=\frac{1}{3}=\frac{1}{3}$

Hence the both given lines are parallel to each other.

So both equations have many solutions.