How do you solve the system of linear equations #x + 3y = 9# and #2x + 6y = 12#?

2 Answers
Jun 1, 2016

There is no solution.


Doubling the first equation #x+3y=9# gives us #2x+6y=18#.

But this is inconsistent with the second equation #2x+6y=12#

Hence, there is no solution.

Drawing graphically these represent two different but parallel lines, who do not intersect and hence there is no solution.

graph{(x+3y-9)(2x+6y-12)=0 [-2.92, 7.08, -0.42, 4.58]}

Jun 1, 2016

Follow the method


x + 3y = 9
2x + 6y = 12

First you check if the two equations are compatible, that is
if they are parallel, they must not intersect.

Parallel means they have the same slope, given by dy/dx in general, and in the case of linear equations by the ratio of the y coefficient to that of x, that is 3 for the first equation, and 3 for the second. Since these two equations are parallel, they are multiple of each other. Now where these straight lines intersect the axis y = 0, that is the x-axis, is the constant term on the right hand side. Try it

Put y = 0 you get x = 9, for the first equation, and y = 0, x = 6 for the second equation, This is geometrically impossible, and therefore algebraically too. So this problem has no solution.