How do you solve for the system of linear equations: x-2y=-1 and 3x-6y=-3?

Answer 1

The equations
x-2y=-1
and
3x-6y=-3
are linearly-dependant, so the system has infinite number of solutions.
The answer is: x and y are random Real numbers(even Complex, if the system is determined in the Complex set)
In other words the answer is
x=2y-1
as in the convention.

Answer 2

[a million] Eqn1 - 2eqn3 : -4x+5y=3 , eqn2 - 3eqn3 : -8x+12y=4 , multily the 1st eqn by way of 2 & subtract the 2d eqn : -2y=2 , y=-a million , -4x+5(-a million)=3 , x=-2 , 3(-2)-2(-a million)+z=0 , z=4 . [2] j+b+c=1480 , j=a hundred and twenty+b then a hundred and twenty+b+b+c=1480 , 2b+c=1360 & b+c=j+280 , j+j+280=1480 , j=six hundred , b+c=880 , b=480 , c=4 hundred

Answer 3

These equations are linear dependent, as they are the same equation. The second equation is just the first equation multiplied by 3. Since they are the same equation, there are many solutions, not just one value for x and one for y. There is no one solution.

Answer 4

3x - 6y = -3
3(1) - 6(1) = -3

and..

x - 2y = -1
(1)x - 2(1) = -1

Both x and y = 1

Answer 5

You can apply Gauss-Jordan method to find x and y.

In this case, there are infinite number of solutions, the answer could be expressed as: (x,y) = (2a - 1, a), being a real.

Answer 6

I've done the work, if you simplify both equations, they are the exact same. they both are:

X = -2y - 1

Answer 7

I find a high school student and ask them to do it for me. Most of the time this kind of problem doesn't come up.

Answer 8

No solution