Question

The lines whose equations are 2x + y = 3z and x + y = 6z intersect at which point? (3z, -3z) (-3z, 9z) (9z, -15z)

Answer 1

Hey there, Xs4etsmith!

The correct answer would be the second option.
To solve for this system of equation, we need to make both equations equal to each other but first we need to make each equation equal to 0.

So, $2x+y=3z$ goes to $2x+y-3z=0$ (I subtracted $3z$ from both sides)
And $x+y=6z$ goes to $x+y-6z=0$ (I subtracted $6z$ from both sides)
Now we can make them equal to each other.
$2x+y-3z=x+y-6z$
$2x-3z=x-6z$ (I subtracted y from both sides to cancel it)
$x-3z=-6z$ (I subtracted x from both sides)
$x=-3z$ (I added 3z to both sides to isolate x)

Now, we replace the value of $x$ for $3z$ and solve
$2(-3z)+y=3z$ 
$-6z+y=3z$ (Distribute)
$y=9z$ (Add $6z$ to both sides)
So $y$ is equal to $9z$

Let's try the second equation
$-3z+y=6z$
$y=9z$ (Add $6z$ to both sides)

The value of $x$ would be $-3z$ and when solving both equations, we get the same value for $y$ which is $9z$

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