Let $x$ and $y$ be numbers satisfying \begin{cases} x+2y=4 \\ 2x + y = 6 \\ x= y +1 \end{cases} Find $x$ and $y$.
$x + 2y = 4$ is equivalent to $x=4-2y$
We substitute $x$ in the second equation
$$ 2x+y=6 \\ 8-4y+y=6 \\ -3y=-2 \\ y=2/3 $$ But $x =4-2y=4-4/3=8/3$ and we have $x = y +2=2/3+1=5/3$ which gives $y=2/3=5/3$.
WHAT?!
I'm in grade 9 and studying equations with two unknowns and thought about this example in my head and that's confusing me.
On
Jose's answer is correct - this system of equations does not have a solution. I thought I would add by giving some more intuition.
Each of the equations you wrote is the equation of a line in the x-y plane. When you take any two lines, unless they are parallel, they must intersect at some point. That point of intersection (x,y) is exactly the solution to the system of the two equations.
The reason (thought of in this way) that your system does not have a solution, is because you have three equations, and the lines they represent do not all intersect at one point. You should find that any two of your three equations will have a solution when put together. All three, however, do not.
What you did proves that the system has no solution. It's just that.