How do you solve the system of equations?

1. \(x - y = 3\)
2. \(x - y = 7\)

\left\{\begin{array}{ccc}x-y=3\\x-y=7&/\cdot(-1)\end{array}\right\\\\+\left\{\begin{array}{ccc}x-y=3\\-x+y=-7\end{array}\right\\---------\\.\ \ \ \ \ \ \ \ 0=-4\ -\ FALSE\\\\Answer:no\ solution;\ x;\ y\in\O

Answered by Anonymous | 2024-06-10

If you are trying to solve those equations using the equal values method, you would first turn both the equations to equal y. So for x-y=3, you would add y to each side to get x=3+y and then subtract 3 from each side to get x-3=y. When you do the same thing to the equation x=y=7, you get x-7=y. When you plug them together you take away the y's and put an equal sign in between: x-3=x-7 -then you would subtract x from any side and do that to the other side. You get -3=-7 which isn't correct so you would get a "no solution" problem.

Answered by 17dreven | 2024-06-10

The system of equations has no solution because the two equations represent parallel lines, which means they will never intersect. Therefore, there are no values of x and y that satisfy both equations.
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Answered by Anonymous | 2024-12-10