How do you solve the system of equations $- 2 x + 4 y = 10$ $-2x + 4y = 10$ and $3 x - 6 y = - 15$ $3x - 6y = - 15$ ?

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Answer 1 · 2018-06-20T09:06:49

$As equations (1), (2) are the same, we cannot solve for x & y.$ $color(crimson)("As equations (1), (2) are the same, we cannot solve for x & y."$

$- 2 x + 4 y = 10$ $-2x + 4y = 10$

$- x + 2 y = 5, Eqn (1)$ $-x + 2y = 5, " Eqn (1)"$

$3 x - 6 y = - 15$ $3x - 6y = -15$

$x - 2 y = - 5, Eqn (2)$ $x - 2y = -5, " Eqn (2)"$

As equations (1), (2) are the same, we cannot solve for x & y.

Answer 2 · 2018-06-20T10:21:36

The system has infinitely many solutions. See explanation.

The system is:

${(-2x+4y=10),(3x-6y=-15):}$

We can divide the first equation by $-2$ and second by $3$ and the equations become the same:

This means that any pair $(x,y)$ fulfilling the equation (1) is also the solution to the initial system.

Example solutions are then:

$(-5;0)$ , $(0;2 1/2)$ , $(5;5)$