solve 3x+y=-2 , x=y-2. this is Systems OF Equations

Question

pleaaaaaaaaase. and tell me how to do it.

Answer 1

3x + y = -2,        [1]

x = y - 2             [2]

 

Since equation [2] is equal to 'x' we can simply put this equation into equation [1]:

So, take equation [1],   3x + y = -2

Put in equation [2],      3(y - 2) + y = -2

Now multiply out the brackets,   3(y) 3x(-2) + y = -2

                                               3y -6 + y = -2 

                                               4y - 6 = -2

Now you can add 6 to both sides of the equation, 4y - 6 + 6 = -2 + 6

                                                                       4y = 4

Now you can divide both sides of the equation by 4,  (4y)/4 = 4/4

                                                                             y = 1

So now you have the value for 'y' you can simply put this value into one of the original equations to find the value of 'x'.

So take equation [2],     x = y - 2

 Substitute in the value of y,         x = 1 - 2

                                               x = -1

 

So the solution to this system of equations is:

(-1, 1)