solve the system by addition or substitution/ I do not understand these.?

Question

1.) -2x - y = 2 y =-2-2x, 2.) 20x -5y = -12 y = 4x + 3 3.) 3x + 6y = 0 x = 2/3

Answer 1

Take 2 equationz first .. then plug the value in the 3rd one and one of the first 2 onez .. then solve !!

Lets take the first one ..

-2x-y(first equation)=2y(second equation)
-2x=3y
x= -3y/2

now

-2(-3y/2) -y=-2-2x(third equation)
3y-y=-2-2x
2y+2=-2x
x= - y - 1

NOW THE ANSWERS ARE EQUAL .. Thus

-3y/2 = -y-2
-3y=-2y-2
3y=2y+2
y=2

Now when u know y equals 2 .. take the 2nd equation and any of the 2 onez !!

2y = -2-2x
2(2)=4
4=-2-2x
6=-2x
x=-3

Y = 2
X = -3

U can check .. it is absolutely correct !!

Answer 2

1) -2x - y = 2
y = -2 - 2x

Your first step is to put BOTH equations in the form ax + by = c.

-2x - y = 2
2x + y = -2

If you add these two equations together, you get:

0 = 0
Which is a true statement. Therefore, we have an infinite number of solutions. {side note, the solutions are x = s and y = -2 - 2s for some real number s.}

2) 20x - 5y = -12
y = 4x + 3

Put both equations in the form ax + by = c

20x - 5y = -12
4x - y = -3

Multiply the second equation by 5.

20x - 5y = -12
20x - 5y = -15

Subtracting the two equations, we get
0 = 3, which is a false statement. Therefore there is no solution.

Answer 3

on account that X=2y-2 replace X contained in the 2d equation with 2y-2 like this: 3y=[2y-2]+6 Then purely sparkling up! So... Y=4 All you do is on account that X is on my own you may replace X contained in the 2d equation with X, lets say it grow to be 2X, then you absolutely ought to ought to get X on my own before substituting! desire this helps!