Solving liner systems using elimination?

i need to solve this:

2x + y = -5
3x + 5y = 3

i keep trying, but i can't seem to eliminate and get the correct answer. you so much. : )

Y = 3
x = -4

(2x + y = -5)3  =  6x + 3y = -15   = 6x + 3y = -15
(3x + 5y = 3)2  = -(6x + 10y = 6) = -6x - 10y = -6
                                                    -7y = -21    :  y = 3
(then we substitute y into equation #1)
2x + 3 = -5
2x = -8
x = -4

2x + y = -5 (i)   
3x + 5y = 3(ii)

Now which do you want us to eliminate first, x or y? I hear you saying y. So we are going to eliminate y first.

But if we look at equations (ii) we see that y has 5y, so we multiply equation (i) by 5, so that it has 5y also.


2x + y = -5 (i)     (x5).
3x + 5y = 3(ii)

5*(2x + y) = 5*(-5) 
10x + 5y = -25(iii).

The new equation becomes (iii), and we can now compare (ii) and (iii)

3x + 5y = 3(ii)
10x + 5y = -25(iii)

Since we have now made 5y on both equations, we can eliminate the y, by subtracting.

Equation  (ii)  Minus (iii).
We perform subtraction before the equality sign and after the equality sign, like in normal subtraction. The xs would subtract the xs and the ys would subtract the ys.

(3x-10x) + (5y -5y) = 3 - (-25)
-7x        +  0y         = 3 +25 .    Minus Minus  is Plus.
-7x                        = 28.    Divide both sides by 7.
x = 28/(-7)
x = -4.

Now we have gotten x, so we can now substitute x = -4, in either equations (i) or (ii). Let us use (i).

2x + y = -5 (i)      x = -4.
2(-4) + y = -5.
-8 + y = -5.
y       = -5 +8. (When -8 crossed the equality sign to other side, sign changed)
y      = 8-5
y =3.

Therefore x = - 4, y =3.  That's it. Cheers.

(answer 2 )i’m very smart that’s how i know the answer.

Aquadratic function is one of the form f(x)=ax2 + bx + c l, where a,b, and c are numbers with are not equal to zero.