algebra system of equations?

Question

The system needs to be solved
5x-2y=-8
4x-3y=-19
I know the answer is (2,9) but I need to know how to solve it. I tried to use substitution and elimination but I kept getting weird fractions. I'm sure it's simple but I'm just not seeing it and need step by step logical steps. is substitution/elimination a sound method of solving this system?

Answer 1

5x-2y=-8
4x-3y=-19

Multiply the top equation by four, and the bottom by negative five...

20x - 8y = -32
-20x + 15y = 95

Add the equations. The 'x' term goes away :)

7y = 63
y = 9

Using y = 9 in either of the starting equations yields x = 2.

HTH

Answer 2

step 1: cancel out y by multiplying the first equation by -3 and multiply the 2nd equation by 2
(5x - 2y = -8) x -3
(4x - 3y = -19) x 2

so you'll get :

-15x + 6y = 24
8x - 6y = -38

step 2: simplify
-15x = 24
8x = -38

step 3: combine like terms.

-7x = -14

step 4: divide both sides by -7

x = 2

step 5: plug in the value of x to solve for y:
5(2) - 2y = -8
4(2) - 3y = -19

so that would be:
10 - 2y = -8
8 - 3y = -19

-2y = -18
-3y = -27

y = 9

hope that helps!

Answer 3

5x - 2y = -8
-2y = -8 - 5x
y = (5/2)x + 4

4x - 3y = -19
4x - 3((5/2)x + 4)) = -19
4x - (15/2)x + 12 = -19
(8/2)x -(15/2)x = -7
-(7/2)x = -7
x = 2

y = (5/2)(2) + 4
y = 5 + 4
y = 9