Can someone do the substitution method for this problem?

Question

3Y+5V=-12

2Y-3V=-8

Answer 1

Sure

3Y + 5V = -12
5V = -12 -3Y
V = -12/5 -3Y/5

REPLACE IN 2ND EQU:
2Y - 3(-12/5-3Y/5) = -8
2Y + 36/5 +9Y/5 = -8
19Y/5 = -76/5
19Y = -76
Y = -76/19 = -4

THEN SOLVE FOR V
V = -12/5 - 3(-4)/5 = -12/5 +12/5 = 0
V = 0

Answer 2

3Y = -5V - 12 .......( subtract 5V from both sides of first equation)

Y= (-5/3)V - 4 ........( divide both sides by 3 to isolate Y)

......

2 ( (-5/3)V -4) -3V = -8 .......( SUBSTITUTE the Y = (-5/3)V -4 from the previous
into the second equation where you see a Y.)

(-10/3)V - 8 - 3V = -8 ...........( distribute the 2 into the parentheses. Basically,
multiply 2 by (-5/3)V and then by (-4). )

(-19/3)V -8 = -8 ....................(Add the V's on the left side)

(-19/3)V = 0 ...........................( Add 8 to both sides)

V = 0 .......................................( divide both sides by (-19/3) to find V)

Y = (-5/3) * (0) - 4 = 4...........( find Y by plugging in V=0 into the 2nd step up
there ^

Answer 3

You can either do the substitution method or solve the equations simultaneously by multiplying equation 1 by 2 and equation 2 by minus 3.
The solution is v = 0 and y = -4

Answer 4

TAKE SECOND EQUATION:
2Y-3V=-8
2Y = 3V-8
Y = (3V-8) / 2

SUBSTITUTE INTO FIRST EQUATION:
3[(3V-8) / 2] + 5V = -12
(9V-24)/2 + 5V = -12
9V -24 + 10V = -24
19V - 24 = -24
19V = 0
V = 0

SUSTITUTE INTO SECOND:
2Y-3V=-8
2Y-3(0)=-8
2Y=-8
Y=-4

V={0}
Y={-4}

Answer 5

3y+5v=-12..................................1
2y-3v=-8........................................2
multiply by 2 in EQUA. 1 and by 3 in EQUA.2
3y+5v=-12 X 2
2y-3v=-8 x3



6y+10V= -24

6Y -9V= -24
- + +

19V=0
V=0
SUBSTITUTE V=0 IN equa.1
3Y+0=-12
Y=-4
VERIFICATION:-
3(-4)+5(0)=-12
-12+0=-12
-12=-12
LHS=RHS