I need some help with these three Algebra problems solve by the substitution method???????????

Question

1.
3x - 5y = -2
6x + 49 = y

2.
6x + 2y = 0
-3x + y = 12

3.
3x + 4y = -10
-4x + y = 26

Answer 1

substitute 6x + 49 = y in the equation 3x - 5y = -2
3x - 5(6x + 49) = -2
3x - 30x - 245 = -2
-27x -245 = -2
add 245 to each side
-27x -245+245 = -2+245
-27x = 243
divide each side by -27
x = -243/27 =-9
substitute x = -9 in y = 6x + 49
y = 6 * -9 + 49
y = -54 + 49
y=-5
(x, y) =(-9, -5)

Answer 2

2) Let 6x + 2y = 0 -------------(1)

         -3x + y = 12--------------(2)

Consider (2)  -3x +y = 12

Add 3x to each side

y = 12 + 3x

Subsistute y = 3x + 12 in (1)

6x + 2( 3x + 12) = 0

Recall : Distributive property a(b + c) = a*b + a*c

6x + 6x + 24 = 0

12x + 24 = 0

Subtract 24 from each side

12x = -24

Divide each side by 12

x = -24/12

x = -2

Substitute x = -2 in (2)

-3(-2) + y = 12

6 + y = 12

Subtract 6 from each side

y = 12 -6

y = 6

Therefore x = -2 , y = 6

 

Answer 3

3) Let 3x + 4y = -10-----------(1)

         -4x + y = 26------------(2)

Consider (2) -4x + y = 26

Add 4x to each side

y = 26 + 4x

Substitute y = 26 + 4x in (1)

3x + 4(26 + 4x) = -10

Recall distributive property a(b+ c) = a*b + a*c

3x + 104 + 16x = -10

Add 10 to each side

3x + 16x + 104 + 10 = 0

19x + 114 = 0

Subtract 114 from each side

19x = -114

Divide each side by 19

x = -114/19

x = -6

Substitute x = -6 in (2)

-4(-6) + y = 26

24 + y = 26

Subtract 24 from each side

y = 26 - 24

y = 2

Therefore x = -6, y = 2