Can someone solve these linear equations using the substitution method?

8x - y= -15 ===> y = 8x + 15

3x + 6y = -12
3x + 6(8x + 15) = -12
3x + 48x + 90 = -12
51x = -102
x = -2
=====
y = 8(-2) + 15
y = -16 + 15
y = -1
=====

Do the other one similarly.

(A)
Given:
8x - y = -15  ................(1)
3x + 6y = -12 ..............(2)

From eqn. (1), we have:

y = 8x + 15  ................(3)

Substituting the value of y from eqn. (3) into eqn. (2), we have:

3x + 6(8x + 15) = -12
=> 3x + 48x + 90 = -12
=> 51x = -102
=> x = -2

Substituting this value of x in eqn. (3), we have:

y = 8*(-2) + 15
=> y = -16 + 15
=> y = -1

Answer: (x, y) = (-2, -1)

(B)
Given:

7x + 7y = -7  ...........(1)
3x + 3y = -3  ...........(2)

Dividing both sides of eqn. (1) by 7, we have:

x + y = -1  .............(3)

Dividing both sides of eqn. (2) by 3, we have:

x + y = -1 ..............(4)

Equations (3) and (4) are the same equation. Hence, the given system of linear equations does not have a unique solution, i.e. there are infinitely many solutions.

Answer: There are infinitely many solutions of the given system of linear equations.

[ A few examples are: (x, y) = (2, -3) or (3, -4) or (4, -5) or (5, -6). You could go on an on like this; there are infinite solutions that satisfy both the given equations. ]

About calculators: I have no idea about any such calculator.