Solving equations y = 2x + 3,3x-2y-8 = 0 by substitution method

Solving equations y = 2x + 3,3x-2y-8 = 0 by substitution method

Substituting y = 2x + 3 into 3x-2y-8 = 0 leads to
3x-2(2x+3)-8=0
3x-4x-6-8=0
-x-14=0
x=-14
Substituting x = - 14 into y = 2x + 3 leads to
y=2*(-14)+3
y=-28+3
y=-25
∴x=-14,y=-25

If the system ax + 2Y = 9,3x-y = 1 has no solution, then a is

ax+2y=9
3x-y=1
That is - 6x + 2Y = - 2
unsolvable
So a = - 6

We know that the system of equations ax + 2Y = 1,3x + y = - 2. (1) when a____ (2) when a____ There is no solution to the equations
Fill in the blanks, the answer should be simple

I do not = 6
II =6

Try to explore whether there are integers a and B such that the equations {ax + by = 53x-y = 1 and {x + 2Y = 5ax-3y = - 7 have the same solution

Simultaneous two equations {ax + by = 5 (1) 3x - y = 1 (2) x + 2Y = 5 (3) ax-3y = - 7 (4) 2 × (2) + (3), 7x = 7, x = 1 into (2), 3-y = 1, y = 2, x, y into (4), a-6 = - 7, a = - 1, a, x, y into (1), A-1 + 2B = 5, B = 3, so when a = - 1, B = 3, the original two squares