Алгебра. Помогите пожалуйста решить системы уравнений

{ x - y = 5 => x= 5+y
{ 2^x \ 6^y = 96 => 2^(5+y) \ 6^y = 96 =>

2^5 * 2^y / 6^y = 96
2^y\6^y = 96 \ 2^5
(2\6)^y = (6*2^4) \ 2^5
(2\6)^y = 6\2 =>
(2\6)^y = (2\6)^(-1) => y = -1 => x = 5+y = 5+(-1) = 4
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{ 5^(x^2) + 7^y = 60
{ 5^(x^2) - 7^y = -38

5^(x^2) = a
7^y = b =>

{ a + b = 60
{ a - b = -38
(a + b) + (a - b) = 60 + (-38)
2a = 22 => a = 11
(a + b) - (a - b) = 60 - (-38)
2b = 98 => b = 49
=>
5^(x^2) = a = 11
7^y = b = 49 = 7^2 => y=2
5^(x^2) - 4y = 11 - 4*2 = 3
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{ 3 * 2^x + 6 * 2^(y-1) = 24
{ 5 * 2^y - 3 * 2^(x+1) = 7

{ 3 * 2^x + 6 * (2^y)\2^1 = 24
{ 5 * 2^y - 3 * 2^x * 2^1 = 7

{ 3 * 2^x + 3 * 2^y = 24
{ 5 * 2^y - 6 * 2^x = 7

{ 2^x +2^y = 8 (* на 2)
{ 6 * 2^x - 5 * 2^y = -7

{ 2^x = 8 - 2^y
{ 6 * (8 - 2^y) - 5 * 2^y = -7

48 - 6 * 2^y - 5 * 2^y = -7
11 * 2^y = 55
2^y = 5 =>
2^x = 8 - 2^y = 8 - 5 = 3
=>
2^(x+y) = 2^x * 2^y = 3 * 5 = 15

№1
x=5+y
2^(5+y)/6ʸ=96
2⁵•2ʸ/(2ʸ•3ʸ)=96
32/3ʸ=96
3ʸ=⅓
3ʸ=3⁻¹
y=-1 ⇒ x=4
2x₀+3y₀=2•4+3•(-1)=...