plz solve the following (show workings) simulteneous equation?

Question

2^(x+1) - 5^y =131


2^(x-4) +5^(y-2)=13

Answer 1

Working with equation 2:

5^2[2^(x-4) +5^(y-2)]=13*25
5^2*2^(x-4) + 5^y = 325

Add this to 1st equation, yielding

5^2*2^(x-4) + 2^(x+1) = 456
2^(x-4)[5^2 + 2^5] = 456
2^(x-4) = 456/(25+32)
2^(x-4) = 8
x-4 = 3
x = 7

Use 1st equation:

2^8 - 5^y = 131
5^y = 256 - 131 = 125
y = 3

Answer 2

2^(x + 1) - 5^y = 131
2^(x - 4) + 5^(y - 2) = 13

2^(x - 4) + 5^(y - 2) = 13
((2^x)/(2^4)) + ((5^y)/(5^2)) = 13
((2^x)/16) + ((5^y)/25) = 13

Multiply everything by 400

25(2^x) + 16(5^y) = 5200

On the 2^(x + 1) - 5^y = 131, you get

2^(x + 1) - 5^y = 131
(2^x * 2) - 5^y = 131

so now you have

2(2^x) - (5^y) = 131
25(2^x) + 16(5^y) = 5200

think of this like this

2x - y = 131
25x + 16y = 5200

2x - y = 131
-y = -2x + 131
y = 2x - 131

25x + 16(2x - 131) = 5200
25x + 32x - 2096 = 5200
57x - 2096 = 5200
57x = 7296
x = 128

y = 2x - 131
y = 2(128) - 131
y = 256 - 131
y = 125

Now you get

2^x = 128
5^y = 125

xln(2) = ln128
x = (ln128)/(ln2)
x = 7

5^y = 125
yln5 = ln125
y = (ln125)/(ln5)
y = 3

so

x = 7 and y = 3

Answer 3

rewrite the second equation as
2^x/2^4 + 5^y/5^2 = 13
25. 2^x + 16. 5^y =13.16.25 =5200
5^y = (5200 - 25.2^x)/16

take the 5^y value and plug it in the first equation,
before that rewrite the first equation as,
2.2^x - 5^y =131
now plug in the 5^y value,
2.2^x - ((5200 - 25.2^x) / 16 ) = 131
solve it,
32.2^x - 5200 + 25.2^x = 2096
57.2^x =7296
2^x = 128
consider the "ln" of the both sides, or you can solve it by using powers, I prefer ln
xln2 = ln128
x = 7

plug in x value to the first equation,
2^(7+1) - 5^y = 131
2^8 - 5^y =131
5^y = 256 - 131
5^y = 125
y = 3

so the values are,
x = 7 and y = 3

Answer 4

first simplify this
(1) 2x+2-5y= 131 or 2x-5y = 129
(2) 2x-8+5y-10= 13 or 2x+ 5y= 31
hence 2x-5y = 129 and 2x+ 5y= 31
add these two,
then 4x= 160 or x= 40.
put this value in any equation you get y= -9.8