Two fish tanks contained of water altogether. 60 degrees % of the water was scooped out from the first fish tank and 0.6L of water was scooped out from the second fish tank. At the end. both fish tanks had the same amount of water. Find the original volume of water in each fish tank at the beginning ?
PLEASE HELP

2.4

Step-by-step explanation:

let a = vol in the 1st tank (in liters)

let b = vol in the 2nd tank

:

"Two fish tanks contained 9L of water altogether."

a + b = 9

:

" 60% of the water was scooped out from the first tank"

a - .6a = .4a amt in the 1st tank now

:

" and 0.6L of the water was scooped out of the 2nd tank."

a - .6 = amt in the 2nd tank

:

" At the end both fish tanks had the same amount of water. "

.4a = b - .6

.4a - b = -.6

use elimination with the first equation

.4a - b = -.6

a + b = 9

addition eliminates b, find a

1.4a + 0 = 8.4

a = 8.4/1.4

a = 6 liters in the 1st tank

then, obviously,

9 - 6 = 3 liters in the 2nd

:

:

Confirm this in the 2nd statement

" 0.6L of the water was scooped out of the 2nd tank. At the end both fish tanks had the same amount of water. "

.4(6) = 3 - .6

2.4 = 2.4

1) the property is   ixi = a ⇔x = a   o x = -a

4(x-5)+3= 15⇒4x-20+3 = 15⇒4x-17= 15⇒4x = 15+17⇒4x = 32⇒x =32/4= 8

4(x-5) +3= -15⇒4x-20 +3 = -15⇒4x-17= -15⇒4x =-15+17⇒4x = 2⇒x =2/4= 1/2

the values are x =8 and x = 0,5