The amount of chlorine in a pool varies directly with the volume of water. the pool has 2.5 mg of chlorine per liter of water. there are 8000g of water in the pool. how much chlorine is in the pool?

For 8000 g of water the amount of chlorine in the pool is 20 mg .

Step-by-step explanation:

Given as :

The pool has 2.5 mg of chlorine per liter of water

I.e The pool has 2.5 mg of chlorine per 1000 g of water

Let The amount of pool in 8000 g of water be x mg

So, From unitary method

∵ For 1000 g of water , The pool has 2.5 mg of chlorine

so,  For 1 g of water , The pool has mg of chlorine

∴ For 8000 g of water , The pool has × 8000 mg of chlorine

I,e For 8000 g of water , The pool has 2.5 × 8 mg of chlorine

Or, For 8000 g of water , The pool has 20 mg of chlorine

Hence for 8000 g of water the amount of chlorine in the pool is 20 mg . Answer

starting with:    

5 + x - 12 = 2x - 7   

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)   

5 - 12 = x - 7 (combine and cancel x's from both sides)   

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)   

0 = x (add numbers together on both sides)       

so in order for the equation to be true, x must be 0.   

trying x = -0.5   

5 + x - 12 = 2x - 7 (original equation)   

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)   

5 + -0.5 - 12 = -1 - 7 (did multiplication)   

-7.5 = -8 (did addition on both sides)       

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.   

trying x = 0   

5 + x - 12 = 2x - 7 (original equation)   

5 + 0 - 12 = 2(0) - 7 (plug in value of x)   

5 + 0 - 12 = 0 - 7 (did multiplication)   

-7 = -7 (did addition on both sides)       

since -7 equals -7, that demonstrates that the value of 0 for x is correct.   

trying x = 1   

5 + x - 12 = 2x - 7 (original equation)   

5 + 1 - 12 = 2(1) - 7 (plug in value of x)   

5 + 1 - 12 = 2 - 7 (did multiplication)   

-6 = -5 (did addition on both sides)       

since -6 is obviously not equal to -5, the value of 1 for x is wrong.