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nignag [31]
3 years ago
7

Help please!! How do I solve this

Mathematics
1 answer:
egoroff_w [7]3 years ago
5 0
This is an exponential equation. We will solve in the following way. I do not have special symbols, functions and factors, so I work in this way
 2 on (2x) - 5 2 on x + 4=0 =>. (2 on x)2 - 5 2 on x + 4=0  We will replace expression ( 2 on x) with variable t => 2 on x=t  =. t2-5t+4=0 => This is quadratic equation and I solve this in the folowing way => t2-4t-t+4=0 =>     t(t-4) - (t-4)=0 => (t-4) (t-1)=0 => we conclude t-4=0 or t-1=0 => t'=4 and t"=1 now we will return t' => 2 on x' = 4 => 2 on x' = 2 on 2 => x'=2 we do the same with t" => 2 on x" = 1 => 2 on x' = 2 on 0 => x" = 0 ( we know that every number on 0 gives 1). Check 1: 2 on (2*2)-5*2 on 2 +4=0 =>                   2 on 4 - 5 * 4+4=0 => 16-20+4=0 =. 0=0 Identity proving solution. 
Check 2:   2 on (2*0) - 5* 2 on 0 + 4=0 => 2 on 0 - 5 * 1 + 4=0 => 
1-5+4=0 => 0=0  Identity provin solution.
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Let the two numbers be x and y.

The difference between the numbers is 4. Therefore
x - y = 4             (1)

The sum of one-half of each number is 18, therefore
x/2 + y/ 2 = 18
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Add equations (1) and (2).
(x - y) + (x + y) = 4 + 36
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x = 20
From (2), obtain
y = 36 - x = 36 - 20 = 16.

Answer: The two numbers are 16 and 20.

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