Question

a positive number is 5 times another number. if 21 is added to both the number than one of the new numbers become twice the other new number. what are the numbers?

Answer 1

Answer:

Step-by-step explanation:

Let x be the first number and let y be the second number. From the first statement, we have:

x = 5y.

From the second statement, we have:

x + 21 = 2*(y + 21).

[Since x = 5y, x must be greater than y, so x + 21 must be greater than y+ 21. That's how we know x + 21 = 2*(y + 21), rather than the other way around, y + 21 = 2*(x + 21).]

So all we have to do is solve this system of equations. This can be done easily by substitution, since the first equation is already solved for x. Plugging in 5y for x in the second equation, we have:

5y + 21 = 2*(y + 21)

5y + 21 = 2y + 42           Distributing on the right-hand side

3y + 21 = 42                   Subtracting 2y from both sides

3y = 21                            Subtracting 21 from both sides

y = 7                                Dividing both sides by 7

Thus, y = 7 and x = 5y = 5*7 = 35, so the two numbers are 7 and 35.  

 

We can also check our answer by verifying that these numbers satisfy the two conditions.

Indeed, 35 is five times seven. Also, if we add 21 to both numbers they become 56 and 28, and 56 is two times 28.

Answer 2

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A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

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Let above number be 'x'

$\leadsto \tt{2 \times x + 21 = 5x + 21}$

$\leadsto \tt{2 x + 42 = 5x + 21}$

$\leadsto \tt{2 x - 5x = 21 - 42}$

$\leadsto \tt{- 3x = 21 }$

$\leadsto \tt{ 3x = 21 }$

$\leadsto \tt{ x = \frac{21}{3} }$

$\leadsto \tt{ x = 7}$

So the two numbers are 7 and 35.

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