Question

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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Answer 1

$\sf{\underline{Let\:the:}}$

$\implies$ First number be x.

$\implies$ Second number be y.

$\sf{\underline{From\:the\:first\:statement:}}$

$\sf{\underline{We\:have:}}$

$\boxed{\sf{x = 5y}}$

$\sf{\underline{From\:the\:second\:statement:}}$

$\sf{\underline{We\:have:}}$

$\boxed{\sf{x + 21 = 2(y + 21)}}$

$\sf{\underline{Since:}}$

x = 5y, x must be greater than y.

$\sf{\underline{So:}}$ (x + 21) must be greater than (y+ 21).

$\sf{\underline{That's\:how\:we\:know:}}$

$\boxed{\sf{x + 21 = 2(y + 21)}}$

$\sf{\underline{Rather\:than\:the\:other\:way\:around:}}$

$\boxed{\sf{y + 21 = 2(x + 21)}}$

$\sf{\underline{So:}}$

This can be done by substitution.

$\sf{\underline{Since:}}$ The first equation is already solved for x.

Plugging in 5y for x in the second equation,

$\sf{\underline{We\:have:}}$

$\implies$ $\sf{5y + 21 = 2(y + 21)}$

$\implies$ $\sf{5y + 21 = 2y + 42}$

$\implies$ $\sf{5y - 2y = 42 - 21}$

$\implies$ $\sf{3y = 21}$

$\implies$ $\sf{y = \frac{21}{3}}$

$\implies$ $\sf{y = 7}$

$\sf{\underline{Thus:}}$ $\boxed{\sf{y = 7}}$

$\sf{\underline{And:}}$

$\implies$ $\sf{x = 5y}$

$\implies$ $\sf{x = 5 (7)}$

$\implies$ $\sf{x = 35}$

$\sf{\underline{Thus:}}$ $\boxed{\sf{x = 35}}$

$\sf{\underline{Therefore:}}$ The two numbers are 7 and 35.

Answer 2

Answer:

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