Question

there are two number such that the sum of the first and three times the second is 54 while the difference between 4 times the first and twice the second is 2 find the number ​

Answer 1

Answer:

$\sf{The \ numbers \ are \ \dfrac{57}{7} \ and}$

$\sf{\dfrac{107}{7} \ respectively.}$

Given:

• There are two number such that the sum of the first and three times the second is 54.

• The difference between 4 times the first and twice the second is 2.

To find:

• The numbers.

Solution:

$\sf{Let \ the \ two \ numbers \ be \ x \ and \ y.}$

$\sf{According \ to \ the \ first \ condition.}$

$\sf{\leadsto{x+3y=54...(1)}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{\leadsto{4x-2y=2}}$

$\sf{\leadsto{2(2x-y)=2}}$

$\sf{\leadsto{2x-y=\dfrac{2}{2}}}$

$\sf{\leadsto{2x-y=1...(2)}}$

$\sf{Multiply \ equation (2) \ by \ 3, \ we \ get}$

$\sf{\leadsto{6x-3y=3...(3)}}$

$\sf{Add \ equations \ (1) \ and \ (3), \ we \ get}$

$\sf{x+3y=54}$

$\sf{+}$

$\sf{6x-3y=3}$

___________________________________

$\sf{7x=57}$

$\sf{\therefore{x=\dfrac{57}{7}}}$

$\sf{Substitute \ x=\dfrac{57}{7} \ in \ equation (2)}$

$\sf{2\times\dfrac{57}{7}-y=1}$

$\sf{\therefore{\dfrac{114}{7}-y=1}}$

$\sf{\therefore{y=\dfrac{114}{7}-1}}$

$\sf{\therefore{y=\dfrac{144-7}{7}}}$

$\sf{\therefore{y=\dfrac{107}{7}}}$

$\sf\purple{\tt{\therefore{The \ numbers \ are \ \dfrac{57}{7} \ and}}}$

$\sf\purple{\tt{\dfrac{107}{7} \ respectively.}}$