Question

If x:y= 3:4, find the value of 7x-4y 3x+y

Answer 1

Given :

$\bullet \: \sf \dfrac{x}{y} = \dfrac{3}{4}$

To Find :

We have to find the value of $\sf\dfrac{7x - 4y}{3x + y}$

Solution :

• Let say x as 3k and y as 4k

$\bigstar\:\underline{\textbf{According to the Question Now :}}$

$:\implies \sf \dfrac{7x - 4y}{3x + y}$

$:\implies \sf \dfrac{7 \times 3k - 4 \times 4k}{3 \times 3k + 4k}$

$:\implies \sf \dfrac{21k - 16k}{9k + 4k}$

$:\implies \sf \dfrac{5k}{13k}$

$:\implies \underline{ \boxed{ \sf \dfrac{5}{13}}}$

Answer 2

$\boxed{\huge{\bf{\star{Correct\:question \:-:}}}}$

• If x:y= 3:4, find , $\implies{\sf{\large { \:\:The\:Value\:of\:=\: \dfrac{7x-4y}{3x+y}\: }}}$

AnswEr-:

• $\boxed{\sf{\large { AnswEr-:\:\:\: \dfrac{5}{13}\: }}}$

$\dag{\sf{\large { EXPLANATION:\:}}}$

• $\underline{\sf{\large { Given-:\:}}}$

• $\implies{\sf{\large { \:\:x:y\:= 3:4\: }}}$

• $\underline{\sf{\large { To\:Find-:\:}}}$

• $\implies{\sf{\large { \:\:The\:Value\:of\:=\: \dfrac{7x-4y}{3x+y}\: }}}$

$\dag{\sf{\large { Solution-:\:}}}$

• $\implies{\sf{\large { \:\:x:y\:= 3:4\: }}}$

• $\dag{\sf{\large { Or,\:}}}$

• $\implies{\sf{\large { \:\:\dfrac{x}{y}\:=\dfrac{3}{4}\: }}}$

$\dag{\sf{\large { Now-:\:}}}$

• $\underline{\sf{\large { Let's\:Assume\:-:}}}$

• $\implies{\sf{\large { \:\:x=3a\:,y =4a\: }}}$

$\star{\sf{\large { According\:To\:The\:Question-:\:}}}$

• $\implies{\sf{\large { \:\:The\:Value\:of\:=\: \dfrac{7x-4y}{3x+y}\: }}}$

$\star{\sf{\large { Substituting \:x =3a\:and\:y=4a-:\:}}}$

• $\implies{\sf{\large { \:\:\: \dfrac{7\times 3a-4\times 4a}{3 \times 3a+4 \times a}\: }}}$

• $\implies{\sf{\large { \:\:\: \dfrac{21a- 16a}{9a+4a}\: }}}$

• $\implies{\sf{\large { \:\:\: \dfrac{5a}{13a}\: }}}$

• $\dag{\sf{\large { Or,\:}}}$

• $\implies{\sf{\large { \:\:\: \dfrac{5\times a}{13\times a}\: }}}$

• $\dag{\sf{\large { A\:will\:be\:eliminated}}}$

• $\implies{\sf{\large { \:\:\: \dfrac{5}{13}\: }}}$

$\dag{\sf{\large { Hence-:\:}}}$

• $\boxed{\sf{\large { AnswEr-:\:\:\: \dfrac{5}{13}\: }}}$

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