Question

guys pls solve this problem fast because my maths exam is tomorrow​

Answer 1

Question

$\sf \bullet\ \; \dfrac{1}{3}x+y=\dfrac{10}{3}\\\\\bullet\ \; \sf 2x+\dfrac{1}{y}4=\dfrac{11}{4}$

To Find

x and y values

Solution

Many methods are there to solve these equations , I am gonna using substitution method .

$\sf \dfrac{1}{3}x+y=\dfrac{10}{3}\\\\\to\ \sf y=\dfrac{10}{3}-\dfrac{x}{3}\\\\\to\ \sf y=\dfrac{(10-x)}{3}$

Now take 2nd equation ,

$\to\ \sf 2x+\dfrac{1}{4}y=\dfrac{11}{4}$

Sub. y value here ,

$\to\ \sf 2x+\dfrac{1}{4}\left( \dfrac{10-x}{3} \right)=\dfrac{11}{4}$

$\to\ \sf 2x+\dfrac{(10-x)}{12}=\dfrac{11}{4}\\\\\to\ \sf 2x+\dfrac{10}{12}-\dfrac{x}{12}=\dfrac{11}{4}\\\\\to\ \sf 2x-\dfrac{x}{12}=\dfrac{11}{4}-\dfrac{10}{12}$

$\to\ \sf \dfrac{23x}{\cancel{12}}=\dfrac{(33-10)}{\cancel{12}}\\\\\to\ \sf 23x=33-10$

$\to\ \sf 23x=23\\\\\to\ \sf x=1\ \; \pink{\bigstar}$

Sub. x value in any of the given equation ,

$\to\ \sf \dfrac{1}{3}(1)+y=\dfrac{10}{3}$

$\to\ \sf y=\dfrac{10}{3}-\dfrac{1}{3}$

$\to\ \; \sf y=\dfrac{9}{3}\\\\\to \sf y=3\ \; \green{\bigstar}$

Answer 2

$\rm \dfrac{x}{3}+y=\dfrac{10}{3}$

$\bf :\implies y=\dfrac{(10-x)}{3}...(1)$

$\rm  2x+\dfrac{y}{4}=\dfrac{11}{4}$

Sub. (1) here ,

$:\implies \rm 2x+\dfrac{1}{4}\left(\dfrac{10-x}{3}\right)=\dfrac{11}{4}$

$:\implies \rm 2x+\dfrac{(10-x)}{12}=\dfrac{11}{4}$

$:\implies \rm \dfrac{24x+10-x}{12}=\dfrac{11}{4}$

$:\implies \rm 23x+10=33$

$:\implies \rm 23x=33-10$

$:\implies \rm 23x=23$

$\bf :\implies x=1$

Sub. x value in (1) ,

$\rm :\implies y=\dfrac{(10-1)}{3}$

$:\implies \rm y=\dfrac{9}{3}$

$:\implies \bf y=3$