Question

Please solve
$\left. \begin{array} { l } { \frac { 3 } { 2 x } + \frac { 2 } { 3 y } = 5 } \\ \\ { \frac { 5 } { x } - \frac { 3 } { y } = 1 } \end{array} \right.$
(Ch- Linear Equations)​

Answer 1

Answer:

• 2x=5N
• [5]-11 z=6 t

• plz follow

Step-by-step explanation:

plz understand brain list answer

Answer 2

Two Equations are :-

$\: \: \: \: \: \: \: \: \: \tt{ \frac{3}{2x} + \frac{2}{3y} = 5 } \\ \: \: \: \: \: \: \: \: \: \: \tt \: \frac{5}{x} - \frac{3}{y} = 1$

Find :-

• Value of x and y

Solution :-

Let $\rm \frac{1}{x}$= A, $\rm \frac{1}{y}$ = B

$\: \: \: \: \: \: \: \: \: \hookrightarrow\orange{\tt{ \frac{3}{2} \sf \: A + \frac{2}{3} B= 5 } } - - (1)\\ \: \: \: \: \: \: \: \: \: \: \hookrightarrow\orange{\tt \: 5 \sf \: A - {3} \sf \: B= 1 }- - (2)$

Multiplying eq (1) by 3 and also multiplying eq (2) by ⅔ to equal the coefficient of B. i. e,

$\dashrightarrow\tt{ \frac{3}{2} \sf \: A + \frac{2}{3} B= 5 } \red{\times 3 }\\ \dashrightarrow\tt \: 5 \sf \: A - {3} \sf \: B= 1 \red{ \times \frac{2}{3} }$

$\tt \frac{9}{2} A \: \: \cancel{ + 2B }= 15 \\ \underline{ \: \: \: \ + \: \: \: \: \: \tt \frac{10}{3} A \: \: \cancel{ - 2B} = ⅔ \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ \rightarrow \bigg(\frac{9}{2} + \frac{10}{3} \bigg) \sf A = \frac{15}{ \red1} + \frac{2}{3} \\ \\ \implies \frac{27 + 20}{6} \sf \: A + \frac{45 + 2}{3} \\ \rightarrow \: \: \: \: \frac{ 47}{6} \sf \: A + \frac{ 47}{3} \\ \sf \: \: \: \: \: \rightarrow \: \: \: A = \cancel{ \frac{6}{3} } \\ \implies \sf \: \large{ \boxed{ \green{ \tt A = 2}}}$

Substituting the Value of A in eq (2) to get the value of B.

$\rightarrow5A - 3B = 1 \\ \rightarrow5 \times 2 - 3B = 1 \\ \rightarrow \: 10 - 3B = 1 \\ \rightarrow - 3B = 1 - 10 \\ \cancel - 3B = \rightarrow \cancel- 9 \\ \rightarrow B = \cancel{\frac{9}{3}} \\ \hookrightarrow \large \boxed{ \green{\tt \: B = 3}}$

Substitute the Value of A and B in our assumption to get final answer value of x and y

$\: \: \: \: \: \: \: \sf \frac{1}{x} = A \implies \frac{1}{x} = 2 \\ \dashrightarrow \boxed{ \underline{ \tt \: x = \frac{1}{2} }}$

Similarly for B

$\: \: \: \: \: \: \: \sf \frac{1}{y} = B \implies \frac{1}{y} = 3 \\ \dashrightarrow \boxed{ \underline{ \tt \: y = \frac{1}{3} }}$