Question

x/3+y/4=11,5x/6-y/3=-7​

Answer 1

${ \huge{ \underline{ \green{Solution}}}}$

${ \sf{ \large{The \: Equations \: Are \longrightarrow}}}$

${ \sf{ \large{ \dfrac{x}{3} + \dfrac{y}{4} = 11 \to \: (i)}}}$

${ \sf{ \large{ \dfrac{5x}{6} - \dfrac{y}{3} = ( - 7)\to \: (ii)}}}$

${ \sf{ \large{Taking \: equation \: \: (i)\longrightarrow}}}$

${ \sf{ { \dfrac{x}{3} + \dfrac{y}{4} = 11 }}}$

${ \sf{ { \implies \: {x}{} = 11 - \dfrac{y \times 3}{4} }}}$

${ \sf{ \large{Taking \: equation \: \: (ii)\longrightarrow}}}$

${ \sf{ { \dfrac{5x}{6} - \dfrac{y}{3} = ( - 7) }}}$

${ \sf{ { \implies\dfrac{5 }{6} ( 11 - \dfrac{y \times 3}{4} ) - \dfrac{y}{3} = ( - 7) }}}$

${ \sf{ { \implies\dfrac{55 }{6} - \dfrac{15y}{4} - \dfrac{y}{3} = ( - 7) }}}$

${ \sf{ { \implies - \dfrac{15y}{4} - \dfrac{y}{3} = ( - 7) - \dfrac{55}{6} }}}$

${ \sf{ { \implies \dfrac{15y}{4} + \dfrac{y}{3} = 7 + \dfrac{55}{6} }}}$

${ \sf{ { \implies \dfrac{45y + 4y}{12} = \dfrac{42 + 55}{6} }}}$

${ \sf{ { \implies \dfrac{49y}{12} = \dfrac{97}{6} }}}$

${ \sf{ { \implies {49y}{} = \dfrac{97 \times 12}{6 } }}}$

${ \sf{ { \implies {49y}{} = \dfrac{97 \times { \not12}^{2} }{ \not6 } }}}$

${ \sf{ { \implies {49y}{} =194 } }}$

${ \sf{ { \implies y = 3.9 } }}$

${ \sf{ \large{Taking \: equation \: \: (i)\longrightarrow}}}$

${ \sf{ { \dfrac{x}{3} + \dfrac{y}{4} = 11 }}}$

${ \sf{ { \implies\dfrac{x}{3} + \dfrac{3.9}{4} = 11 }}}$

${ \sf{ { \implies\dfrac{x}{3} = 11 - \dfrac{39}{40} }}}$

${ \sf{ { \implies\dfrac{x}{3} = \dfrac{440 - 39}{40} }}}$

${ \sf{ { \implies\dfrac{x}{3} = \dfrac{ { \not401}^{10.02} }{ \not40} }}}$

${ \sf{ { \implies{x}{} = {10.02} \times 3}}}$

${ \sf{ { \implies{x}{} = 30.06}}}$

${ \sf{ { So \: The \: Value \: Of \: X = 3.9 \: nd \: Y = 30.06}}}$

Answer 2

Answer:

x = 15 ; y = 117

Step-by-step explanation:

given,

$\frac{x}{3} + \frac{y}{4} = 11$      ⇒ (A)

$\frac{5x}{6} - \frac{y}{3} = -7$   ⇒ (B)

Considering the equation (A),

$\frac{x}{3} + \frac{y}{4} = 11$

$\frac{x+y}{12} = 11$

$x + y = 132$  ⇒ (1)

Considering the equation (B),

$\frac{5x}{6} - \frac{y}{3} = -7$

$\frac{5x-y}{6} = -7$

$5x - y = -42$  ⇒ (2)

Considering equation (1) and (2),

$x + y = 132$

$5x - y = -42$ (+)      // adding the equation (1) and (2)  

-------------------

$6x = 90$

-------------------

→ $6x = 90$

$x = \frac{90}{6}$

→ $x = 15$

Substitute $x = 15$ in equation (1),

$x + y = 132$

$15 + y = 132$

$y = 132 - 15$

→ $y = 117$

Therefore, $x = 15$

                  $y = 117$

....

Hope it helps,

Thank you!!!