Question

(vi)
$\frac{3x}{2} - \frac{5y}{3 } = - 2 \: \: and \: \frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
solve using substitution method​

Answer 1

$\sf\huge\bold{Solution}$

$\sf \longmapsto \dfrac{3x}{2} - \dfrac{5y}{3} = - 2$

$\sf \longmapsto \dfrac{9x - 10y}{6} = - 2$

$\sf \longmapsto 9x - 10y = - 12$

$\sf \boxed{x = \dfrac{ - 12 + 10y}{9}}$

Now, substitute X's value into another equation:

$\sf \longmapsto \dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$

Now, putting X's value :

$\sf \longmapsto \dfrac{ - 12 + 10y}{9} + \dfrac{y}{2} = \frac{13}{6}$

$\sf \longmapsto \dfrac{2( - 12 + 10y) + 9y}{18} = \dfrac{13}{6}$

$\sf \longmapsto \dfrac{ - 24 + 20y + 9y}{18} = \dfrac{13}{6}$

$\sf \longmapsto6( - 24 + 29y) = 13 \times 18$

$\sf \longmapsto - 144 + 174y = 234$

$\sf \longmapsto174y = 234 + 144$

$\sf \longmapsto174y = 378$

$\sf \boxed{ y = \dfrac{378}{174} = \dfrac{189}{87} = \dfrac{63}{29} }$

Now,put y's value into first Equation again to find X's value :

$\sf \longmapsto \boxed{ x = \dfrac{ - 12 + 10y}{9}}$

$\sf \longmapsto \: x = \dfrac{ - 12 + 10 \times \dfrac{63}{29} }{9}$

$\sf \longmapsto \: x = \dfrac{ - 348 + 630}{ \dfrac{29}{9} }$

$\sf \boxed{ x = \dfrac{282}{261} = \dfrac{94}{87}}$

Check:-

$\sf \longmapsto \dfrac{3x}{2} - \dfrac{5y}{3} = - 2$

$\sf \longmapsto \dfrac{3 \times \dfrac{94}{87} }{2} - \dfrac{5 \times \dfrac{63}{29} }{3}$

$\sf \longmapsto \dfrac{282 - 2(315)}{174}$

$\sf \longmapsto \dfrac{282 - 630}{174}$

$\sf \longmapsto \dfrac{ - 348}{174} = - 2$

Answer 2

$\huge{\underline{\red{\mathfrak{♡ANSWER♡}}}}$

hope his answer help u