Question

❤ĀHØŁĂ BŘĀÏÑŁĪÀǧ❤

CLASS 10...

Solve by substitution method :-

$2(ax - by) + a + 4b = 0$
and
$2(bx + ay) + b - 4a = 0$

COME ON ....WRONG NHI HONA CHAHIYE...


✨ALL THE BEST✨


⭕FOLLOW ME⭕

​

Answer 1

Answer:

x = - 1/2

y = 2

Explanation :

Refer the attached picture.

Answer 2

SOLUTION ☺️

The given system of Equation:

$= > 2(ax - by) + a + 4b = 0.....(1) \\ = > 2(bx + ay) + b - 4b = 0......(2) \\ = > now \: from \: (1) \: we \: get \\ = > 2ax - 2by = - 4b - a \\ = > 2ax = 2by - 4b - a \\ = > x = \frac{2by - 4b - a}{2a} .....(3) \\ = > substituting \: the \: value \: of \: x \: from \: (3) \: in \: (2) \: we \: get \\ = > 2b( \frac{2by - 4b - a}{2a} ) + 2ay = 4a - b \\ = > \frac{2b {}^{2} y - 4b {}^{2} - ab }{a} + 2ay = 4a - b \\ = > ( \frac{2b {}^{2} }{a } + 2a)y - \frac{4b {}^{2} }{a} - b = 4a - b \\ = > ( \frac{2a {}^{2} + 2b {}^{2} }{a} )y = 4a + \frac{4b {}^{2} }{a} \\ = > ( \frac{2a {}^{2} + 2b {}^{2} }{a} )y = ( \frac{4a {}^{2}4b {}^{2} }{a} ) \\ = > 2(a {}^{2} + b {}^{2} )y = 4(a {}^{2} + b {}^{2} ) \\ = > 2y = 4 \\ = > y = 2 \\ \\ = > now \: substituting \: the \: value \: of \: y \: in \: (3) \: we \: get \\ = > x = \frac{2b \times 2 - a - 4b}{2a} = \frac{4b - a - 4b}{2a} \\ = > - \frac{1}{2} \\ = > so \: x = - \frac{1}{2} \: and \: y \: = 2$

hope it helps ✔️❣️