Question

Solve this problem guyzzzz



By using substitution method​

Answer 1

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Your answer is attached.....☺

There are three pictures attached...✌

Answer 2

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★FIND:

The value of 'x' and 'y'

Of 1st and 2nd questions.

★GIVEN,

1)4x+y/3=8/3,x/2+3y/4=-5/2

★SOLUTION:1

$1)4x + \frac{y}{3} = \frac{8}{3} \\ \ \frac{ x}{2} + \frac{3y}{4} = \frac{ - 5}{2} \\ \frac{12x + y}{3} = \frac{8}{3} \\ 12x + y = 8..eq1 \\ \frac{2x + 3y}{4} = \frac{ - 5}{2} \\ 2x +3 y = - 10..eq2 \\ solving \: eq1 \: and \: eq2 \\3 \times eq1 = > 36x + 3y = 24..eq3 \\ 1 \times eq2 = > 2x + 3y = - 10 ..eq4\\ subtracting \: eq3\: and \: eq4\\ = > 36x + 3y - 2x - 3y = 24 - ( - 10) \\ = > 34x = 24 + 10 \\ = > x = \frac{34}{34} \\ = > x = 1$

$sub.x = 1 \: in \: eq1 \: or \: eq2 \\ eq1 = > 12x + y = 8 \\ = > 12(1) + y = 8 \\ = > 12 + y = 8 \\ = > y = 8 - 12 \\ = > y = - 4$

★VERIFICATION:

$sub.(x, y) = (1,- 4) \: in \: eq1 \: or \: eq2 \\ eq1 = > 12x + y = 8 \\ = > 12(1) + ( - 4) = 8 \\ = > 12 - 4 = 8 \\ = > 8 = 8 \\ you \: can \: also \: substitute \: in \: eq2 \: also \: it \: will \: be \: equal \: to \: one \: of \: the \: value.$

★SOLUTION:2

Similarly u can do the 2nd answer also..

But, u guys have to change 0.4x+3y=1.2 as

=>$\bold{4x/10+3y=12/10}$

$\bold{=>4x+30y=12..EQ1}$

$\bold{=>7x-2y=17/6}$

$\bold{=>42x-12y=17..EQ2}$

SOLVE EQ1 AND EQ2 U GET THE (X, Y) VALUES

Hope it helps u