Question

x-5y+4=0 2x+y-8=0 by elimination method by equating the co-efficient method​

Answer 1

There are two equations provided here.

3x – 5y – 4 = 0 or 3x -5y = 4 ……. (1).

9x = 2y + 7 or 9x -2y = 7 ……(2).

Now only focusing on eq. numbered 1 and 2 …

3x – 5y = 4………(1)

9x – 2y = 7 ……(2)

Multiply eq. (1) by 3, we get….

3 (3x – 5y = 4) or 9x – 15y = 12…… (3).

Now we can easily subtract (2) and (3) to get…..

9x – 2y = 7

– 9x – 15y = 12.

– +13 y = -5

Or we get 13 y = -5 or y = -5/13.

Now we substitute the value of Y = -5/13 in equation (1) we get the value of x.

3x – 5(-5/13) = 4

3x +25/13 = 4

39x + 25 = 52

Or 39x = 52 – 25

39x = 27

Or x = 27/39 = 9 / 13

So x = 9/13 and y = -5/13.

Answer 2

Question -

x-5y+4=0 2x+y-8=0 by elimination method by equating the co-efficient method.

Solution -

$x-5y+4=0 \: \: ...(1)$

$2x+y-8=0 \: \: ...(2)$

Now, multiply by 2 in equation (1).

$= > 2x - 10y + 8 = 0 \: \: ...(3)$

Now subtract equation (2) from equation (3).

$(2x +10 y + 8 = 0) \\ - (2x + y - 8 = 0)$

$= > 9y + 16 = 0$

$= > 9y = - 16$

$= > \boxed{ y = \frac{16}{9} }$

Put the value of y in equation (1).

$= > x - 5 \times \frac{16}{9} + 4 = 0$

$= > x - \frac{80}{9} + 4 = 0$

$= > \frac{9x - 80 + 36}{9} = 0$

$= > \frac{9x - 44}{9} = 0$

$= > 9x - 44 = 0$

$= > 9x = 44$

$= > \boxed{x = \frac{44}{9} }$

ﾟ･*☆¸¸.•*¨*•♪¸¸.•*¨*•.•*¨*•.¸¸☆*･ﾟ

➻❥Iռɕɾεɖıɓɭε Kɦυꜱɦı࿐ཽ༵