Question

47x+31y=63 9x=2y+7 by elimination method by equating the co-efficient method​

Answer 1

Answer:

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 -

47x+31y=63 9x=2y+7 by elimination method by equating the co-efficient method.

Solution -

$47x+31y=63 \: \: ...(1) \\ 9x - 2y = 7 \: \: ...(2)$

Now, multiply by 2 in equation (1) and 31 in equation (2).

$94x + 62y = 126 \: \: ...(3) \\ 279x - 62y = 217 \: \: ...(4)$

Add both the equations,

$= > 373x = 343 \\ = > x = \frac{343}{373}$

Put the value of X in equation (2)

$= > 9 \times \frac{343}{373} - 2y = 7$

$= > \frac{3087}{373} - 2y = 7$

$= > \frac{(3087 - 746y)}{373} = 7$

$= > 3087 - 746y = 7 \times 373$

$= > 3087 - 746y = 2611$

$= > - 746y = - 476$

$= > y = \frac{119}{189}$