Question

find the values of candy y in the following rectangle​

Answer 1

Answer:

$x + 3y = 13 - - - - - (1)\\ rectangle \: me \: aamne \: samne \: ki \: side \: eqaul \: hoti \: h \: \\ \\ 3x + y = 7 - - - - - - (2) \\ \\ 3 x + 9y = 39 \\ (3 \ \times eq. (1))(equation1 \: me \: 3 \: ka \: \\ multiply) \\ \\ equation \: (1) \: me \: equation(2) \: ko \: \\ subtract \: krne \: pr \\ 3x + 9y = 39 - - - (1) \\ 3x + y = 7 - - - (2) \\ \: \: \: - \\ - - - - - - - - - - - - - \\ 8y = 32 \\ = > y = \frac{32}{8} \\ (y = 4) \\ y \: ki \: value \: ko \: equation \\ (2) \: me \: put \: krne \: pr \: \\ 3x + y = 7 \\ = > 3x + 4 = 7 \\ = > 3x = 3 \\ \\ (x = 1)$

Step-by-step explanation:

x=1. y=4

Answer 2

The value of x and y in the following rectangle is 1 and 4

Solution :

Since we know that two lengths and breadths of the rectangle is equal. Hence ,

$x + 3y = 13.......(1) \\ 3x + y = 7........(2)$

Let's evaluate (1)

$x + 3y = 13 \\ x = 13 - 3y$

So now substitute the value of x in (2)

$3x + y = 7 \\ 3(13 - 3y) + y = 7 \\ 39 - 9y + y = 7 \\ 39 - 8y = 7 \\ - 8y = 7 - 39 \\ - 8y = - 32 \\ y = \frac{ - 32}{ - 8} \\ y = 4$

Now let's substitute the value of y in (1)

$x + 3y = 13 \\ x + 3(4) = 13 \\ x + 12 = 13 \\ x = 13 - 12 \\ x = 1$