Question

Find the values of x and y if x + 3y, 7, 13,3x +y are sides of a
rectangle taken in order.

Answer 1

Answer:

We know that opposite sides of a rectangle are equal

Then,let x+3y=13

              x=13-3y............(i)equation

And,

               3x+y=7.............(ii)equation

Putting x=13-3y in (ii)equation,

                3(13-3y)+y=7

                39-9y+y=7

                 39-8y=7

                 39-7=8y

                  32=8y

                  y=32/8

                  y=4  

Putting, y=4 in (i)equation

x=13-3y

x=13-3(4)

x=13-12

x=1

Put the values you'ill get the sides

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Sides of rectangle= x+3y, 7, 13, 3x+y

Consider it to be the sides of a rectangle ABCD.

Since the sides are in order, they are:

AB=x+3y

BC=7

CD=13

AD=3x+y

Opposite sides of a rectangle are equal......

Hence, AB=CD

x+3y=13

3y=13-x

y=13-x/3.......(Eq.1)

Hence, y= 13-x/3

AD=BC

3x+y=7

3x+13-x/3=7   (From Eq.1)

9x/3+13-x/3=7

9x-x+13/3=7

8x+13/3=7

8x+13=7 x 3

8x+13=21

8x=21+13

8x=8

x=8/8

x=1/1=1

Hence, value of x=1

            value of y=13-x/3  (From Eq.1)

            value of y=13-1/3

            value of y=12/3=4

            value of y=4