Question

a rectangle figure has length 3x-y and 2x+y and width is 2x-3 and perimeter of rectangle is 120 cm. find the area of rectangle and values of x and y involving simultaneous equation. plxx help....

Answer 1

Wait, is this a three dimensional shape or 2 dimensional? Because a rectangle can't have 2 different lengths...

Answer 2

this rectangle has 2 different lengths

Answer 3

here
we have
3x-y= 2x+y or x= 2y
given
3x-y + 2x+y +2(2x-3) = 120
5x +4x -6 =120
9x = 126
x= 126/9
x=34

Answer 4

this isnt correct answer the value of x is14cm and y is 7 cm and the area is 875 cm square did u involve smultaneous equation? i know the answers not the procedure... i will give u the shape wait

Answer 5

the procedure of matricked is correct,. may be its typo for the value of x..

Answer 6

yup you are correct
here i claculated wrong
x= 126/9 x=34

but actually x=126/9 =14
hence y=14/2
=7

Answer 7

x=14
y= x/2 =14/2 =
7 thus width = 2*14 -3 = 28-3 =25

Answer 8

y u subtracted 3 when finding the width?

Answer 9

length = 3*14-7 =35
hence reqd area = 35*25 = 875

Answer 10

can u help me understand it? plx

Answer 11

@Luckyyyy thanks for correcting my claculation error

Answer 12

width is
2x-3
2*14-3 =25

Answer 13

oh yeah and what r the equations? i mean two equations? u did involve simultaneous equation right?

Answer 14

x= 2y

3x-y + 2x+y +2(2x-3) = 120

Answer 15

3x-y= 2x+y or x= 2y

Answer 16

thnx alotttt i have become your fan now..

Answer 17

yw