Question

one rectangle has a length of (x +2 )cm and a breadth of 2cm, and another, of equal area, has a length of 5cm and a breadth of (x-3)
1 write down an equation in x and solve it
2 what is the area of each rectangle

Answer 1

please you have to solve this equation:
\[(x+2)*2=5*(x-3)\]

Answer 2

help?

Answer 3

I'm going to simpify the above equation:
\[2x+4=5x-15\]
, ading to both sides 15 we have:
\[2x+19=5x\]
now I add to both sides of last equation -2x, so:
\[19=3x\]
finallyividing by 3, the last equation, I get:
\[x=\frac{ 19 }{ 3 }\]

Answer 4

what about part 2

Answer 5

next step is to insert the value of x, namely 19/3 into the left side, for example, of the first equation, so, we get:
\[(x+2)*2=(\frac{ 19 }{ 3 }+2)*2=\frac{ 50 }{ 3 }\]
so your area is 50/3 cm^2

Answer 6

why do we add 2 to 19/3

Answer 7

because your rectangle has a length equals to x+2, and, solving our equation, we got x=19/3, so the length of first rectangle is 19/3+2=25/3

Answer 8

and is this equal to the other rectangle (x-3)

Answer 9

not precisely, the area is equal for both rectangles, the single dimensions not, for example the breadth of the second rectangle is x-3=19/3-3=10/3, the breadth of the first rectangle is 2 cm instead, similar for the length

Answer 10

for example length of the first rectangle is x+2=19/3+2=25/3, whereas length of the second rectangle is 5 cm

Answer 11

so the area of the rectangle with the length (x+2) is 50/3cm squared
the area of the other rectangle is 20/3 cm squared ?

Answer 12

I think no, cause boh areas ae equal to 50/3. Area of secon rectangle is:
length*breadth=
\[5*(x-3)=5*(\frac{ 19 }{ 3 }-3)=5*\frac{ 10 }{ 3 }=\frac{ 50 }{ 3 }\]

Answer 13

please, note that from the text of your problem I read that both rectangles have the same area

Answer 14

yep thank you ever so much for your help michele !!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 15

thank you!

Answer 16

no, thank uuuuuuuuuuuuuuuuuuuuuuuuuu, hope u have a good night, bye