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Mathematics 16 Online
OpenStudy (anonymous):

how do i solve 2x+2y= for a rectangle

Parth (parthkohli):

I see. Are you calculating the perimeter?

OpenStudy (anonymous):

yes but i dont know how to work this

Parth (parthkohli):

Okay.. this one is easy ^.^ I'd try to explain my best right there

Parth (parthkohli):

So, x is the length of the rectangle and y is the width of the rectangle. If you are given an equation to solve, you must be given two of these - perimeter, length, width

OpenStudy (anonymous):

x an y and the prem is 34

Parth (parthkohli):

I'd give you an example. The perimeter is 40 cm and the width is 8 cm. Find the length. You'd have this equation. \( \color{Black}{\Rightarrow 2(x + 8) = 40}\) You'd now solve it \( \color{Black}{\Rightarrow x + 8 = 20}\) \( \color{Black}{\Rightarrow x = 12}\)

Parth (parthkohli):

But we can't solve it straight away, or this may have many solutions.

OpenStudy (anonymous):

???? lost

Parth (parthkohli):

Umm, are you sure you are given just the perimeter?

Parth (parthkohli):

What do you have to solve for exactly?

Parth (parthkohli):

Do you want a solution of x or y or just for the sake of isolation?

OpenStudy (anonymous):

dua... perm is 34 and x is 1 more than y

Parth (parthkohli):

Oh lol, now we can solve it ;)

Parth (parthkohli):

If you see it, x is 1 more than y. We can easily replace x with 1 + y because x is 1 more than y! \( \color{Black}{\Rightarrow 2((y + 1) + y) = 34 }\) \( \color{Black}{\Rightarrow 2(2y + 1 ) = 34}\) \( \color{Black}{\Rightarrow 2y + 1 = 17 }\) Can you continue from here?

OpenStudy (anonymous):

x= 18? y= 17

Parth (parthkohli):

Eek

Parth (parthkohli):

\( \color{Black}{\Rightarrow 2y = 17 - 1}\) \( \color{Black}{\Rightarrow 2y = 16}\) \( \color{Black}{\Rightarrow y = 8 }\) If you recall, x was 1 more than y.. so one more than 8 is 9. x = 9 y = 8

Parth (parthkohli):

If you try to solve it with the values, \(2(9 + 8) = 34\) \( \color{Black}{\Rightarrow 2(17) = 34}\) \( \color{Black}{\Rightarrow 34 = 34 }\) \(\Huge \checkmark \mathfrak{\text{CORRECT!}} \)

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