how do i solve 2x+2y= for a rectangle
I see. Are you calculating the perimeter?
yes but i dont know how to work this
Okay.. this one is easy ^.^ I'd try to explain my best right there
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
x an y and the prem is 34
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 \( \color{Black}{\Rightarrow x + 8 = 20}\) \( \color{Black}{\Rightarrow x = 12}\)
But we can't solve it straight away, or this may have many solutions.
???? lost
Umm, are you sure you are given just the perimeter?
What do you have to solve for exactly?
Do you want a solution of x or y or just for the sake of isolation?
dua... perm is 34 and x is 1 more than y
Oh lol, now we can solve it ;)
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?
x= 18? y= 17
Eek
\( \color{Black}{\Rightarrow 2y = 17 - 1}\) \( \color{Black}{\Rightarrow 2y = 16}\) \( \color{Black}{\Rightarrow y = 8 }\) If you recall, x was 1 more than y.. so one more than 8 is 9. x = 9 y = 8
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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