I am trying to solve a problem and get down to w2+5w-4w-20=0 they simplified it to w(w+5)-4(w+5) cant figure out how the first came to the second equation
Hi, what exactly are we doing here?
well the full question is "a rectangles length is 2 units more than twice its width. its area is 40 square units. the equation w(2w+2)=40 can be used to find w, the width of the rectangle what is the width
The distributive property says \(a(b+c) = a(b) + a(c)\) and \(a(b-c) = a(b) - a(c)\) From ^ I think you can see how you came from the first equation to the second
okay i get it to that point
From there, you can simplify it from \(2(w^2+w-20)\) to \(2(w+5)(w-4)\)
Which makes your answer either -5 or 4 But since lengths can't be negative, it must be 4
okay so the 2 just disapears?
Well, it's \(2(w-4)(w+5)=0\) and as you should know \(ab=c\\a=c\\b=c\) So, yeah, we just get rid of the 2
gotch ya okay i think that helps!
thanks a lot everyone
yw~
i also have one more question "a rectangle's width is one-fourth of its length. its area is 9 square units. the equation (1/4)*L=9 can be used to find L (length). what is length
i thought it would be L=36
Well, the equation is actually \[L(\frac{1}{4}L)=9\]
ohhh that would help!!! so u square root it duh
didn't see the other L
square root it-?
like L= square root of 36
ohhh yeah I'm slow
nono its ok. maybe i need my eyes checked
thanks!
yw~
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