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Mathematics 18 Online
OpenStudy (anonymous):

The sum of the length and width of a rectangle is 29. The length is 15 less than 3 times the width. Find the dimensions of the rectangle

OpenStudy (anonymous):

okay, so you have two equations here, we'll let the length be the variable L, and the width the variable W. \[L+W=29\]\[L=3W-15\] There are many many ways to find out the value of both of them, the simplest in my mind is to solve one equation in terms of one variable then replace it in the other equation with what you've solved for, luckly that's already done for us in the 2nd equation: \[W+(3W-15)=29\] so now we can solve this to see what W is equal to, then sub that value back into the 1st equation. \[4W-15=29~~\rightarrow~~~4W=34~\rightarrow~~~W=\frac{34}{4}=\frac{17}{2}\] now sub that into our other equation: \[L=3\frac{17}{2}-15~~~\rightarrow~~~L=\frac{51-30}{2}=\frac{21}{2}\]

OpenStudy (anonymous):

L+W= 29 where L= 3W-15; => 3W-15+W=29 => 4W-15+15=29+15 => 4W= 44 => W=11 L=3W-15 => 3(11)- 15 => 33-15 => L= 18

OpenStudy (anonymous):

thank u both! :)

OpenStudy (anonymous):

hahahaha oh man, bad adding on mine...

OpenStudy (anonymous):

No worries, I got 34 the first time I did it as well. Always have to check answers! :)

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