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shortie1515
Translate this problem to an equation and solve. The width of a rectangle is 6 inches less than twice its length. Find the dimension of the rectangle if its area is 108 square inches. @johnweldon1993
let length be l width= 2 * l - 6 let width be w l * w = 108 I think you can solve this now
Just as @sampatho2 has shown W = 2L - 6 A = L x W We know area 108 = L x W We know W = 2L - 6 so substitute that in for 'W' 108 = L x (2L - 6) can you take it from here...?
where are you facing the problem. You just need to solve the equation for l
but i still dont get it
Let x be the length and y the width. Then : \[2x-6=y\\xy=108\] So from the 2nd equation we have : \[x=\frac{108}{y}\] And from the 1st equation : \[2\frac{108}{y}-6=y\] We multiply by y to get : \[216-6y=y^2\] And then : \[y^2+6y-216=0\] so : \[\Delta=36-4\times(-216)=900\] So : \[y=\frac{-6+\sqrt{900}}{2}=\frac{-6+30}{2}=12 \text{ inches}\] And then : \[x=\frac{108}y=\frac{108}{12}=9 \text{ inches}\]
i dont get it still @johnweldon1993 @Noura11
Tell me what didn't understand in my solution and I will try to explain it to you ?
the whole thing tht u did its suppost to be like this type the answer in this from:6*7
Can you explain more , please ?