A rectangular city lot has an area of 1000 square meters. If the length of the lot is 10 meters more than twice its width, how do you find its length and width?

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Answer 1 · 2017-04-27T06:54:46

Length of the Area $= 50 m t r s$ $=50 mtrs$

Width of the Area $= 20 m t r s$ $=20 mtrs$

Let the $x$ $x$ be the length and $y$ $y$ be the width of the Area

So we can write

$x = 2 y + 10$ $x=2y+10$

and

$x y = 1000$ $xy=1000$

or

$(2 y + 10) y = 1000$ $(2y+10)y=1000$

or

$2 y^{2} + 10 y = 1000$ $2y^2+10y=1000$

or

$2 y^{2} + 10 y - 1000 = 0$ $2y^2+10y-1000=0$

or

$y^{2} + 5 y - 500 = 0$ $y^2+5y-500=0$

or

$y^{2} + 25 y - 20 y - 500 = 0$ $y^2+25y-20y-500=0$

or

$y (y + 25) - 20 (y + 25) = 0$ $y(y+25)-20(y+25)=0$

or

$(y + 25) (y - 20) = 0$ $(y+25)(y-20)=0$

or

$y + 25 = 0$ $y+25=0$ or $y - 20 = 0$ $y-20=0$

or

$y = - 25$ $y=-25$ Invalid value contains negative sign

or

$y = 20$ $y=20$

So Width of the Area is $20$ $20$ ----------Ans

By plugging the value of $y = 20$ $y=20$ in the eqn $x = 2 y + 10$ $x=2y+10$

we get

Length of the Area $= 2 \times 20 + 10 = 50$ $=2times20+10=50$ -------------Ans