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Topic: **How to solve problem solving math****Question:**
Wel, my teacher discussed it but I dunno what I did But I Can't seemingly figure it out now..
Anyways, Here is the problem:
Solve for the Value of "X"
Here is the pic for you..
http://imageshack.us/photo/my-images/856/unledhkz.png/
What you're gonna is solve for the value of X and then Get the width and length and that is equal to the given area.
Can you Explain how you do it?

June 20, 2019 / By Layton

(x+1)(x-2) = 18 x^2 - x - 2= 18 x^2 - x - 20 = 0 (x+4)(x-5) = 0 x = -4, 5 but length cannot be negative so x = 5

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Did you like the answer? We found more questions related to the topic: **How to solve problem solving math**

-y + 3y - 6y + 12y 2y - 6y + 12y -4y + 12y 8y ====== free to e-mail if have a question

I do not think that you are correct. In an attempt to show you how this is done, I will eliminate on term at a time and I will start with the negative terms first: -y+3y-y+12y = 2y-y+12y = y+12y = 13y I hope that this helps. :)

x = 5 (a+b)x2=Area a=x+1 b=x-2 {( x+1) + (x-2)} x 2 = Area (6 + 3) x 2 = 18

👍 90 | 👎 6

(x+1)(x-2) = 18 x^2 - x - 2= 18 x^2 - x - 20 = 0 (x+4)(x-5) = 0 x = -4, 5 but length cannot be negative so x = 5

x = 5 (a+b)x2=Area a=x+1 b=x-2 {( x+1) + (x-2)} x 2 = Area (6 + 3) x 2 = 18

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