I am struggling with algebraic equations like...

(W)(w-5)-w-22=(w+5)(w-5) and solving for w

Answers
First thing you need to do is expand the brackets. Thusw^2-5w-w-22=w^2-25 Subtract w^2 from both sides. This gives you-6w-22=-25 Add 22 on both sides-6w=-3 Divide both sides by -6 (always divide with the coefficient of yhe variable -the number in front of the letter-)w=1/2
marias
16 November 2018
you will have to expand the brackets first, then take all the Ws to one side and then solve for w
Paul T.
17 November 2018
thanks!
y_do_i_do_school
17 November 2018
thanks!
y_do_i_do_school
17 November 2018
we have to open brackets firstw^2-5w-w-22 = w^2-25similar terms get cancelled-6w= -3so w= 3/6=1/2
deepakgupta
18 November 2018
W^2 -5w-w-22= w^2-25-6w=-3w=0.5 0r 1/2
Suhail A.
21 November 2018
(w2-5w)-w-22 = w2+5w-5w-25 = w2-25w2-6w-22 = w2 - 25-6w-22 = -25      -6w = -25+22-6w = -3w = 3/6 or .5
Kaustuv B.
21 November 2018
w×(w-5)-w-22=(w+5)×(w-5)> w^2 -5w-w-22= w^2 - 25> 6w=25-22=3; ( w^2 will be cancel from both side)> w=3/6=0.5
saikat1458007
28 November 2018
(w^2-5w)-w-22 = w^2+5w-5w-25 = w^2-25w^2-6w-22 = w^2 - 25-6w-22 = -25      -6w = -25+22-6w = -3w = 0.5
Joe W.
28 November 2018
its simple :) we want to find the values of the unknown variables: in this case it is w..so we need to find a way to get w on its own..Steps:1. Expand2. Group like terms (w)3. Solve
Hannah Y.
28 November 2018
I can help
kumar55nishant
28 December 2018
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