SOLUTION: w(w+2)=a
(w+2)(w+4)=a+20
W^2+2w=a
w^2+6w+8=a+20
4w+8=20
4w=12
w=3
Please could someone show me how to check this problem I did ,I have to show a che
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Question 636429: w(w+2)=a
(w+2)(w+4)=a+20
W^2+2w=a
w^2+6w+8=a+20
4w+8=20
4w=12
w=3
Please could someone show me how to check this problem I did ,I have to show a check and I am lost please help
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
w(w+2)=a
(w+2)(w+4)=a+20
W^2+2w=a
w^2+6w+8=a+20
4w+8=20
4w=12
w=3
Please could someone show me how to check this problem I did ,I have to show a check and I am lost please help
Your w = 3 solution is correct. Good work!!
However, you now need to find the value of a, by substituting 3 for w in any of the equations with the 2 variables, a and w. Having found a and w, you can then substitute each value for each variable into any or all equations with the two variables to determine if those variable-values make those equations true.
Send comments and “thank-yous” to “D” at MathMadEzy@aol.com
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