Let x be the 1st odd number, and x+2 the second odd consecutive number:
(x)(x + 2) = 6[((x) + (x+2)] -1
x² + 2x = 6(2x + 2) - 1
x² + 2x = 12x +12 - 1
And x² - 10x - 11=0
Solve this quadratic expression:
x' = [+10 +√(10²- 4.(1)(-11)]/2 and x" = [+10 -√(10²- 4.(1)(-11)]/2
x' = [10 + √144]/2 and x" = [10 - √64]/2
x' = (10+12)/2 and x" = (10-12)/2
x = 11 and x = -1
We have 2 solutions that satisfy the problem:
1st for x = 11, the numbers at 11 and 13
2nd for x = - 1 , the numbers are -1 and +1
If you plug each one in the original equation :(x)(x + 2) = 6[((x) + (x+2)] -1
you will find that both generates an equlity
Answer:
64= 39+x
Step-by-step explanation:
x represents the unknown or in this case the money made to get to 64
39 is what she started with and 64 is the final total
Answer:
a = -6
Step-by-step explanation:
-5a-9+a=15
Combine like terms
-5a+a -9 = 15
-4a -9 = 15
Add 9 to each side
-4a -9+9 = 15+9
-4a = 24
Divide by -4
-4a/-4 = 24/-4
a = -6
200x 9, nine times two hundred,
Answer:
-2d-8
Step-by-step explanation:
