<span>2x-4y=32
2x-8y=48
--------------subtract
4y = - 16
y = -4
</span>2x-4y=32
2x- 4(-4)=32
2x + 16 = 32
2x = 16
x = 8
answer
(8, -4)
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
The answer is 1.79166666667.Hope it helps and have a great day!
You could not use as much as your using or save more to have more.
They would be more down and not as high
3,450
