Problem Solving

x (x- 5x +6 )=0
------
x

I'm not understanding how the denominator x is cancelled out and the is x = 6 or -1.


(4^y + 4^y + 4^y + 4^y)(3^y+3^y+3^y) = (12)^y+1

Not sure how you get y+1 as the ecponent

Thanks!

I'm not sure what your question is.
Nonetheless, to solve the equation x (x- 5x +6 )=0
we can factorise it:
x(x-2)(x-3)= 0
So x = 0, 2 and 3

Mathsbuddy wrote:I'm not sure what your question is.
Nonetheless, to solve the equation x (x- 5x +6 )=0
we can factorise it:
x(x-2)(x-3)= 0
So x = 0, 2 and 3

Ah, I see the left side is divided by x.

x/x = 1, so the equation is just (x-2)(x-3)= 0
So x = 2 and 3

For the second part of your question:

Here is an Index Law formula:
x^a x x^b = x^(a+b)

So:
(4^y + 4^y + 4^y + 4^y)= 4^y x 4 = 4^y x 4^1 = 4^(y+1)

and (3^y+3^y+3^y) = 3^y x 3^1 = 3^(y+1)

Multiply these together:
4^(y+1) x 3^(y+1) = 4 x 4 x 4 x 4... [(y+1) times] x 3 x 3 x 3 x 3 ... [(y+1) times] = 12 x 12 x 12 x 12 x 12 ... [(y+1) times]

= 12^(y+1)

I hope this helps

[email protected] wrote:x (x- 5x +6 )=0
------
x

I'm not understanding how the denominator x is cancelled out and the is x = 6 or -1.

Thanks!

Please note that Denominator x can be cancelled out here as it's an EQUATION and not an Inequation

In case of Inequation X can be cancelled without any change only if X is positive otherwise Inequation sign changes while cancelling x if x to be cancelled is Negative

[email protected] wrote:

(4^y + 4^y + 4^y + 4^y)(3^y+3^y+3^y) = (12)^y+1

Not sure how you get y+1 as the ecponent

Thanks!

[email protected] wrote:
(4^y + 4^y + 4^y + 4^y)(3^y+3^y+3^y) = (12)^y+1

Not sure how you get y+1 as the ecponent

Thanks!

(4^y + 4^y + 4^y + 4^y)(3^y+3^y+3^y) = (12)^y+1

Take 4^y common from all terms in first bracket and 3^y from all terms in other bracket on left hand side

4^y(1 + 1 + 1 + 1)3^y (1 + 1 + 1)
= 4^y(4)3^y (3)
= 4^(y+1)3^(y+1)
= 12^(y+1)