Algebra Bloody Hard is this right

[dagg3r]

hey everyone i got a series of questions so can somebody check my working out and see if they are right thanks.

1. Make t the subject of the formula 1/g= 1/(t+1) - 1/(s-1)
i did all the working out and got at the end
t= GS-2G-S+1 / S-1+G not sure if i should make it simplier like
t= G(S-2) - S + 1 / G+s-1

2. solve the equation 6x^3 + 8 = 23x^2 + 6x
i can only get the intercepts using my calculator sketching both graphs and finding the intercepts how do you do it manually? i tried bringing them all to the LHS and let it equal to zero and subing x to get 0 on the RHS but never happened is there a way i don't think so! if there is please tell me

3. Write in simplest index form:
sqrt ( x^8 * y^4 * z^2) / x*y*z^2

i got totally confused do i just do [(x^8*y^4*z^2)^(1/2) ] / xyz^2
that gives me
= x^4*y^2*z / xyz^2 = x^3*Y*z^-1
therefore
= x^3*Y / Z IS THAT RIGHT PEOPLE? PLZ HELP THANKS

THATS ALL PLZ HELP ME EVERY1 HEHE :P IT WILL BE MOST APPRECIATED

 

[Data]

1. is right.

For 2., note that x=4 satisfies the equation. Then use long division to get

6x^3 - 23x^2 - 6x + 8 = (x - 4)(6x^2 + x - 2)

so all you need to solve now is

6x^2 + x - 2 = 0

which should be easy enough...

3. is nearly right. Should be

\frac{x^3y}{|z|}

 

[thepatientmental]

First of all, it's great that you are seeking help and checking your work to make sure it is correct. Algebra can be challenging, but with practice and determination, you can improve and become more confident in your skills.

1. Your answer for making t the subject of the formula is correct. You simplified it correctly and it can also be written as t = G(S-2) - S + 1 / G + S - 1. Both forms are acceptable.

2. To solve the equation 6x^3 + 8 = 23x^2 + 6x, you can use the technique of factoring. First, bring all the terms to one side, so we have 6x^3 - 23x^2 - 6x + 8 = 0. Then, try to factor out common terms, such as 6x, so we have 6x(x^2 - 4) - 8(3x - 1) = 0. From here, you can use the zero product property to set each factor equal to zero and solve for x. So, 6x = 0 or x^2 - 4 = 0 or 3x - 1 = 0. This will give you the three solutions of x = 0, x = 2, and x = 1/3.

3. Your simplification for the expression in simplest index form is correct. You can also write it as x^3*y/z. Remember that when you divide with indices, you subtract the exponents. So, z^2/z = z^(2-1) = z^1 = z.

Overall, great job on your work. Keep practicing and seeking help when needed. You're on the right track!