# Math questions pre algebra?

My Math teacher gave us some math problems and told us to google it so see what other people came up with.... here goes....

Find the Domain
1. g(x)= 1/3x-3

2. f(x)= (the square root sign) -x-2

Solve...
1. x^3=x

2. x^2=-5x=24

Simplify
1. x^2+2x-3/x-5

2. x^3-27/x-3

3. (2x^-5y^-1)-3/(-5x^-3y^-4)-2

4. (x^-2y^3/-2x^-5y^7)^-3

1.

g(x) = 1 / 3x - 3

We can find the domain then

3x - 3 = 0

3x - 3 + 3 = 3 + 0

3x = 3

Divide 3 to each side

3x / 3 = 3 / 3

x  = 1

There the domain is x = 1.

Domain of is all real numbers.

Domain of is

f(x) = the square root sign -x  - 2

f(x) =sqrt(-1(x + 2))

Substitute i^2 = -1 in the f(x)

f(x) = sqrt(i^2 ( x + 1)

f(x) = isqrt (x + 1)

We can find the Domain sqrt(x + 1) = 0

Take square each side

x + 1 = 0

Subtract from 1 to each side

x + 1 -  1 = 0 - 1

x + 0 = -1

x = -1

The domain is x = -1.

Domain of is

Solve x3 = x

Subtract x to each side

x3 - x = 0 - x

x3 - x = 0

Take out common term x in x3 - x = 0

x(x2 - 1) = 0

x(x2 - 12) = 0

Recall : a2 - b2 = (a - b)(a + b)

Substitute a = x  and b = 1 in the x2 - 12

x(x - 1)(x + 1) = 0

It have 3 roots

x = 0 or x - 1 = 0 or x + 1 = 0

x - 1 = 0

x - 1 + 1 = 0 + 1

x - 0 = 1

x = 1

x + 1 = 0

Subtract 1 from each side

x + 1 - 1 = 0 - 1

x + 0 = -1

x = -1

The solution of the equation : -1 , 0 , 1.

Solve x3 = x

Subtract x to each side

x3 - x = 0 - x

x3 - x = 0

Take out common term x in x3 - x = 0

x(x2 - 1) = 0

x(x2 - 12) = 0

Recall : a2 - b2 = (a - b)(a + b)

Substitute a = x  and b = 1 in the x2 - 12

x(x - 1)(x + 1) = 0

It have 3 roots

x = 0 or x - 1 = 0 or x + 1 = 0

x - 1 = 0

x - 1 + 1 = 0 + 1

x - 0 = 1

x = 1

x + 1 = 0

Subtract 1 from each side

x + 1 - 1 = 0 - 1

x + 0 = -1

x = -1

The solution of the equation : -1 , 0 , 1.

x2 -5x = 24

Subtract 24 to each side

x2 + 5x  -24= 24 - 24

x2 + 8x - 3x  -24 = 0

Take out common term x  and -3 in the above equation

x(x + 8) -3(x +8) = 0

Take out common term x + 8 in the above equation

(x + 8)(x - 3) = 0

x + 8 = 0 or x - 3 = 0

If x + 8 = 0

Subtract 8 from each side

x + 8 - 8 = -8

x + 0 = -8

x = -8

If x - 3 = 0

x -3 + 3 = 0 + 3

x - 0 = 3

x = 3.

The solution of the equation : -8 , 3.

Solutions of the equation x2 -5x = 24 are x = -3 , 8.

Simplify 1. x^2 + 2x - 3 / x - 5 (x^2 + 3x - x - 3) / x - 5 Take out common term x and -1 in the above expression x(x + 3) -1(x + 3) / x - 5 Take out common term x + 3 x + 3[x - 1] / x - 5 (x + 3)(x - 1) / x - 5.

Simplify

2.

x3  - 27 / x - 3

=x3 - 33 / x - 3

Recall : (a3 - b3) = ( a - b)(a2 + ab + b2)

Substitute x = a  and b = 3 in the above formula

(x - 3)(x2 + 3x + 32) / (x - 3)

= (x2 + 3x + 9).

4.

(x-2y3 / -2x-5y7)-3

Apply the power rule (a / b)-3 = a-3 / b-3

(x-2y3)-3 / (-2x-5y7)-3

Apply the power rule (ab)-3 = a-3 b-3

x6y-9 / -2x15y-21

Apply the power rule

y-9+21 / -2x15-6

y12 / -2x9

-1/2(y12 / x9)

-1/2(y12 x-9).

3.

(2x-5y-1) -3 / (-5x-3y-4) - 2

Apply the power rule

((2 / x5y) - 3) / (-5 / x3y4) - 2

(2 - 3x5y / x5y) / (-5 - 2x3y4 / x3y4)

Simplify

(x3y4 / x5y)(2 - 3x5y / -(5 + 2x3y4))

Apply the power rule

(y4-1 / x5-3)(2 - 3x5y / -(5 + 2x3y4))

-(y3 / x2)(2 - 3x5y / 5 + 2x3y4)

[-y3(2 - 3x5y) / x^2(5 + 2x3y4)] .

Siimplify

4) Given ( x ^-2 y ^ 3 / -2x  ^-5 y ^ 7 ) ^ -3

Apply the power rule (a / b) ^ -3 = a ^ -3 / b ^ -3

( x ^ -2 y ^ 3 ) ^ -3 / ( -2 x ^ -5 y ^ 7 ) ^ -3

Apply the power rule ( ab )^ -3 = a ^ -3 * b ^ -3

x ^ 6 y ^ -9 / ( ( -2 ) ^ -3 x ^ 15 y ^ -21 )

Apply the power rule

x ^ 6 y ^ -9 / ( -1/8) x ^ 15 y ^ -21

- 8 y ^ - 9 + 21 / x ^ 15 - 6

- 8 y ^ 12 / x ^ 9

- 8 y ^ 12 x ^ -9