Mathematics, 20.12.2019 00:31 journeyhile5

Pl he m. this is my third time asking for . show all your work so i undersatnd how you did it alng with the answer

Pl he m. this is my third time asking for . show all your work so i undersatnd how you did it alng

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Answer from: khalaflaf9381

The solutions are $x=0 \text{ or } x=1$ .

I had no extraneous solutions (solutions that were contradictory).

Step-by-step explanation:

$\frac{x}{x-2}+\frac{x-1}{x+1}=-1$

First step: Notice the domain of the equation.

What values can definitely not take on.

$x \neq 2 \text{ or } -1$ since that would make the denominators of x-2 and x+1 zero.

That is x-2=0 when x=2 (added 2 on both sides).

That is x+1=0 when x=-1 (subtracted 1 on both sides).

So neither one of those can be solutions to the equation.

Second step: Multiply both sides by the least common multiple of the denominators, or a common multiple which happens to be the product of the denominators.

So I'm multiplying both sides by (x-2)(x+1) .

This gives me:

$(x-2)(x+1)\frac{x}{x-2}+(x-2)(x+1)\frac{x-1}{x+1}=(x-2)(x+1)(-1)$

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

Third step: We are going to do some distributive property here.

(x^2+x)+(x^2-2x-1x+2)=-1(x^2-2x+1x-2)

(x^2+x)+(x^2-2x-1x+2)=-x^2+2x-1x+2

Fourth step: Combine all like terms on left hand side and then do the right hand side.

2x^2-2x+2=-x^2+x+2

Fifth step: Move everything to one side so one of the sides is just equal to 0.

Subtract 2 on both sides:

2x^2-2x+2-2=-x^2+x+2-2

Simplify:

2x^2-2x=-x^2+x

Subtract on both sides:

2x^2-2x-x=-x^2+x-x

Simplify:

2x^2-3x=-x^2

Add x^2 on both sides:

2x^2-3x+x^2=-x^2+x^2

Simplify:

3x^2-3x=0

Sixth step: I'm going to factor and set any factors equal to right hand side,0. I will then solve those equations.

Factoring the left hand side:

I notice the left hand side has terms that each contain a factor of so I will factor that common factor out:

3x(x-1)=0

This gives us either: 3x=0 or x-1=0

The first equation can be solved by dividing both sides by 3 giving us x=0 .

The second equation can be solved by adding 1 on both sides giving us x=1 .

These do not contradict the 's we said could definitely not be so these do appear to be our solutions if we have made no mistake.

Just to verify I'm going to add one more step.

Seventh step: Let's verify our answer.

x=0 ?

$\frac{x}{x-2}+\frac{x-1}{x+1}=-1$ with x=0 :

$\frac{0}{0-2}+\frac{0-1}{0+1}=-1$

$\frac{0}{-2}+\frac{-1}{1}=-1$

0+-1=-1

-1=-1 is a true equation so we have verified x=0 is a solution.

x=1 ?

$\frac{x}{x-2}+\frac{x-1}{x+1}=-1$ with x=1 :

$\frac{1}{1-2}+\frac{1-1}{1+1}=-1$

$\frac{1}{-1}+\frac{0}{1}=-1$

-1+0=-1

-1=-1 is a true equation so we have verified x=1 is a solution.

Answer from: Quest

i believe that the answer is

b. y-1=-2(x-4)

hope this !

Answer from: Quest

22, 28, 34, 40, 46

b) f(n) = 6n + 16

n = what is the number of the row ( 22 would be row 1) so if we replaced n with 1, we would get 22

c) the 14th row will have 100 seats

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14)

22 | 28 | 34 | 40 | 46 | 52 | 58 | 64 | 70 | 76 | 82 | 88 | 94 | 100

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