Question

Can someone help me with algebra 2 with an explanation?

Answer 1

Answer: This took me a long time to type, hope you read it and find it helpful and easy to understand...

1.
`x=1`

2.
`x=(3)/(2)`

3.
`x=1`

4.
`x=(6)/(5)`

Explanation:

1.

`(1)/(x)-(x-2)/(3x)=(4)/(3x)`

To make it easier to add / subtract fractions, we are looking for common denominators. I can see that if I move the fraction on the left with a denominator 3x to the right, I'll be able to add two equations with like denominator.

Let's add
`(x-2)/(3x)`

`(1)/(x)=(4)/(3x)+(x-2)/(3x)`

Add the numerators and keep the same denominator.

`(1)/(x)=(4+x-2)/(3x)`

`(1)/(x)=(2+x)/(3x)`

To get rid of the denominator x on the left side, we can multiply by x.

`(x)(1)/(x)=(2+x)/(3x)(x)`

`1=(2+x)/(3)`

Now multiply by 3.

`(3)1=(2+x)/(3)(3)`

`3=2+x`

Subtract 2.

`3-2=x\\1=x`

The value of x is 1.

Proof.

`(1)/(x)-(x-2)/(3x)=(4)/(3x)`

`(1)/(1)-(1-2)/(3(1))=(4)/(3(1))`

`1-(-1)/(3)=(4)/(3)`

`1+(1)/(3)=(4)/(3)`

`(1*3+1)/(3) =(4)/(3)\\(4)/(3)=(4)/(3)`

I can't show the proof to the following problems because I have a 5000 character limit.

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2.

`(5x-5)/(x^2-4x) -(5)/(x^2-4x)=(1)/(x)`

Again, we have like denominators, therefore, we can simply subtract the numerators and keep the same denominator.

`(5x-5-5)/(x^2-4x)=(1)/(x)`

`(5x-10)/(x^2-4x)=(1)/(x)`

Again, we can multiply by x to get rid of the x in the denominator.

`(x)(5x-10)/(x^2-4x)=(1)/(x)(x)`

Do not distribute the x in the numerator yet because we're gonna eliminate it later.

`((x)(5x-10))/(x^2-4x)=1`

Factor the denominator.

`((x)(5x-10))/((x)(x-4))=1`

Simplify x's.

`(5x-10)/(x-4) =1`

Multiply by x-4 to get rid of the denominator.

`(x-4)(5x-10)/(x-4) =1(x-4)`

`5x-10=x-4`

Add 10

`5x=x-4+10`

Subtract x

`5x-x=-4+10`

Combine like terms;

`4x=6`

Divide by 4.

`(4x)/(4)=(6)/(4)`

`x=(6)/(4)`

Simplify by 2.

6/2 = 3

4/2 = 2

`x=(3)/(2)`

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3.

`(x^2-7x+10)/(x)+(1)/(x)=x+4`

Same denominator, add numerators.

`(x^2-7x+10+1)/(x)=x+4`

`(x^2-7x+11)/(x)=x+4`

Multiply by x to get rid of the x.

`(x)(x^2-7x+11)/(x)=(x+4)(x)`

`x^2-7x+11=x^2+4x`

Subtract
`-x^2-4x`

`x^2-7x+11-x^2-4x=0`

Combine like terms;

`-11x+11=0`

Subtract 11.

`-11x=-11`

Divide by -11

`x=(-11)/(-11) \\x=1`

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4.

`(x^2+7x+10)/(5x-30)+(x)/(x-6)=(x^2-13x+40)/(5x-30)`

It's easier if you move the operation with denominator x+6 to the right side with negative sign and bring the operation on the right side to the left side with negative sign as well.

`(x^2+7x+10)/(5x-30)-(x^2-13x+40)/(5x-30)=-(x)/(x-6)`

Now, since we have the same denominators, we can simply subtract numerators.

`(x^2+7x+10-(x^2-13x+40))/(5x-30) =-(x)/(x-6)`

Distribute the negative sign.

`(x^2+7x+10-x^2+13x-40)/(5x-30) =-(x)/(x-6)`

Combine like terms;

`(20x-30)/(5x-30)=-(x)/(x-6)`

Factor.

I can see that I can factor a 5 on both numerator and numerator. This will allow me to simplify them.

`((5)(4x-6))/((5)(x-6)) =-(x)/(x-6)`

Simplify.

`(4x-6)/(x-6)=-(x)/(x-6)`

Multiply by x-6

`(x-6)(4x-6)/(x-6)=-(x)/(x-6)(x-6)`

This will simplify the denominators.

`4x-6=-x`

Add x and 6.

`4x+x=6\\`

Combine like terms;

`5x=6`

Divide by 5.

`x=(6)/(5)`