solve for x: 1/2-x + 3/1-2x = 1

[susan pearson]

i am doing something wrong here. i have reviewed my text several times for this type of problem and i just can't seem to figure it out. can someone help? where am i going wrong?

1/2-x + 3/1-2x = 1

LCD is (1-2x)(2-x)

1/2-x (1-2x/1)(2-x/1) = 1-2x

3/1-2x (1-2x/1)(2-x) = 6-3x

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

so...1-2x+6-3x=2x^2-5x+2

which is 7-5x=2x^2-5x+2

then 0=2x^2-5

this does not solve for x...what am i doing wrong?

 

[tkhunny]

Re: solve for x

1/2-x + 3/1-2x = 1
LCD is (1-2x)(2-x)

What are you doing?

1) Add parentheses to clarify meaning.
2) You don't have to do everything in one swath.

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

Multiply by (2-x)

1 + 3*(2-x)/(1-2x) = (2-x)

Multiply by (1-2x)

(1-2x) + 3*(2-x) = (2-x)*(1-2x)

Can you continue?

 

[susan pearson]

Re: solve for x

1/2-x + 3/1-2x = 1
LCD is (1-2x)(2-x)

What are you doing?

1) Add parentheses to clarify meaning.
2) You don't have to do everything in one swath.

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

Multiply by (2-x)

1 + 3*(2-x)/(1-2x) = (2-x)

Multiply by (1-2x)

(1-2x) + 3*(2-x) = (2-x)*(1-2x)

Can you continue?

This is where I think I get lost. Do I leave the numbers factored out or do I multiply which brings me to 1-2x+6-3x=2-4x-1+2x^2. (7-3x=2x^2-4x+1) Yeah, this is where I get lost.

 

[stapel]

This is where I think I get lost. Do I leave the numbers factored out or do I multiply...?

This should have been covered in class. To solve a quadratic, you need to get "(quadratic) = 0". Only then do you factor. So, yes, you must first multiply everything out, combine the "like" terms, and get everything together on one side. Then apply whatever method you prefer (the Quadratic Formula, completing the square, etc) to solve the equation.

Note: (2 - x)(1 - 2x) = (2)(1) + (-x)(1) + (2)(-2x) + (-x)(-2x) = 2 - x - 4x + 2x² = 2 - 5x + 2x².

Eliz.