Simplify x + 5 / x^2 + 3x + 2 plus x – 7 / x^2 + x - 2
State the restrictions.
2006-11-23 00:53:46 · 5 answers · asked by thomasgraham880 1 in Science & Mathematics ➔ Mathematics
Factor the denominators
x^2 + 3x + 2 = (x+2)(x+1)
x^2 + x - 2 = (x+2)(x-1)
The common denominator must have each factor that occurs in either denominator, so it is (x+2)(x+1)(x-1)
The first fraction is missing the (x-1) so multiply top and bottom by (x-1). Do not multiply out the bottom, just the top:
(x+5)(x-1) = x^2 + 4x - 5
The second fraction is missing the (x+1) so multiply top and bottom by (x+1) (again just do out the top)
(x-7)(x+1) = x^2 - 6x - 7
Now since both fractions have the same denominator, add the like terms on top. This will be 2x^2 - 2x - 12
This simplifies to 2(x^2 - x - 6) which is 2(x-3)(x+2)
The (x+2) on top cancels the (x+2) on the bottom.
Final answer: [2(x-3)] / [(x+1)(x-1)]
The restrictions are any values for x that cause the denominator to be zero. This is before reducing! So x cannot be 1, -1, or -2.
2006-11-23 01:12:34 · answer #1 · answered by hayharbr 7 · 1⤊ 0⤋
((x + 5)/(x^2 + 3x + 2)) + ((x - 7)/(x^2 + x - 2))
((x + 5)/((x + 2)(x + 1))) + ((x - 7)/((x + 2)(x - 1)))
Multiply everything by (x + 2)(x + 1)(x - 1)
(((x + 5)(x - 1)) + ((x - 7)(x + 1)))/((x + 2)(x + 1)(x - 1))
((x^2 - x + 5x - 5) + (x^2 + x - 7x - 7)) / ((x + 2)(x + 1)(x - 1))
(x^2 + 4x - 5 + x^2 - 6x - 7)/((x + 2)(x + 1)(x - 1))
(2x^2 - 2x - 12)/((x + 2)(x + 1)(x - 1))
(2(x^2 - x - 6))/((x + 2)(x + 1)(x - 1))
(2(x - 3)(x + 2)))/((x + 2)(x + 1)(x - 1))
(2(x - 3))/((x + 1)(x - 1))
(2x - 6)/(x^2 - 1)
The restrictions are -2, -1, and 1, although the (x + 2) gets taken out, -2 still won't work.
2006-11-23 11:02:28 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
From left to right combine like terms. So add the x with the 3x and x, then add the two fractions to get -2/x^2 and then the numbers cancel each other. That's it, unless your teacher wants you to get the x^2 off the bottom. If so multiply each term by x^2, then you are done.
2006-11-23 01:07:55 · answer #3 · answered by Dave & T 1 · 0⤊ 0⤋
(x + 5 / x^2 + 3x + 2 )+( x – 7 / x^2 + x - 2)
collect like terms
6x -2x^(-2)
2006-11-23 01:03:24 · answer #4 · answered by mathman241 6 · 0⤊ 0⤋
step 1:x^2+3x+2 cannot be zero so (x+1)(x+2)!=0so x cannot be -1 or -2thenx^2 + x - 2 cannot be zero so (x-1)(x+2)!=0so x cannot be 1 or -2,
statment eQuals (x+5)(x-1)+(x-7)(x+1)/(x-1)(x+1)(x+2)==2x^2-12-12x/(x-1)(x+1)(x+2)== [2(x-3)(x+2)] / [(x+1)(x-1)(x+2)]== [2(x-3)] / [(x+1)(x-1)]
2006-11-23 01:37:28 · answer #5 · answered by Lady 2 · 0⤊ 0⤋