Question 696438: 3x/ x - 2 = 3x + 5/ x -1
Answer by persian52(161)   (Show Source):
You can put this solution on YOUR website! To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is x. Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions.
(3x)/(x)-2=3x*(x)/(x)-(5)/(x)-1*(x)/(x)
Multiply 3x by x to get 3x^(2).
(3x)/(x)-2=(3x^(2))/(x)-(5)/(x)-1*(x)/(x)
Multiply -1 by x to get -x.
(3x)/(x)-2=(3x^(2))/(x)-(5)/(x)-(x)/(x)
Combine the numerators of all expressions that have common denominators.
(3x)/(x)-2=(3x^(2)-5-x)/(x)
Reorder the polynomial 3x^(2)-5-x alphabetically from left to right, starting with the highest order term.
(3x)/(x)-2=(3x^(2)-x-5)/(x)
Remove the common factors that were cancelled out.
3-2=(3x^(2)-x-5)/(x)
Subtract 2 from 3 to get 1.
1=(3x^(2)-x-5)/(x)
Find the LCD (least common denominator) of 1+((3x^(2)-x-5))/(x).
Least common denominator: x
Multiply each term in the equation by x in order to remove all the denominators from the equation.
1*x=(3x^(2)-x-5)/(x)*x
Multiply 1 by x to get x.
x=(3x^(2)-x-5)/(x)*x
Cancel the common factor of x from the denominator of the first expression and the numerator of the second expression.
x=(3x^(2)-x-5)/(x)*x
Cancel the common factor of x from the denominator of the first expression and the numerator of the second expression.
x=(3x^(2)-x-5)
Remove the parentheses around the expression 3x^(2)-x-5.
x=3x^(2)-x-5
Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.
3x^(2)-x-5=x
Since x contains the variable to solve for, move it to the left-hand side of the equation by subtracting x from both sides.
3x^(2)-x-5-x=0
According to the distributive property, for any numbers a, b, and c, a(b+c)=ab+ac and (b+c)a=ba+ca. Here, x is a factor of both -x and -x.
3x^(2)+(-1-1)x-5=0
Subtract 1 from -1 to get -2.
3x^(2)+(-2)x-5=0
Remove the parentheses.
3x^(2)-2x-5=0
For a polynomial of the form ax^(2)+bx+c, find two factors of a*c (-15) that add up to b (-2).In this problem 1*-(5)/(3)=-(5)/(3) (which is (c)/(a)) and 1-(5)/(3)=-(2)/(3) (which is ((b)/(a)) , so insert 1 as the right hand term of one factor and -(5)/(3) as the right-hand term of the other factor.
(x+1)(x-(5)/(3))=0
Remove the fraction by multiplying the first term of the factor by the denominator of the second term.
(x+1)(3x-5)=0
Set each of the factors of the left-hand side of the equation equal to 0.
x+1=0_3x-5=0
Since 1 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 1 from both sides.
x=-1_3x-5=0
Set each of the factors of the left-hand side of the equation equal to 0.
x=-1_3x-5=0
Since -5 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 5 to both sides.
x=-1_3x=5
Divide each term in the equation by 3.
x=-1_(3x)/(3)=(5)/(3)
Cancel the common factor of 3 in (3x)/(3).
x=-1_(3x)/(3)=(5)/(3)
Remove the common factors that were cancelled out.
x=-1_x=(5)/(3)
The complete solution is the set of the individual solutions.
ANSWER:
x=-1,(5)/(3)
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