Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
5. Solving Rational Equations
Continue to next subchapter

Exercise 46 Page 697

Multiply both sides of the rational equation by the least common denominator.

9

Practice makes perfect
We want to solve the given rational equation. x-6/x+3+2x/x-3=4x+3/x+3 Let's begin by highlighting all of the different factors in the denominators. This will help us find the least common denominator (LCD). x-6/x+3+2x/x-3=4x+3/x+3We will solve the equation by multiplying each side by the LCD to clear denominators.
x-6/x+3+2x/x-3=4x+3/x+3
(x-6/x+3+2x/x-3)* (x+3) (x-3)=4x+3/x+3* (x+3) (x-3)
x-6/x+3* (x+3)(x-3)+2x/x-3* (x+3)(x-3)=4x+3/x+3* (x+3)(x-3)
(x-6)(x+3)(x-3)/x+3+2x(x+3)(x-3)/x-3=(4x+3)(x+3)(x-3)/x+3
(x-6)(x+3)(x-3)/x+3+2x(x+3)(x-3)/x-3=(4x+3)(x+3)(x-3)/x+3
(x-6)(x-3)+2x(x+3)=(4x+3)(x-3)
x(x-3)-6(x-3)+2x(x+3)=4x(x-3)+3(x-3)
â–Ľ
Distribute x & -6 & 2x & 4x & 3
x^2-3x-6(x-3)+2x(x+3)=4x(x-3)+3(x-3)
x^2-3x-6x+18+2x(x+3)=4x(x-3)+3(x-3)
x^2-3x-6x+18+2x^2+6x=4x(x-3)+3(x-3)
x^2-3x-6x+18+2x^2+6x=4x^2-12x+3(x-3)
x^2-3x-6x+18+2x^2+6x=4x^2-12x+3x-9
â–Ľ
Simplify
3x^2-3x+18=4x^2-9x-9
-3x+18=x^2-9x-9
18=x^2-6x-9
0=x^2-6x-27
x^2-6x-27=0
We obtained a quadratic equation. Let's identify the values of a, b, and c. x^2-6x-27=0 ⇕ 1x^2+( - 6)x+( -27)=0 We see that a = 1, b = - 6, and c = -27. Next, we will substitute these values into the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
x=- ( - 6)±sqrt(( - 6)^2-4( 1)( -27))/2( 1)
â–Ľ
Solve using the quadratic formula
x=6±sqrt((- 6)^2-4(1)(-27))/2(1)
x=6±sqrt(36-4(1)(-27))/2(1)
x=6±sqrt(36-4(-27))/2
x=6±sqrt(36+108)/2
x=6±sqrt(144)/2
x=6± 12/2
We will find the values for x by using the positive and the negative signs.
x=6± 12/2
x=6+ 12/2 x=6- 12/2
x=18/2 x=-6/2
x=9 x=-3

We found that x=9 and x=-3 are possible solutions. Now we have to check them!

Checking the Solutions

We check our solutions to see whether any of them are extraneous. To do so, we will substitute x=9 and x=-3 into the given equation. Let's start with x=9.
x-6/x+3+2x/x-3=4x+3/x+3
9-6/9+3+2( 9)/9-3? =4( 9)+3/9+3
â–Ľ
Simplify
9-6/9+3+18/9-3? =36+3/9+3
3/12+18/6? =39/12
3/12+36/12? =39/12
39/12=39/12 âś“
Since we obtained a true statement, x=9 is a solution of the equation. Let's now substitute x=-3 into the equation.
x-6/x+3+2x/x-3=4x+3/x+3
-3-6/-3+3+2( -3)/-3-3? =4( -3)+3/-3+3
â–Ľ
Simplify
-3-6/-3+3+-6/-3-3? =-12+3/-3+3
-9/0+-6/-6=-9/0 *
Note that a fraction with denominator equal to 0 is undefined, so we did not obtain a true statement. Therefore, x=-3 is an extraneous solution.