Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
5. Solving Rational Equations
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Exercise 14 Page 695

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

1, 4

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We want to solve the given rational equation. 2/c-2=2-4/c For simplicity, we will start by replacing the variable c with the variable x. 2/c-2=2-4/c c= x âź¶ 2/x-2=2-4/x Let's begin by highlighting all of the different factors in the denominators. This will help us find the least common denominator (LCD). 2/x-2=2-4/x We will solve the equation by multiplying each side by the LCD to clear denominators.
2/x-2=2-4/x
2/x-2 * x (x-2) = (2-4/x) * x (x-2)
2/x-2 * x(x-2) = 2x(x-2)-4/x * x(x-2)
2x(x-2)/x-2 = 2x(x-2)-4x(x-2)/x
2x(x-2)/x-2 = 2x(x-2)-4x(x-2)/x
2x=2x(x-2)-4(x-2)
â–Ľ
Distribute 2x & 4
2x=2x^2-4x-4(x-2)
2x=2x^2-4x-4x+8
â–Ľ
Simplify
2x=2x^2-8x+8
0=2x^2-10x+8
0=x^2-5x+4
x^2-5x+4=0
Note that we have a quadratic equation now. Let's identify the values of a, b, and c. x^2-5x+4=0 ⇔ 1x^2 +( -5)x+4=0 We have that a= 1, b= -5, and c=4. Let's substitute these values into the Quadratic Formula and solve for x.
x=- b±sqrt(b^2-4ac)/2a
x=- ( -5)±sqrt(( -5)^2-4( 1)(4))/2( 1)
â–Ľ
Simplify right-hand side
x=5±sqrt((-5)^2-4(1)(4))/2(1)
x=5±sqrt(25-4(1)(4))/2(1)
x=5±sqrt(25-4(4))/2
x=5±sqrt(25-16)/2
x=5±sqrt(9)/2
x=5± 3/2
Let's calculate both solutions by using the positive and negative signs.
x=5± 3/2
x=5+ 3/2 x=5- 3/2
x=8/2 x=2/2
x=4 x=1
Therefore, the solutions are x=4 and x=1. Recall that, at the beginning of the solution, we substituted x for c. Therefore, the solutions to the original equation are c=4 and c=1. x=4 &x=c âź¶ c=4 [0.69em] x=1 &x=c âź¶ c=1 Let's check them to see if we have any extraneous solutions. To do this, we need to substitute 4 and 1 for c in the original equation. Let's start with c=4.
2/c-2=2-4/c
2/4-2? =2-4/4
â–Ľ
Simplify
2/2? =2-4/4
1? =2-1
1=1 âś“
Now, we will check our second solution, c=1.
2/c-2=2-4/c
2/1-2? =2-4/1
â–Ľ
Simplify
2/-1? =2-4/1
-2/1? =2-4/1
-2? =2-4
-2=-2 âś“
Neither of our solutions is an extraneous solution. Therefore, the solutions to the given equation are c=4 and c=1.