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

Exercise 31 Page 695

Can you rewrite the right-hand side of the equation so that it is a rational expression?

See solution.

Practice makes perfect
We are asked how we could use the Cross Products Property to solve the given equation. 1/x-2=2x-6/x+6+1 The Cross Products Property can be used when both sides of the equation are rational expressions. Recall that a rational expression is a fraction whose numerator and denominator are polynomials. In our case, the left-hand side of the equation is a rational expression but the right-hand side is not.
Expression Is It a Rational Expression?
1/x-2 Yes âś“
2x-6/x+6+1 No *
Before we can use the Cross Products Property, we have to rewrite the right-hand side of our equation as a rational expression. We can do it by rewriting 1 as a rational expression with a denominator of x+6. In doing so, we will obtain a sum of two rational expressions with a common denominator. Then we can add the expressions.
2x-6/x+6+1
2x-6/x+6+x+6/x+6
2x-6+x+6/x+6
3x/x+6
We rewrote the right-hand side of the equation as a rational expression. 1/x-2=2x-6/x+6+1 ⇔ 1/x-2=3x/x+6 The newly obtained form of our original equation allows us to use the Cross Products Property to solve it.

Extra

Solving the Equation
In case you were interested, here we have included the process of solving the given equation using the Cross Products Property. We will use the form of our equation with the right-hand side already rewritten as a rational expression.
1/x-2=3x/x+6
1(x+6)=3x(x-2)
x+6=3x(x-2)
x+6=3x^2-6x
â–Ľ
Simplify
6=3x^2-7x
0=3x^2-7x-6
3x^2-7x-6=0
We obtained a quadratic equation. To solve it, we need to use the Quadratic Formula. x=- b±sqrt(b^2-4 a c)/2 a The values a, b, and c correspond with the values of a quadratic equation written in standard form, ax^2+ bx+ c=0. In our case, a= 3, b= - 7, and c= - 6. We can substitute these values into the above formula and simplify.
x=- b±sqrt(b^2-4ac)/2a
x=-( - 7)±sqrt(( - 7)^2-4( 3)( - 6))/2( 3)
â–Ľ
Simplify right-hand side
x=7±sqrt((- 7)^2-4(3)(- 6))/2(3)
x=7±sqrt(49-4(3)(- 6))/2(3)
x=7±sqrt(49-12(- 6))/6
x=7±sqrt(49+72)/6
x=7±sqrt(121)/6
x=7±11/6
We got that the solutions for this equation are x= 7±116. Next, we should separate them into the positive and negative cases.
x=7±11/6
x_1=7+11/6 x_2=7-11/6
x_1=18/6 x_2=- 4/6
x_1=3 x_2=-2/3

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=3 and x_2=- 23. Note that since the original equation contained rational expressions, we have to check for extraneous solutions.

x_1=3 x_2=- 23
Substitute 1/3-2? =2( 3)-6/3+6+1 1/- 23-2? =2( - 23)-6/- 23+6+1
Simplify 1=1 -3/8=-3/8
Is Valid? Yes âś“ Yes âś“

Neither x_1=3 nor x_2=- 23 are extraneous, so these are the solutions of the given equation.