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 28 Page 695

Multiply the numerator on the left-hand side by the denominator on the right-hand side, and the denominator on the left-hand side by the numerator on the right-hand side.

4

Practice makes perfect

We will solve the given equation and then check the solutions. 2x+4/x-3=3x/x-3 Let's do these two things one at a time!

Solving the Equation

We will solve the rational equation by using the Cross Products Property. If a numerator or denominator contains addition or subtraction, be sure to treat each one as a parenthetical factor in the cross multiplication process.
2x+4/x-3=3x/x-3
(2x+4)(x-3)=3x(x-3)
2x(x-3)+4(x-3)=3x(x-3)
â–Ľ
Distribute 2x & 4 & 3x
2x^2-6x+4(x-3)=3x(x-3)
2x^2-6x+4x-12=3x(x-3)
2x^2-6x+4x-12=3x^2-9x
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Simplify
- 6x +4x-12=x^2-9x
- 2x-12=x^2-9x
-12=x^2-7x
0=x^2-7x+12
x^2-7x+12=0
We obtained a quadratic equation. Let's identify the values of a, b, and c. x^2-7x+12=0 ⇕ 1x^2+( - 7)x+ 12=0 We see that a = 1, b = - 7, and c = 12. Next, we will substitute these values into the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
x=- ( - 7)±sqrt(( - 7)^2-4( 1)( 12))/2( 1)
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Solve using the quadratic formula
x=7±sqrt((- 7)^2-4(1)(12))/2(1)
x=7±sqrt(49-4(1)(12))/2(1)
x=7±sqrt(49-4(12))/2
x=7±sqrt(49-48)/2
x=7±sqrt(1)/2
x=7± 1/2
We will find the values for x by using the positive and the negative signs.
x=7± 1/2
x=7+ 1/2 x=7- 1/2
x=8/2 x=6/2
x=4 x=3

We found that x=4 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=4 and x=3 into the given equation. Let's start with x=4.
2x+4/x-3=3x/x-3
2( 4)+4/4-3? =3( 4)/4-3
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Simplify
8+4/4-3? =12/4-3
12/1=12/1 âś“
Since we obtained a true statement, x=4 is a solution of the equation. Let's now substitute x=3 into the equation.
2x+4/x-3=3x/x-3
2( 3)+4/3-3? =3( 3)/3-3
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Simplify
6+4/3-3? =9/3-3
10/0=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.