# RATL 2 | Lesson 4 | Practice (Solving Rational Equations) Solutions

1. To solve simple rational equations, you can use cross multiplication. Use cross multiplication to solve the equations below.
1. $$\dfrac{x-1}{5}=\dfrac{9}{7}$$

$$7(x-1) = 5•9$$

$$7x-7 = 45$$

$$7x = 52$$  or $$x = \dfrac{52}{7}$$

1. $$\dfrac{x+2}{3}=\dfrac{4}{9}$$

$$9x+18 = 12$$

$$9x = -6$$

$$x = \dfrac{-2}{3}$$

1. What about more complicated equations?

Try finding common denominators as a first step in solving the next equations. Once all of the terms have the same denominator, solve the equation of the numerators.

1. $$\dfrac{3}{x+4}+\dfrac{1}{4}=5$$

LCD is the product of the two denominators or $$4x+16$$ because the denominators have no common factors.

Multiply the equation by $$4x+16$$. The resulting equation is

$$12+x+4 = 20x+80$$ or $$19x =-64$$

$$x = -\dfrac{64}{9}$$ or $$x = -3\dfrac{7}{19}$$

1. $$\dfrac{x}{x-5}+\dfrac{3}{x+2}=\dfrac{7x}{x^2-3x-10}$$

In order to find the LCD we need to factor $$x^2-3x-10$$, the factors are $$x-5$$ and $$x+2$$ therefore in looking at the three denominators we determine that the LCD is

$$(x-5)(x+2)$$.   Multiplying through by the LCD, the equation becomes:

$$x^2+2x+3x-15 = 7x$$.  Solving for $$x$$:  $$x^2-2x-15 = 0$$

Factoring: $$(x+3)(x-5) = 0$$ or $$x = -3$$ or $$x = 5$$.   But the restricted values are 5 and -2, so the only solution is $$x = -3$$.  See next page as to a brief explanation as to why the the equation had restricted values.