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

1) To solve simple rational equations, you can use cross multiplication. Use cross multiplication to solve the equations below.

a) $$\Large \frac{x-1}{5}=\frac{9}{7}$$

$$7(x-1) = 5•9$$

$$7x-7 = 45$$

$$7x = 52$$  or $$x = \frac{52}{7}$$

b) $$\Large \frac{x+2}{3}=\frac{4}{9}$$

$$9x + 18 = 12$$

$$9x = -6$$

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

2) 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.

a) $$\Large \frac{3}{x+4}+\frac{1}{4}=\normalsize 5$$

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 $$19x = -64$$

$$x = -64/9$$ 0r $$x = -3 7/19$$

b) $$\Large \frac{x}{x-5}+\frac{3}{x+2}=\frac{7x}{x^2-3x-10}$$

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.