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# Rational Equations Using Proportions

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What if you had a rational equation like $\frac{x + 2}{x} - 2 = \frac{1}{x + 3}$ ? How could you solve it for x ? After completing this Concept, you'll be able to solve rational equations like this one using cross products and lowest common denominators.

### Guidance

A rational equation is one that contains rational expressions. It can be an equation that contains rational coefficients or an equation that contains rational terms where the variable appears in the denominator.

An example of the first kind of equation is: $\frac{3}{5}x+\frac{1}{2}=4$ .

An example of the second kind of equation is: $\frac{x}{x-1}+1=\frac{4}{2x+3}$ .

The first aim in solving a rational equation is to eliminate all denominators. That way, we can change a rational equation to a polynomial equation which we can solve with the methods we have learned this far.

Solve Rational Equations Using Cross Products

A rational equation that contains just one term on each side is easy to solve by cross multiplication. Consider the following equation:

$\frac{x}{5}=\frac{x+1}{2}$

Our first goal is to eliminate the denominators of both rational expressions. In order to remove the 5 from the denominator of the first fraction, we multiply both sides of the equation by 5:

$5 \cdot \frac{x}{5} &= 5 \cdot \frac{x+1}{2}\\x &= \frac{5(x+1)}{2}$

Now, we remove the 2 from the denominator of the second fraction by multiplying both sides of the equation by 2:

$2 \cdot x &= 2 \cdot \frac{5(x+1)}{2}\\2x &= 5(x+1)$

Then we can solve this equation for $x$ .

Notice that this equation is what we would get if we simply multiplied each numerator in the original equation by the denominator from the opposite side of the equation. It turns out that we can always simplify a rational equation with just two terms by multiplying each numerator by the opposite denominator; this is called cross multiplication.

#### Example A

Solve the equation $\frac{2x}{x+4}=\frac{5}{x}$ .

Solution

$\text{Cross-multiply. The equation simplifies to:} \qquad \qquad \quad 2x^2=5(x+4)\!\\\\\text{Simplify:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 2x^2=5x+20\!\\\\\text{Move all terms to one side of the equation:} \qquad \qquad \quad \ 2x^2-5x-20=0\!\\\\\text{Solve using the quadratic formula:} \qquad \qquad \qquad \qquad \quad x=\frac{5 \pm \sqrt{185}}{4} \Rightarrow \underline{\underline{x=-2.15}} \ \text{or} \ \underline{\underline{x=4.65}}$

It’s important to plug the answer back into the original equation when the variable appears in any denominator of the equation, because the answer might be an excluded value of one of the rational expressions. If the answer obtained makes any denominator equal to zero, that value is not really a solution to the equation.

Check: $\frac{2x}{x+4} = \frac{5}{x} \Rightarrow \frac{2(-2.15)}{-2.15+4} {\overset{?}=} \frac{5}{-2.15} \Rightarrow \frac{-4.30}{1.85} {\overset{?}=} -2.3 \Rightarrow -2.3=-2.3.$ The answer checks out.

$\frac{2x}{x+4} = \frac{5}{x} \Rightarrow \frac{2(4.65)}{4.65+4} {\overset{?}=} \frac{5}{4.65} \Rightarrow \frac{9.3}{8.65} {\overset{?}=} 1.08 \Rightarrow 1.08=1.08.$ The answer checks out.

Solve Rational Equations Using Lowest Common Denominators

Another way of eliminating the denominators in a rational equation is to multiply all the terms in the equation by the lowest common denominator. You can use this method even when there are more than two terms in the equation.

#### Example B

Solve $\frac{3}{x+2}-\frac{4}{x-5}=\frac{2}{x^2-3x-10}$ .

Solution

$\text{Factor all denominators:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{3}{x+2}-\frac{4}{x-5}=\frac{2}{(x+2)(x-5)}\!\\\\\text{Find the lowest common denominator:} \qquad \qquad \qquad \qquad \text{LCD} = (x+2)(x-5)\!\\\\\text{Multiply all terms in the equation by the LCD:}$

$(x+2)(x-5) \cdot \frac{3}{x+2}-(x+2)(x-5) \cdot \frac{4}{x-5}=(x+2)(x-5) \cdot \frac{2}{(x+2)(x-5)}$

$\text{The equation simplifies to:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad 3(x-5)-4(x+2)=2\!\\\\\text{Simplify:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 3x-15-4x-8=2\!\\\\{\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \underline{\underline{x=-25}}$

Check: $\frac{3}{x+2}-\frac{4}{x-5}=\frac{2}{x^2-3x-10} \Rightarrow \frac{3}{-25+2}-\frac{4}{-25-5} {\overset{?}=} \frac{2}{(-25)^2-3(-25)-10} \Rightarrow .003=.003.$ The answer checks out.

#### Example C

Solve $\frac{2x}{2x+1}+\frac{x}{x+4}=1$ .

Solution

$\text{Find the lowest common denominator:} \qquad \qquad \quad \text{LCD} = (2x+1)(x+4)\!\\\\\text{Multiply all terms in the equation by the LCD:}$

$(2x+1)(x+4) \cdot \frac{2x}{2x+1}+(2x+1)(x+4) \cdot \frac{x}{x+4}=(2x+1)(x+4)$

$\text{Cancel all common terms.} \qquad \qquad \qquad \qquad \qquad \ 2x(x+4)+x(2x+1)=(2x+1)(x+4)\!\\\text{The simplified equation is:} \!\\\!\\\text{Eliminate parentheses:} \quad \ \qquad \qquad \qquad \qquad \ \qquad 2x^2+8x+2x^2+x=2x^2+9x+4\!\\\!\\\text{Collect like terms:} \qquad \quad \ \qquad \qquad \qquad \qquad \ \qquad 2x^2=4\!\\\!\\{\;} \qquad \qquad \qquad \quad \qquad \qquad \qquad \qquad \qquad \qquad \ \qquad x^2=2 \Rightarrow \underline{\underline{x=\pm\sqrt{2}}}$

Check: $\frac{2x}{2x+1}+\frac{x}{x+4} = \frac{2\sqrt{2}}{2\sqrt{2}+1}+\frac{\sqrt{2}}{\sqrt{2}+4}=0.739+0.261=1.$ The answer checks out.

$\frac{2x}{2x+1}+\frac{x}{x+4} = \frac{2\left(-\sqrt{2}\right)}{2\left(-\sqrt{2}\right)+1}+\frac{-\sqrt{2}}{-\sqrt{2}+4}=1.547-0.547=1.$ The answer checks out.

Watch this video for help with the Examples above.

### Vocabulary

• For a quadratic equation in standard form, $ax^2 + bx + c = 0$ , the quadratic formula looks like this:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

### Guided Practice

Solve the following rational equations :

1. $\frac{3}{5}x+\frac{1}{2}=4$ .

2. $\frac{x}{x-1}+1=\frac{1}{x+3}$ .

Solutions:

1.

$\text{Start with the originl equation:} \qquad \qquad \qquad \quad \frac{3}{5}x+\frac{1}{2}=4\!\\\\\text{Multiply by the LCD:} \qquad \qquad \qquad 10 \cdot \left(\frac{3}{5}x+\frac{1}{2}\right)=10\cdot 4\!\\\\\text{Simplify:} \qquad \qquad \qquad \qquad \qquad 6x+5=40\!\\\\\text{Isolate x first by subtracting 5 from each side:} \qquad \qquad \quad \ 6x+5-5=40-5\!\\\\\text{Simplify:} \qquad \qquad \quad \ 6x=35\!\\\\\text{Isolate x by dividing by 6:} \qquad \qquad \quad \ \frac{6x}{6}=\frac{35}{6}\!\\\\\text{Simplify:} \qquad \qquad \quad \ x=5 \frac{5}{6}$

2.

$\text{Start with the originl equation:} \qquad \qquad \qquad \quad \frac{x}{x-1}+1=\frac{1}{x+3}.\!\\\\\text{Multiply by the LCD:} \qquad \qquad \qquad \quad (x-1)(x+3)\cdot\left(\frac{x}{x-1}+1\right)=(x-1)(x+3)\cdot\frac{1}{2x+3}.\!\\\\\text{Simplify:} \qquad \qquad \qquad \quad x(x+3)+1(x-1)(x+3)=1(x-1)\!\\\\\text{Distribute:} \qquad \qquad \qquad \quad x^2+3x+x^2+3x-1x-3=x-1\!\\\\\text{Combine like terms:} \qquad \qquad \qquad \quad 2x^2+5x-3=x-1\!\\\\\text{Set one side equal to 0:} \qquad \qquad \qquad \quad 2x^2+4x+-2=0$

Now we have a quadratic equation. Since it's not factorable (check for yourself!), we must use the quadratic formula.

$\text{Start with the quadratic formula.} \qquad x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\text{Substitute in the appropriate values.} \qquad x &= \frac{-4 \pm \sqrt{4^2 - 4(2)(-2)}}{2(2)}\\\text{Simplify.} \qquad x &= \frac{-4 \pm \sqrt{32}}{4}=\frac{-4 \pm 4\sqrt{2}}{4}=-1 \pm \sqrt{2}$

This means that $x=-1+\sqrt{2} \approx 0.4$ or $x=-1-\sqrt{2} \approx -2.4$

### Practice

Solve the following equations.

1. $\frac{2x+1}{4}=\frac{x-3}{10}$
2. $\frac{4x}{x+2}=\frac{5}{9}$
3. $\frac{5}{3x-4}=\frac{2}{x+1}$
4. $\frac{7}{x+3}=\frac{x+1}{2x-3}$
5. $\frac{7x}{x-5}=\frac{x+3}{x}$
6. $\frac{2}{x+3}-\frac{1}{x+4}=0$
7. $\frac{3}{2x-1}+\frac{2}{x+4}=2$
8. $\frac{2x}{x-1}-\frac{x}{3x+4}=3$
9. $\frac{x+1}{x-1}+\frac{x-4}{x+4}=3$
10. $\frac{x}{x-2}+\frac{x}{x+3}=\frac{1}{x^2+x-6}$
11. $\frac{2}{x^2+4x+3}=2+\frac{x-2}{x+3}$
12. $\frac{1}{x+5}-\frac{1}{x-5}=\frac{1-x}{x+5}$
13. $\frac{x}{x^2-36}+\frac{1}{x-6}=\frac{1}{x+6}$
14. $\frac{2x}{3x+3}-\frac{1}{4x+4}=\frac{2}{x+1}$
15. $\frac{-x}{x-2}+\frac{3x-1}{x+4}=\frac{1}{x^2+2x-8}$