Solving Absolute Value Equations »

Equations with no parentheses

Example 1

Solve 5x - 4 - 2x + 3 = -7 - 3x + 5 + 2x

Solution 1

Step 1: Combine the similar terms	$5x - 4 - 2x + 3 = - 7 - 3x + 5 + 2x$
	$3x - 1 = - x - 2$
Step 2: Add x to both sides.	$3x - 1 + x = - x - 2 + x$
	$4x - 1 = - 2$
Step 3: Add 1 to both sides.	$4x - 1 + 1 = - 2 + 1$
	$4x = - 1$
Step 4: Divide both sides by 4:	$\frac{{4x}}{4} = \frac{{ - 1}}{4}$
Solution is:	$x = - \frac{1}{4}$

Exercise 1: Solve equations

Equations with parentheses

Example 2

Solve $2 (x - 4) + 4 (2 - x) = 5x - 4 (x + 1)$

Solution 2

Step 1: Simplify both sides:	$2 (x - 4) + 4 (2 - x) = 5x - 4 (x + 1)$
	$2x - 8 + 8 - 4x = 5x - 4x - 4$
	$- 2x = x - 4$
Step 2: Subtract x from both sides.	$- 2x - x = x - 4 - x$
	$- 3x = - 4$
Step 3: Divide both sides by -3:	$\frac{{ - 3x}}{{ - 3}} = \frac{{ - 4}}{{ - 3}}$
Solution is:	$x = \frac{4}{3}$

Exercise 2: Solve equations

Equations containing fractions

Example 3

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

Solution 3

Step 1: Multiply both sides by the LCD. Lowest common multiple of 2 and 3 is 6. So, we multiply both sides by 6.	$6 \cdot \left( {x + \frac{1}{2}} \right) = 6 \cdot \left( {\frac{x}{2} - \frac{2}{3}} \right)$
Step 2: Remove brackets:	$6 \cdot x + 6 \cdot \frac{1}{2} = 6 \cdot \frac{x}{2} - 6 \cdot \frac{2}{3}$
Step3: This problem is similar to the previous	$6x + 3 = 3x - 4$
	$6x + 3 - 3x = 3x - 4 - 3x$
	$3x + 3 = - 4$
	$3x + 3 - 3 = -4 - 3$
	$3x = - 7$
	$\frac{{3x}}{3} = \frac{{ - 7}}{3}$
Solution is:	$x = - \frac{7}{3}$

Exercise 3: Solve equations

More Complicated Example

Example 4

$\frac{{2x}}{{x + 2}} = \frac{4}{{x - 10}} + 2$

Solution 4

In this case the LCD is (x + 2)(x - 10). Here is the complete solution to this problem.

$$\frac{{2x}}{{x + 2}} = \frac{4}{{x - 10}} + 2$$ $$\cancel{{(x + 2)}}(x - 10)\frac{{2x}}{{\cancel{{x + 2}}}} = (x + 2)\cancel{{(x - 10)}}\frac{4}{{\cancel{{x - 10}}}} + (x + 2)(x - 10) \cdot 2$$ $$(x - 10) \cdot 2x = (x + 2) \cdot 4 + 2({x^2} + x - 10x - 20)$$ $$\cancel{{2{x^2}}} - 20x = 4x + 8 + \cancel{{2{x^2}}} + 2x - 20x - 40$$ $$- 20x = 4x + 8 + 2x - 20x - 40$$ $$- 20x = - 14x - 32$$ $$- 20x + 14x = - 14x - 32 + 14x$$ $$- 6x = - 32$$ $$x = \frac{{ - 32}}{{ - 6}}$$ $$x = \frac{{16}}{3}$$

Much more Complicated Example

Example 5

$\frac{1}{{x + 3}} = \frac{{ - 2x}}{{{x^2} + 5x + 6}}$

Solution 4

The first step is to factor the denominators

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

The LCD is (x + 3)(x + 2). The solution is:

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

Exercise 4: Solve equations