Math Calculators, Lessons and Formulas

It is time to solve your math problem

mathportal.org
Solving Equations: (lesson 1 of 4)

Solving Linear 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

Level 1

$$ \color{blue}{2x - 3 - 4x = 6 + x} $$ $ x = - 1 $
$ x = - 4 $
$ x = 7 $
$ x = - 3 $

Level 2

$$ \color{blue}{- 12x - 13 + 5x = - 3x + 11 - 2x} $$ $ x = 3 $
$ x = 2 $
$ x = 1 $
$ x = 0 $

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

Level 1

$$ \color{blue}{2 ( 3 - x ) = 3 - 3 ( x + 1 )} $$ $ x = - 4 $
$ x = - 2 $
$ x = - 6 $
$ x = - 8 $

Level 2

$$ \color{blue}{- 2 - 2 ( - 3 - x ) = - 6 - 4 ( - x - 2 )} $$ $ x = 2 $
$ x = 3 $
$ x = 1 $
$ x = 0 $

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

Level 1

$$ \color{blue}{\frac{x}{3} + 2 = \frac{x}{2} - 1} $$ $ x = 18 $
$ x = 16 $
$ x = 14 $
$ x = 12 $

Level 2

$$ \color{blue}{2\left( {\frac{x}{4} + 3} \right) = \frac{x}{4} - 5 + \frac{x}{3}} $$ $ x = 130 $
$ x = 128 $
$ x = 134 $
$ x = 132 $

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

Level 1

$$ \color{blue}{\frac{4}{{x + 2}} = \frac{{3x}}{{x - 1}} - 3} $$ $ x = - 2 $
$ x = - 3 $
$ x = - 4 $
$ x = - 5 $

Level 2

$$ \color{blue}{\frac{3}{{x + 1}} = \frac{9}{{{x^2} + 3x + 2}}} $$ $ x = 3 $
$ x = 2 $
$ x = 1 $
$ x = 0 $