Solving equations and inequalities is a fundamental topic in algebra. Here are some key concepts and techniques for solving equations and inequalities:

## Solving Linear Equations

I’d be happy to help you solve linear equations. Linear equations are mathematical equations that involve variables raised to the power of 1 (i.e., they have no exponents other than 1) and can be written in the form of:

ax + b = 0

where “a” and “b” are constants, and “x” is the variable we’re solving for. To solve linear equations, we typically aim to isolate the variable “x” on one side of the equation.

### A step-by-step process for solving a linear equation:

- Simplify both sides of the equation by combining like terms.
- Move the constant term (the term without the variable) to the other side of the equation.
- Combine any like terms on each side of the equation.
- Divide both sides of the equation by the coefficient of the variable to isolate “x.”
- Check the solution by substituting the value of “x” back into the original equation.

Let’s go through an example. Consider the equation:

2x + 5 = 11

- We start by simplifying both sides: 2x + 5 – 5 = 11 – 5, which becomes 2x = 6.
- Next, we move the constant term to the other side: 2x = 6.
- There are no like terms to combine in this case.
- To isolate “x,” we divide both sides by the coefficient of “x,” which is 2: (2x)/2 = 6/2, resulting in x = 3.
- Finally, we substitute the value of “x” (3) back into the original equation: 2(3) + 5 = 6 + 5, which simplifies to 6 + 5 = 11. Since both sides are equal, we can conclude that x = 3 is the correct solution.

Remember, this is a general method for solving linear equations. Depending on the specific equation, there may be slight variations in the steps required.

## Solving Quadratic Equations

Quadratic equations are second-degree polynomial equations in the form of:

ax^2 + bx + c = 0,

where a, b, and c are constants, and x represents the variable. To solve a quadratic equation, you can use several methods, such as factoring, completing the square, or using the quadratic formula.

**Factoring:**If the quadratic equation can be factored, you can solve it by setting each factor equal to zero and solving for x. Here’s an example:

**Example:** Solve the equation x^2 + 5x + 6 = 0.

We can factor this equation as (x + 2)(x + 3) = 0.

Setting each factor equal to zero gives us two equations: x + 2 = 0 –> x = -2 x + 3 = 0 –> x = -3

Therefore, the solutions to the equation are x = -2 and x = -3.

**Completing the Square:**Completing the square is another method to solve quadratic equations. Here’s an example:

**Example:** Solve the equation x^2 + 6x + 8 = 0.

First, move the constant term to the other side of the equation: x^2 + 6x = -8.

Next, complete the square by adding the square of half the coefficient of x to both sides of the equation. Half of 6 is 3, so we add 3^2 = 9 to both sides: x^2 + 6x + 9 = -8 + 9, (x + 3)^2 = 1.

Taking the square root of both sides gives us: x + 3 = ±√1.

Simplifying further: x + 3 = 1 –> x = -2, x + 3 = -1 –> x = -4.

The solutions to the equation are x = -2 and x = -4.

**Quadratic Formula:**The quadratic formula is a general method to solve quadratic equations. It is given by:

x = (-b ± √(b^2 – 4ac)) / (2a).

**Example:** Solve the equation 2x^2 – 5x + 2 = 0.

Comparing this equation with the standard quadratic form, we have a = 2, b = -5, and c = 2.

Applying the quadratic formula, we get: x = (-(-5) ± √((-5)^2 – 4(2)(2))) / (2(2)), x = (5 ± √(25 – 16)) / 4, x = (5 ± √9) / 4.

Simplifying further: x = (5 ± 3) / 4, x = 8/4 = 2 or x = 2/4 = 1/2.

The solutions to the equation are x = 2 and x = 1/2.

These are the basic methods for solving quadratic equations. Depending on the equation, one method may be more suitable than another. Remember to always check your solutions by substituting them back into the original equation to verify their accuracy. Note that the golden ratio, which is a very typical number, is a solution to the following quadratic equation: a^2 – a – b = 0. Here

## Solving Rational Equations

To solve rational equations, follow these steps:

- Identify any restrictions on the variables. Look for values that would make the denominators zero since division by zero is undefined. These values are called “extraneous solutions” and must be excluded from the solution set.
- Clear the equation of fractions by multiplying every term by the least common denominator (LCD) of all the fractions involved. This step eliminates the denominators.
- Simplify the resulting equation by distributing and combining like terms if necessary.
- Solve the simplified equation by isolating the variable on one side of the equation.
- Check the solutions obtained to ensure they satisfy the original equation. Substitute each solution back into the original equation and see if both sides are equal.

### Let’s work through an example to illustrate these steps:

**Example:** Solve the rational equation: 1/x + 1/(x + 2) = 1/3

**Step 1:** Identify restrictions In this case, there are no restrictions since the denominators x and (x + 2) are never zero.

**Step 2:** Clear the equation of fractions To eliminate the denominators, we multiply every term by the LCD, which is 3x(x + 2): 3x(x + 2) * (1/x) + 3x(x + 2) * (1/(x + 2)) = 3x(x + 2) * (1/3)

After simplification, the equation becomes: 3(x + 2) + 3x = x(x + 2)

**Step 3: **Simplify the equation Distribute and combine like terms: 3x + 6 + 3x = x^2 + 2x

**Step 4:** Solve the equation Rearrange the equation to have all terms on one side: x^2 + 2x – 6x – 6 = 0

Combine like terms: x^2 – 4x – 6 = 0

Now we can either factor this quadratic equation or use the quadratic formula to solve for x. In this example, we’ll use the quadratic formula:

x = (-b ± √(b^2 – 4ac)) / (2a)

Substituting the values a = 1, b = -4, and c = -6 into the formula, we have: x = (-(-4) ± √((-4)^2 – 4(1)(-6))) / (2(1)) x = (4 ± √(16 + 24)) / 2 x = (4 ± √40) / 2 x = (4 ± 2√10) / 2 x = 2 ± √10

**Step 5:** Check the solutions Substitute each solution back into the original equation and verify if both sides are equal:

For x = 2 + √10: 1/(2 + √10) + 1/((2 + √10) + 2) = 1/3 1/(2 + √10) + 1/(4 + √10) = 1/3

[common denominator: (2 + √10)(4 + √10)]

(4 + √10 + 2 + √10) / [(2 + √10)(4 + √10)] = 1/3 (6 + 2√10) / [(2 + √10)(4 + √10)] = 1/3 (6 + 2√10) / (8 + 6√10 + 2√10 + 10)

## Solving Radical Equations

To solve radical equations, follow these general steps:

- Isolate the radical term: Move all other terms to the opposite side of the equation, leaving the radical term by itself on one side.
- Square both sides of the equation: Raise both sides of the equation to the power of 2 (or whatever power corresponds to the index of the radical). This eliminates the radical.
- Solve the resulting equation: Simplify the equation and solve for the variable.
- Check for extraneous solutions: Substitute the obtained solutions back into the original equation and verify that they satisfy the original equation. Sometimes, extraneous solutions can be introduced when squaring both sides.

### An example to illustrate the steps:

Solve the equation √(x + 5) = 3.

**Step 1:** Isolate the radical term: √(x + 5) = 3

**Step 2:** Square both sides: (√(x + 5))^2 = 3^2 x + 5 = 9

**Step 3:** Solve the resulting equation: x = 9 – 5 x = 4

**Step 4:** Check for extraneous solutions: Substitute x = 4 back into the original equation: √(4 + 5) = 3 √9 = 3 3 = 3 (satisfied)

Therefore, the solution to the equation √(x + 5) = 3 is x = 4.

It’s important to note that when solving radical equations, you need to check for extraneous solutions introduced by squaring both sides. In some cases, additional solutions may arise that do not satisfy the original equation.

## Solving Absolute Value Equations

To solve absolute value equations, we need to consider two cases: when the expression inside the absolute value bars is positive and when it is negative. Let’s go through the steps for solving absolute value equations:

### Case 1: Expression inside the absolute value bars is positive

- Set the expression inside the absolute value bars equal to the right-hand side of the equation. |x| = a, where ‘a’ represents a positive number.
- Solve the equation for ‘x’ by removing the absolute value bars. x = a

### Case 2: Expression inside the absolute value bars is negative

- Set the expression inside the absolute value bars equal to the negation of the right-hand side of the equation. |x| = -a, where ‘a’ represents a positive number.
- There are no solutions in this case because the absolute value of any real number cannot be negative. In other words, the equation has no solutions when the expression inside the absolute value bars is negative.

Let’s go through an example to illustrate this:

Example: Solve the absolute value equation |3x – 2| = 4.

We’ll consider two cases:

Case 1: 3x – 2 is positive: 3x – 2 = 4 3x = 6 x = 2

Case 2: 3x – 2 is negative: 3x – 2 = -4 3x = -2 x = -2/3

So, the solutions to the absolute value equation |3x – 2| = 4 are x = 2 and x = -2/3.

Remember to always check your solutions by substituting them back into the original equation to ensure they satisfy the equation.

## Solving Linear Inequalities

Solving linear inequalities involves finding the set of values that satisfy the given inequality. Here are the general steps to solve a linear inequality:

- Start with the given inequality. It should be in the form of “expression <, >, ≤, or ≥ constant.”
- Treat the inequality sign as an equal sign and solve the corresponding equation. This will give you the boundary line.
- If the inequality sign is “<” or “>”, the boundary line should be dashed to indicate that the points on the line are not included in the solution set. If the inequality sign is “≤” or “≥”, the boundary line should be solid to indicate that the points on the line are included in the solution set.
- Choose a test point that is not on the boundary line and substitute it with the original inequality. If the inequality is true, shade the region containing the test point. If it is false, shade the other region.
- Repeat step 4 for any other inequalities if you have a system of linear inequalities.
- The solution to the inequality is the shaded region that satisfies all the given inequalities.

Let’s go through an example to illustrate the process:

**Example:** Solve the inequality 2x – 3 > 5.

- Start with the inequality 2x – 3 > 5.
- Treat the inequality sign as an equal sign and solve the equation: 2x – 3 = 5. Add 3 to both sides: 2x = 8. Divide both sides by 2: x = 4.
- The boundary line is x = 4. Since the inequality sign is “>”, the boundary line should be dashed.
- Choose a test point, for example, x = 0. Substitute it into the original inequality: 2(0) – 3 > 5. Simplify -3 > 5. Since this is false, shade the region on the other side of the dashed line.
- The solution is the shaded region to the right of x = 4.

Please note that this example involves a single linear inequality. If you have a system of linear inequalities, you would need to repeat steps 4 and 5 for each inequality to find the solution.

## Solving Quadratic Inequalities

To solve quadratic inequalities, follow these steps:

- Move all terms to one side of the inequality, so you have a quadratic expression on one side and zero on the other side.
- Factor the quadratic expression if possible. If the quadratic expression cannot be factored easily, you can use the quadratic formula to find the roots of the quadratic equation.
- Once you have factored in the quadratic expression or found its roots, plot these points on a number line. The points where the expression equals zero are important because they represent the x-values where the quadratic expression changes sign.
- Determine the sign of the quadratic expression in each of the intervals defined by the points on the number line.
- Finally, based on the sign of the quadratic expression in each interval, determine the solution to the quadratic inequality.

### Here are a few examples to illustrate the process:

**Example 1:** Solve the inequality x^2 – 4x > 0.

Step 1: Move all terms to one side: x^2 – 4x > 0.

Step 2: Factor the quadratic expression: x(x – 4) > 0. The roots are x = 0 and x = 4.

Step 3: Plot the points on a number line: —|—|—|— -1 0 4 5

Step 4: Determine the sign of the quadratic expression in each interval:

- In the interval (-∞, 0), the expression is positive (because both factors are negative or positive).
- In the interval (0, 4), the expression is negative (because x is positive, but x – 4 is negative).
- In the interval (4, ∞), the expression is positive (because both factors are positive).

Step 5: Based on the sign of the quadratic expression in each interval, we can see that the solution is x < 0 or x > 4.

**Example 2:** Solve the inequality x^2 + 3x – 4 ≤ 0.

Step 1: Move all terms to one side: x^2 + 3x – 4 ≤ 0.

Step 2: Factor the quadratic expression: (x – 1)(x + 4) ≤ 0. The roots are x = 1 and x = -4.

Step 3: Plot the points on a number line: —|—|—|— -5 -4 1 2

Step 4: Determine the sign of the quadratic expression in each interval:

- In the interval (-∞, -4), the expression is positive (because both factors are negative or positive).
- In the interval (-4, 1), the expression is negative (because x + 4 is positive, but x – 1 is negative).
- In the interval (1, ∞), the expression is positive (because both factors are positive).

Step 5: Based on the sign of the quadratic expression in each interval, we can see that the solution is -4 ≤ x ≤ 1.

Remember to include or exclude the endpoints based on whether the inequality is strict (< or >) or inclusive (≤ or ≥).

## General Remark about solving equations and inequalities

Remember to always perform equivalent operations on both sides of an equation or inequality to maintain balance and find the correct solutions. Additionally, check your solutions by substituting them back into the original equation or inequality to ensure they are valid.