Math Examples: Step-by-Step Solutions (No AI)
Hey guys! Let's dive into some math problems with detailed solutions. No AI here, just good ol' fashioned math! We will solve four different mathematical problems.
1. Solving Linear Equations
Let's kick things off with solving linear equations. Linear equations are the foundation of algebra, and mastering them is super important. We'll break down the steps so you can tackle any linear equation that comes your way. Remember, the goal is to isolate the variable, usually 'x', by performing the same operations on both sides of the equation.
Example 1: Simple Linear Equation
Solve for x: 3x + 5 = 14
Step 1: Subtract 5 from both sides to isolate the term with 'x'.
3x + 5 - 5 = 14 - 5
3x = 9
Step 2: Divide both sides by 3 to solve for 'x'.
3x / 3 = 9 / 3
x = 3
So, the solution is x = 3. Easy peasy!
Example 2: Linear Equation with Distribution
Solve for x: 2(x - 1) + 5 = 3x - 1
Step 1: Distribute the 2 into the parentheses.
2x - 2 + 5 = 3x - 1
Step 2: Combine like terms on the left side.
2x + 3 = 3x - 1
Step 3: Subtract 2x from both sides to get 'x' terms on one side.
2x + 3 - 2x = 3x - 1 - 2x
3 = x - 1
Step 4: Add 1 to both sides to isolate 'x'.
3 + 1 = x - 1 + 1
4 = x
So, the solution is x = 4. Remember to take it step by step!
Example 3: Linear Equation with Fractions
Solve for x: (x / 2) + 3 = 7
Step 1: Subtract 3 from both sides to isolate the term with 'x'.
(x / 2) + 3 - 3 = 7 - 3
x / 2 = 4
Step 2: Multiply both sides by 2 to solve for 'x'.
(x / 2) * 2 = 4 * 2
x = 8
So, the solution is x = 8. Fractions are no match for us!
Example 4: Linear Equation with Decimals
Solve for x: 0.5x - 1.2 = 0.8
Step 1: Add 1.2 to both sides to isolate the term with 'x'.
- 5x - 1.2 + 1.2 = 0.8 + 1.2
0.5x = 2
Step 2: Divide both sides by 0.5 to solve for 'x'.
0.5x / 0.5 = 2 / 0.5
x = 4
So, the solution is x = 4. Decimals? We got this!
2. Quadratic Equations
Next up, quadratic equations! These equations involve a variable raised to the power of 2 (x²). Solving them might seem tricky, but with the right methods, you'll nail it. The most common methods are factoring, completing the square, and using the quadratic formula.
Example 1: Factoring
Solve for x: x² - 5x + 6 = 0
Step 1: Factor the quadratic expression.
(x - 2)(x - 3) = 0
Step 2: Set each factor equal to zero and solve for 'x'.
x - 2 = 0 => x = 2
x - 3 = 0 => x = 3
So, the solutions are x = 2 and x = 3. Factoring is your friend!
Example 2: Quadratic Formula
Solve for x: 2x² + 3x - 5 = 0
Step 1: Identify a, b, and c in the quadratic equation ax² + bx + c = 0.
a = 2, b = 3, c = -5
Step 2: Use the quadratic formula:
x = [-b ± √(b² - 4ac)] / (2a)
x = [-3 ± √(3² - 4 * 2 * -5)] / (2 * 2)
x = [-3 ± √(9 + 40)] / 4
x = [-3 ± √49] / 4
x = [-3 ± 7] / 4
Step 3: Calculate the two possible values for 'x'.
x = (-3 + 7) / 4 = 4 / 4 = 1
x = (-3 - 7) / 4 = -10 / 4 = -2.5
So, the solutions are x = 1 and x = -2.5. The quadratic formula to the rescue!
Example 3: Completing the Square
Solve for x: x² + 6x + 5 = 0
Step 1: Rewrite the equation by moving the constant term to the right side.
x² + 6x = -5
Step 2: Complete the square by adding (b/2)² to both sides. Here, b = 6, so (6/2)² = 9.
x² + 6x + 9 = -5 + 9
Step 3: Factor the left side as a perfect square.
(x + 3)² = 4
Step 4: Take the square root of both sides.
x + 3 = ±2
Step 5: Solve for x.
x = -3 + 2 = -1
x = -3 - 2 = -5
So, the solutions are x = -1 and x = -5. Completing the square – you've got this!
Example 4: Quadratic Equation with No Real Solutions
Solve for x: x² + 2x + 5 = 0
Step 1: Use the quadratic formula:
x = [-b ± √(b² - 4ac)] / (2a)
x = [-2 ± √(2² - 4 * 1 * 5)] / (2 * 1)
x = [-2 ± √(4 - 20)] / 2
x = [-2 ± √(-16)] / 2
Since we have a negative number under the square root, there are no real solutions. Complex numbers, anyone?
3. Systems of Equations
Alright, let's tackle systems of equations. These are sets of two or more equations with the same variables. The goal is to find values for the variables that satisfy all equations simultaneously. Common methods include substitution and elimination.
Example 1: Substitution
Solve the system:
- x + y = 5
- y = 2x - 1
Step 1: Substitute the expression for 'y' from the second equation into the first equation.
x + (2x - 1) = 5
Step 2: Simplify and solve for 'x'.
3x - 1 = 5
3x = 6
x = 2
Step 3: Substitute the value of 'x' back into either equation to solve for 'y'.
y = 2(2) - 1
y = 4 - 1
y = 3
So, the solution is x = 2 and y = 3. Substitution is a winner!
Example 2: Elimination
Solve the system:
- 2x + y = 7
- x - y = 2
Step 1: Add the two equations together to eliminate 'y'.
(2x + y) + (x - y) = 7 + 2
3x = 9
Step 2: Solve for 'x'.
x = 3
Step 3: Substitute the value of 'x' back into either equation to solve for 'y'.
3 - y = 2
-y = -1
y = 1
So, the solution is x = 3 and y = 1. Elimination, mission accomplished!
Example 3: System with No Solution
Solve the system:
- x + y = 3
- 2x + 2y = 8
Step 1: Multiply the first equation by -2.
-2(x + y) = -2(3)
-2x - 2y = -6
Step 2: Add the modified first equation to the second equation.
(-2x - 2y) + (2x + 2y) = -6 + 8
0 = 2
Since 0 ≠ 2, there is no solution to this system. Parallel lines, maybe?
Example 4: System with Infinite Solutions
Solve the system:
- x + y = 4
- 2x + 2y = 8
Step 1: Multiply the first equation by -2.
-2(x + y) = -2(4)
-2x - 2y = -8
Step 2: Add the modified first equation to the second equation.
(-2x - 2y) + (2x + 2y) = -8 + 8
0 = 0
Since 0 = 0, there are infinite solutions to this system. These are the same line!
4. Basic Trigonometry
Last but not least, basic trigonometry! This involves the relationships between angles and sides of triangles. We'll focus on the sine, cosine, and tangent functions in right triangles.
Example 1: Finding Sine, Cosine, and Tangent
In a right triangle, the opposite side is 3, the adjacent side is 4, and the hypotenuse is 5. Find sin(θ), cos(θ), and tan(θ).
Step 1: Use the definitions:
sin(θ) = Opposite / Hypotenuse = 3 / 5
cos(θ) = Adjacent / Hypotenuse = 4 / 5
tan(θ) = Opposite / Adjacent = 3 / 4
So, sin(θ) = 3/5, cos(θ) = 4/5, and tan(θ) = 3/4. That was a breeze!
Example 2: Finding an Angle Using Inverse Trigonometric Functions
In a right triangle, the opposite side is 7 and the hypotenuse is 14. Find the angle θ.
Step 1: Use the sine function: sin(θ) = Opposite / Hypotenuse = 7 / 14 = 0.5
Step 2: Use the inverse sine function (arcsin) to find the angle.
θ = arcsin(0.5)
θ = 30°
So, the angle is θ = 30°. Inverse trig functions to the rescue!
Example 3: Finding a Side Length
In a right triangle, one angle is 60°, the hypotenuse is 10, and we need to find the length of the opposite side.
Step 1: Use the sine function: sin(60°) = Opposite / Hypotenuse
Step 2: Solve for the opposite side.
Opposite = Hypotenuse * sin(60°)
Opposite = 10 * (√3 / 2)
Opposite = 5√3
So, the length of the opposite side is 5√3. Trig power!
Example 4: Using Trigonometry in a Real-World Scenario
A ladder leaning against a wall forms a 70° angle with the ground. If the base of the ladder is 5 feet from the wall, how high up the wall does the ladder reach?
Step 1: Use the tangent function: tan(70°) = Opposite / Adjacent
Step 2: Solve for the opposite side (height).
Opposite = Adjacent * tan(70°)
Opposite = 5 * tan(70°)
Opposite ≈ 5 * 2.747
Opposite ≈ 13.735
So, the ladder reaches approximately 13.735 feet up the wall. Practical trig, right?
That's all for now, folks! Hope these examples helped you brush up on your math skills. Keep practicing, and you'll become a math whiz in no time!