Two-step equations are an essential part of algebra that help students understand more complex problems in mathematics. Whether you're a student studying for a test or someone looking to refresh their algebra skills, this guide will provide quick solutions to ten common two-step equations, complete with explanations. Let's dive into solving these equations step-by-step, ensuring you have the tools to tackle any two-step equation you might encounter.
How to Solve Two-Step Equations
Before we jump into specific examples, here’s a brief overview of the method:
- Isolate the Variable: Perform operations to get the variable on one side of the equation.
- Apply Inverse Operations: Use the opposite operation (addition with subtraction, multiplication with division) to cancel out coefficients and constants.
Let’s explore this method through examples:
Example 1: 3x + 5 = 17
Here’s how to solve:
- Subtract 5 from both sides to isolate the variable: 3x + 5 - 5 = 17 - 5, which simplifies to 3x = 12.
- Divide both sides by 3 to find x: 3x/3 = 12⁄3, so x = 4.
📝 Note: Always check your answer by plugging it back into the original equation to ensure accuracy.
Example 2: x/4 - 6 = 9
Solving this equation:
- Add 6 to both sides: x/4 - 6 + 6 = 9 + 6, simplifying to x/4 = 15.
- Multiply both sides by 4: (x/4) * 4 = 15 * 4, so x = 60.
Example 3: 2x - 8 = 0
Here’s the process:
- Add 8 to both sides to isolate the variable: 2x - 8 + 8 = 0 + 8, resulting in 2x = 8.
- Divide both sides by 2: 2x/2 = 8⁄2, so x = 4.
Example 4: -3y + 7 = -11
Let’s solve it:
- Subtract 7 from both sides: -3y + 7 - 7 = -11 - 7, simplifying to -3y = -18.
- Divide both sides by -3: -3y/-3 = -18/-3, so y = 6.
Example 5: x/5 + 2 = 8
The steps are:
- Subtract 2 from both sides: x/5 + 2 - 2 = 8 - 2, which simplifies to x/5 = 6.
- Multiply both sides by 5: (x/5) * 5 = 6 * 5, so x = 30.
Example 6: 7 - 2z = 1
Here’s how to solve it:
- Subtract 7 from both sides to isolate the term with the variable: 7 - 7 - 2z = 1 - 7, simplifying to -2z = -6.
- Divide both sides by -2: -2z/-2 = -6/-2, so z = 3.
Example 7: 5a + 10 = 35
Solving this equation:
- Subtract 10 from both sides: 5a + 10 - 10 = 35 - 10, resulting in 5a = 25.
- Divide both sides by 5: 5a/5 = 25⁄5, so a = 5.
Example 8: 6 + k/3 = 8
Steps to solve:
- Subtract 6 from both sides: 6 + k/3 - 6 = 8 - 6, which simplifies to k/3 = 2.
- Multiply both sides by 3: (k/3) * 3 = 2 * 3, so k = 6.
Example 9: 15 - 3p = 0
Here’s the process:
- Add 15 to both sides: 15 - 15 - 3p = 0 - 15, simplifying to -3p = -15.
- Divide both sides by -3: -3p/-3 = -15/-3, so p = 5.
Example 10: 4q - 12 = 4
Here are the steps:
- Add 12 to both sides: 4q - 12 + 12 = 4 + 12, simplifying to 4q = 16.
- Divide both sides by 4: 4q/4 = 16⁄4, so q = 4.
📝 Note: For two-step equations, keeping track of the operations performed is crucial for understanding and verifying your solutions.
In mastering two-step equations, the key is to maintain a clear thought process, follow the steps logically, and always check your answers for accuracy. Understanding these foundational algebraic concepts not only enhances problem-solving skills but also prepares one for more complex mathematical challenges. Remember, consistent practice with these equations will improve your speed and accuracy, making algebra an exciting and rewarding subject.
Why are two-step equations important in mathematics?
+
Two-step equations are foundational in algebra because they introduce students to the concept of solving equations by isolating the variable through multiple steps. They are a stepping stone to understanding more complex algebraic manipulations.
Can I solve a two-step equation in one step?
+
Generally, you need at least two operations to solve a two-step equation. However, if one operation cancels out an opposite operation, you might find yourself simplifying an equation in one step, but this is rare and more advanced.
What are common mistakes when solving two-step equations?
+
Common mistakes include:
- Not performing the same operation on both sides of the equation.
- Incorrectly applying inverse operations.
- Not checking the solution by substituting back into the original equation.
