Mastering Two-Step Linear Equations: Your Guide to Solving ax + b = c
Unlock the secrets of two-step linear equations and boost your algebra prowess. Learn efficient techniques to solve ax + b = c with confidence, preparing you for advanced math concepts.

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Now Playing:Solve two step linear equations– Example 0
Intros
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  1. How to turn a word problem into an equation?
    • ex. 1: "revenue" problem
    • ex. 2: "area" problem
Examples
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  1. Solve.
    1. 45+54x=13\frac{4}{5} + \frac{5}{4}x = \frac{1}{3}

    2. 34+2x=513\frac{3}{4} + 2x = 5\frac{1}{3}

    3. 23x2=47\frac{2}{3} - \frac{x}{2} = \frac{4}{7}

    4. 334=614+18x - 3\frac{3}{4} = - 6\frac{1}{4} + \frac{1}{8}x

Solving linear equations using multiplicattion and division
Notes
Solving two-step linear equations literally means solving the equations in two major steps. First, we need to isolate the unknown "x" to one side of the equation. You can then solve the "x". To complete these two steps, you may need to perform addition, subtraction, division, multiplication, cross multiplication, and so on. When the equation has fractions, you may also need to find the common denominator before proceeding further. Seems complicated? No worries! You will learn all the tricks in this section.
Concept

Introduction to Solving Two-Step Linear Equations

Solving two-step linear equations is a fundamental skill in algebra that builds upon the basics of one-step equations. This topic introduces students to more complex problem-solving techniques, preparing them for advanced mathematical concepts. The introduction video serves as a crucial starting point, offering a visual and auditory explanation of the process. It demonstrates how to isolate the variable by performing inverse operations in algebra in a specific order. Mastering two-step equations is essential for progressing in mathematics, as it forms the foundation for solving more complex equations, inequalities, and word problems. Students who grasp this concept will find it easier to tackle systems of equations preparation, quadratic equations foundation, and even calculus in the future. The skills learned here, such as logical thinking and step-by-step problem-solving, are transferable to many areas of mathematics and real-world applications, making it a vital topic in any algebra curriculum.

Understanding the importance of mastering two-step equations cannot be overstated. It is a stepping stone to more advanced topics, including systems of equations preparation and quadratic equations foundation. By learning how to perform inverse operations in algebra, students develop a deeper understanding of mathematical principles that will aid them in future studies and practical applications.

Example

Solve. 45+54x=13\frac{4}{5} + \frac{5}{4}x = \frac{1}{3}

Step 1: Identify the Equation and Isolate the Variable

In this problem, we are given the equation 45+54x=13\frac{4}{5} + \frac{5}{4}x = \frac{1}{3}. The goal is to solve for xx. To do this, we need to isolate xx on one side of the equation. The first step is to remove the term that is furthest away from xx, which is 45\frac{4}{5}.

Step 2: Subtract 45\frac{4}{5} from Both Sides

To isolate the term with xx, subtract 45\frac{4}{5} from both sides of the equation: 54x=1345 \frac{5}{4}x = \frac{1}{3} - \frac{4}{5} This step ensures that we are left with the term containing xx on one side of the equation.

Step 3: Find a Common Denominator

To subtract the fractions on the right-hand side, we need a common denominator. The denominators are 3 and 5, so the least common denominator (LCD) is 15. Convert each fraction to have the LCD: 13=1×53×5=515 \frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5}{15} 45=4×35×3=1215 \frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15} Now the equation looks like this: 54x=5151215 \frac{5}{4}x = \frac{5}{15} - \frac{12}{15}

Step 4: Subtract the Fractions

Subtract the fractions on the right-hand side: 5151215=51215=715 \frac{5}{15} - \frac{12}{15} = \frac{5 - 12}{15} = \frac{-7}{15} Now the equation is: 54x=715 \frac{5}{4}x = \frac{-7}{15}

Step 5: Cross Multiply to Solve for xx

To solve for xx, we can use cross multiplication. Cross multiply the fractions: 5x×15=4×(7) 5x \times 15 = 4 \times (-7) Simplify the multiplication: 75x=28 75x = -28

Step 6: Solve for xx

Finally, solve for xx by dividing both sides by 75: x=2875 x = \frac{-28}{75} This is the solution for xx.

FAQs

Here are some frequently asked questions about solving two-step linear equations:

1. What are simple 2-step equation examples?

Simple 2-step equation examples include: 3x + 5 = 20, 2y - 7 = 11, and 4z + 3 = 15. These equations require two operations to solve for the variable.

2. How do you solve two linear equations?

To solve two-step linear equations: First, isolate the variable term by adding or subtracting the constant from both sides. Then, divide or multiply both sides by the coefficient of the variable to solve for x.

3. What is the first step to solve a two-step equation?

The first step is to isolate the variable term. This usually involves adding or subtracting the constant from both sides of the equation to get the variable term alone on one side.

4. How to solve a two-step equation step by step?

Step 1: Add or subtract to isolate the variable term. Step 2: Multiply or divide both sides by the coefficient of the variable. Step 3: Simplify to get the solution. Always perform the same operation on both sides of the equation.

5. What is the rule for solving two-step equations?

The rule is to isolate the variable by performing inverse operations in the correct order. First, add or subtract to get the variable term alone, then multiply or divide to solve for the variable. Always maintain equation balance by performing the same operation on both sides.

Prerequisites

Understanding the foundations of algebra is crucial when tackling more complex problems like solving two-step linear equations of the form ax + b = c or x/a + b = c. To master this topic, it's essential to have a solid grasp of several prerequisite concepts.

First and foremost, a strong understanding of order of operations (PEMDAS) is vital. This fundamental principle guides the sequence in which we perform mathematical operations, ensuring consistency and accuracy in our calculations.

Building on this, solving two-step linear equations using addition and subtraction forms the basis for more complex equation solving. This skill directly applies to our main topic, as it introduces the concept of isolating variables through inverse operations.

Speaking of inverse operations, understanding inverse operations in algebra is crucial. While our focus isn't on matrices, the principle of using inverse operations to solve equations is directly applicable to two-step linear equations.

When dealing with equations involving fractions, such as x/a + b = c, proficiency in simplifying complex fractions becomes invaluable. This skill helps in manipulating equations to isolate the variable efficiently.

Additionally, multiplying improper fractions and mixed numbers is often necessary when solving these types of equations, especially when dealing with fractional coefficients.

For more advanced applications, distance and time related questions in linear equations demonstrate real-world applications of two-step linear equations, enhancing understanding through practical context.

While not directly related to our main topic, concepts like solving systems of linear equations by substitution and using the quadratic formula to solve quadratic equations build upon the skills learned in solving two-step linear equations, showcasing the progression of algebraic problem-solving.

Lastly, understanding least common multiples (LCM) and greatest common factors can be helpful when dealing with equations involving fractions, aiding in simplification and solution processes.

By mastering these prerequisite topics, students will be well-equipped to tackle two-step linear equations with confidence, understanding not just the how, but also the why behind each step in the solution process.