Your AssignmentIs X = 6 A Solution To The Equation 5(x 3) = X + 13?Follow The Steps To Find Out.1. Rewrite
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Introduction
When tackling algebraic equations, understanding the fundamental steps involved in solving for an unknown variable is crucial. One common type of problem involves linear equations, where the goal is to isolate the variable and determine its value. In this article, we will explore the equation:
\[ 5(x + 3) = x + 13 \]
and examine whether X = 6 is a solution. To do so, we'll walk through each step meticulously, starting with rewriting the equation, simplifying it, and then checking the proposed solution.
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Understanding the Equation
Before diving into solving, let's understand what the equation represents:
\[ 5(x + 3) = x + 13 \]
- The left side involves distributing the 5 across the expression inside the parentheses.
- The right side is a simple linear expression.
The goal is to find the value of \( x \) that makes both sides equal, i.e., satisfies the equation.
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Step 1: Rewrite the Equation
Rewriting the equation is the first crucial step in solving for \( x \). This involves expanding and simplifying the expression to make solving more straightforward.
Distribute the 5 on the left side
Using the distributive property:
\[ 5(x + 3) = 5 \times x + 5 \times 3 = 5x + 15 \]
Rewrite the original equation
So, the equation now becomes:
\[ 5x + 15 = x + 13 \]
This rewritten form makes it easier to isolate \( x \) by combining like terms.
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Step 2: Isolate the Variable \( x \)
The goal now is to get all terms containing \( x \) on one side and constant terms on the other.
Subtract \( x \) from both sides
\[
5x + 15 - x = x + 13 - x
\]
which simplifies to:
\[
(5x - x) + 15 = 13
\]
\[
4x + 15 = 13
\]
Subtract 15 from both sides
\[
4x + 15 - 15 = 13 - 15
\]
\[
4x = -2
\]
Divide both sides by 4
\[
x = \frac{-2}{4} = -\frac{1}{2}
\]
Final solution
\[
x = -\frac{1}{2}
\]
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Step 3: Verify the Solution \( x = 6 \)
The question asks whether X = 6 is a solution to the given equation. To verify, substitute \( x = 6 \) into the original equation:
\[ 5(x + 3) = x + 13 \]
Substitute \( x = 6 \):
\[ 5(6 + 3) = 6 + 13 \]
Calculate each side:
- Left side:
\[
5 \times 9 = 45
\]
- Right side:
\[
6 + 13 = 19
\]
Since:
\[
45 \neq 19
\]
Conclusion: \( x = 6 \) does not satisfy the equation.
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Step 4: Summarize Findings
- The solution to the equation \( 5(x + 3) = x + 13 \) is \( x = -\frac{1}{2} \).
- The proposed solution \( x = 6 \) is incorrect because substituting it back into the original equation results in unequal sides.
- Therefore, X = 6 is not a solution for the equation.
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Additional Tips for Solving Similar Equations
To enhance your algebraic problem-solving skills, keep these tips in mind:
- Always Distribute Properly
When parentheses are involved, distribute coefficients evenly to eliminate parentheses.
- Combine Like Terms
Gather all variables on one side and constants on the other to simplify the equation.
- Check Your Solution
Substitute the found value of \( x \) back into the original equation to verify its correctness.
- Be Careful with Fractions
When dividing, ensure to simplify fractions properly to avoid errors.
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Conclusion
Solving algebraic equations involves a systematic process: rewriting the equation, simplifying, isolating the variable, and verifying solutions. In the case of the equation \( 5(x + 3) = x + 13 \), the solution is \( x = -\frac{1}{2} \), not \( x = 6 \). Understanding these steps thoroughly allows you to approach similar problems confidently and accurately.
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FAQs
Q1: How do I know when I've solved an equation correctly?
A: When both sides of the equation are equal after substituting your solution back into the original equation, you've correctly solved it.
Q2: Can an equation have more than one solution?
A: Yes, some equations can have multiple solutions, especially quadratic or higher-degree equations. Linear equations like the one discussed typically have only one solution.
Q3: What should I do if I get a fraction as a solution?
A: Ensure to verify the solution by substitution. If it satisfies the original equation, it is valid. Simplify fractions where possible for clarity.
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By following these detailed steps and understanding the principles involved, you can confidently solve similar algebraic equations and determine whether a specific value is a solution.
