Problem 74
Question
Solve each equation, and check the solution. \(0.09 x+0.13(x+300)=61\)
Step-by-Step Solution
Verified Answer
x = 100
1Step 1 - Distribute and Combine Like Terms
First, distribute the 0.13 across the term \(x+300\). This gives \(0.09x + 0.13x + 0.13 \cdot 300 \). Combine like terms: \[ 0.09x + 0.13x = 0.22x \] and \[ 0.13 \cdot 300 = 39 \]. Therefore, the equation becomes: \[ 0.22x + 39 = 61 \]
2Step 2 - Isolate the Variable Term
Next, subtract 39 from both sides of the equation: \[ 0.22x + 39 - 39 = 61 - 39 \] This simplifies to: \[ 0.22x = 22 \]
3Step 3 - Solve for x
To solve for \(x\), divide both sides of the equation by 0.22: \[ x = \frac{22}{0.22} = 100 \]
4Step 4 - Check the Solution
Substitute \(x = 100\) back into the original equation to ensure it holds true: \[ 0.09(100) + 0.13(100 + 300) = 61 \] Calculate each term: \[ 0.09 \cdot 100 = 9 \] \[ 0.13 \cdot 400 = 52 \] Adding these gives: \[ 9 + 52 = 61 \] Since both sides of the equation are equal, \(x = 100\) is correct.
Key Concepts
distributive propertycombining like termsisolating the variable
distributive property
The distributive property is a fundamental principle used in algebra. It allows you to multiply a single term by each term inside a set of parentheses. In the equation given, we applied the distributive property like this:
First, focus on the part of the equation \(0.13(x+300)\). To distribute 0.13, multiply it by each term inside the parentheses: \[0.13 \times x = 0.13x \] and \[0.13 \times 300 = 39 \]. This step helps to simplify the equation by removing the parentheses.
Therefore, the equation updates from \(0.09x + 0.13(x + 300) = 61\) to \[0.09x + 0.13x + 39 = 61\].
This is the precise step of using the distributive property in action.
First, focus on the part of the equation \(0.13(x+300)\). To distribute 0.13, multiply it by each term inside the parentheses: \[0.13 \times x = 0.13x \] and \[0.13 \times 300 = 39 \]. This step helps to simplify the equation by removing the parentheses.
Therefore, the equation updates from \(0.09x + 0.13(x + 300) = 61\) to \[0.09x + 0.13x + 39 = 61\].
This is the precise step of using the distributive property in action.
combining like terms
Combining like terms is an important skill when solving linear equations. Several terms must be simplified into a single term when they are alike. Here, the equation includes two terms with the variable \(x\).
Looking at \[0.09x + 0.13x \], these are 'like terms' because they both have an \(x\). Combine them by adding their coefficients: \[0.09x + 0.13x = 0.22x \].
Combining like terms helps to reduce the equation into a simpler form. So, the equation transitions from \[0.09x + 0.13x + 39 = 61\] to \[0.22x + 39 = 61\].
Simplifying the equation this way makes it easier to solve.
Looking at \[0.09x + 0.13x \], these are 'like terms' because they both have an \(x\). Combine them by adding their coefficients: \[0.09x + 0.13x = 0.22x \].
Combining like terms helps to reduce the equation into a simpler form. So, the equation transitions from \[0.09x + 0.13x + 39 = 61\] to \[0.22x + 39 = 61\].
Simplifying the equation this way makes it easier to solve.
isolating the variable
Isolating the variable is the process of getting the variable (such as \(x\)) alone on one side of the equation to solve for it. This is done through inverse operations.
In this case, to isolate \(0.22x\) in the equation \[0.22x + 39 = 61\], first subtract 39 from both sides: \[0.22x + 39 - 39 = 61 - 39\], simplifying to \[0.22x = 22\].
Next, to isolate \(x\), divide both sides by 0.22: \[x = \frac{22}{0.22} = 100\].
The result \(x = 100\) is the solution to the equation. By isolating the variable, you've essentially unlocked the value that balances the original equation.
In this case, to isolate \(0.22x\) in the equation \[0.22x + 39 = 61\], first subtract 39 from both sides: \[0.22x + 39 - 39 = 61 - 39\], simplifying to \[0.22x = 22\].
Next, to isolate \(x\), divide both sides by 0.22: \[x = \frac{22}{0.22} = 100\].
The result \(x = 100\) is the solution to the equation. By isolating the variable, you've essentially unlocked the value that balances the original equation.
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