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Chapter 2: Problem 49

Solve each equation, and check your solution. $$ 0.2(60)+0.05 x=0.1(60+x) $$

Short Answer

Expert verified
x = 120

Step by step solution

01

Distribute constants

Distribute the constants in the parentheses to simplify the equation: \[ 0.2 \times 60 + 0.05x = 0.1 \times 60 + 0.1x \] Simplifying the multiplication gives: \[ 12 + 0.05x = 6 + 0.1x \]
02

Rearrange equation to isolate variables

Subtract \(0.05x\) from both sides to move the variable terms to one side of the equation: \[ 12 + 0.05x - 0.05x = 6 + 0.1x - 0.05x \] Simplifying gives: \[ 12 = 6 + 0.05x \]
03

Isolate the variable term

Subtract 6 from both sides to isolate the term with the variable: \[ 12 - 6 = 6 + 0.05x - 6 \] Simplifies to: \[ 6 = 0.05x \]
04

Solve for the variable

Divide both sides by \(0.05\) to solve for \(x\): \[ \frac{6}{0.05} = \frac{0.05x}{0.05} \] Simplifies to: \[ x = 120 \]
05

Check your solution

Substitute \(x = 120\) back into the original equation to verify: \[ 0.2(60) + 0.05(120) = 0.1(60 + 120) \] Simplifies to: \[ 12 + 6 = 18 \] The right side simplifies to \(0.1(180) = 18\), which confirms the solution is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

distributive property
When solving algebra equations, the **distributive property** is a crucial step. This property allows us to multiply a number by a group of numbers added together inside parentheses. For example, in the equation \[ 0.2(60) + 0.05x = 0.1(60 + x) \], we use the distributive property to spread the numbers outside the parentheses across the terms inside.

Here's how it happens: \begin{align*} 0.2 \times 60 + 0.05x &= 0.1 \times 60 + 0.1x \ 12 + 0.05x &= 6 + 0.1x \ \ \text{These simple multiplications make the equation cleaner and simpler to work with.}

Remember:
  • Apply multiplication to each term inside the parentheses separately.
  • Simplify the result before moving to the next steps.
Practice makes perfect, so try distributing on various problems to get comfortable.
isolating variables
The core idea of **isolating variables** in algebra is to get the variable on one side of the equation by itself. This makes it easier to solve.

After using the distributive property, we had: \[ 12 + 0.05x = 6 + 0.1x \]
To isolate variables, follow these steps:
  • Move variable terms to one side by subtracting or adding. In our case, subtracting 0.05x from both sides: \[ 12 + 0.05x - 0.05x = 6 + 0.1x - 0.05x\ 12 = 6 + 0.05x \]
  • Next, eliminate constant terms from the variable side. Subtract 6 from both sides: \[ 12 - 6 = 6 + 0.05x - 6 \][/li>
  • Which simplifies to: \[ 6 = 0.05x \]
Now the variable is isolated, ready for solving. It's important to:
  • Keep equations balanced by performing the same operation on both sides.
  • Simplify step by step to avoid mistakes.
solving equations
Now, let's talk about the final step: **solving equations**. This involves finding the value of the variable that makes the equation true.

Our isolated equation was: \[ 6 = 0.05x \]
To solve for \(x\), follow these simple steps:
  • Divide both sides by the coefficient of the variable: \[ \frac{6}{0.05} = \frac{0.05x}{0.05} \]
  • Simplify the division: \[ 120 = x \]
We've found the value of \(x\)! But always remember to:
  • Check your solution by substituting it back into the original equation.
  • Verify both sides are equal to confirm correctness.
For our problem: \[ 0.2(60) + 0.05(120) = 0.1(60 + 120)\ 12 + 6 = 18 \ 18 = 18 \] The solution checks out!

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