Problem 122
Question
determine whether each statement makes sense or does not make sense, and explain your reasoning. Because I know how to clear an equation of fractions. I decided to clear the equation \(0.5 x+8.3=12.4\) of decimals by multiplying both sides by 10
Step-by-Step Solution
Verified Answer
The method of clearing decimals by multiplying both sides of the equation by 10 is valid, as it simplifies the equation while preserving the relationship between quantities.
1Step 1: Verifying the Validity of the Method
1. Write down the given equation, which is \(0.5 x+8.3=12.4\). 2. Apply the method used by the person, that is, multiply each term of the equation by 10. This gives: \(10 \times 0.5 x+10 \times 8.3=10 \times 12.4\) which simplifies to \(5x + 83 = 124\).
2Step 2: Cross-checking the Outcome
1. Solve the simplified equation from Step 1, which is \(5x + 83 = 124\). Subtract 83 from both sides of the equation and divide by 5 to solve for x. This gives \(x = \frac{124-83}{5}\) which is \(x=8.2\). 2. Substitute this value of \(x=8.2\) into the original equation to check the validity. Substituting \(x=8.2\) into \(0.5 x+8.3=12.4\) gives \(0.5 \times 8.2 + 8.3 = 12.4\), verifying the solution.
Key Concepts
Decimals in EquationsMultiplication to Clear DecimalsEquation VerificationFraction Elimination
Decimals in Equations
When dealing with equations that contain decimals, it can sometimes feel a bit daunting, especially if you're more familiar with whole numbers. Decimals, however, are just another form of representing numbers and can often pop up in real-world problems or scientific calculations.
You might find decimals attached to variables like in the equation \(0.5x + 8.3 = 12.4\). Here, not only the coefficient of \(x\) (which is 0.5), but also the constant terms (8.3 and 12.4) are in decimal form.
Handling these decimals directly in calculations can be tricky because they require precise arithmetic manipulation. That's where techniques like clearing decimals become handy, simplifying the process and making calculations easier to follow.
You might find decimals attached to variables like in the equation \(0.5x + 8.3 = 12.4\). Here, not only the coefficient of \(x\) (which is 0.5), but also the constant terms (8.3 and 12.4) are in decimal form.
Handling these decimals directly in calculations can be tricky because they require precise arithmetic manipulation. That's where techniques like clearing decimals become handy, simplifying the process and making calculations easier to follow.
Multiplication to Clear Decimals
One effective method to handle decimals in an equation is to "clear" them by multiplying every term by the power of 10 that can eliminate the decimal points. In the given equation \(0.5x + 8.3 = 12.4\), each term includes decimals up to one decimal place (tenths). By multiplying through by 10, each term becomes a whole number.
This transformation gives us:
This technique not only simplifies arithmetic operations but also helps in visualizing the solution clearly without the added complexity of decimals.
This transformation gives us:
- \(10 \times 0.5x = 5x\)
- \(10 \times 8.3 = 83\)
- \(10 \times 12.4 = 124\)
This technique not only simplifies arithmetic operations but also helps in visualizing the solution clearly without the added complexity of decimals.
Equation Verification
Verifying a solution to an equation is essential to ensure that your calculations are correct. Once you've found a potential solution to a transformed equation, substitute it back into the original or adjusted equation to check if it holds true.
In our example, after clearing decimals and solving \(5x + 83 = 124\), you find \(x = 8.2\). Substitute \(x = 8.2\) back into the original equation \(0.5x + 8.3 = 12.4\) to verify:
This step is crucial as it assures your transformation and calculations didn't inadvertently alter the solution.
In our example, after clearing decimals and solving \(5x + 83 = 124\), you find \(x = 8.2\). Substitute \(x = 8.2\) back into the original equation \(0.5x + 8.3 = 12.4\) to verify:
- Calculate \(0.5 \times 8.2 = 4.1\)
- Add \(4.1 + 8.3 = 12.4\)
This step is crucial as it assures your transformation and calculations didn't inadvertently alter the solution.
Fraction Elimination
Although our problem dealt primarily with decimals, it's worth understanding that a similar process applies to equations with fractions. Clearing an equation of fractions involves multiplying by the least common denominator (LCD) of all fractions present.
This process effectively turns fractional coefficients or terms into whole numbers, simplifying the arithmetic and reducing potential errors.
The principle is akin to clearing decimals: by multiplying all terms by an appropriate number (here, the LCD applies), you simplify the character of the equation.
Therefore, when you encounter a fraction-heavy equation, multiply through by the LCD, solve the simplified equation, and verify by substituting back to ensure no errors occurred in the process.
This process effectively turns fractional coefficients or terms into whole numbers, simplifying the arithmetic and reducing potential errors.
The principle is akin to clearing decimals: by multiplying all terms by an appropriate number (here, the LCD applies), you simplify the character of the equation.
Therefore, when you encounter a fraction-heavy equation, multiply through by the LCD, solve the simplified equation, and verify by substituting back to ensure no errors occurred in the process.
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