Problem 16
Question
Solve each equation. $$\frac{1}{7} x+2.87=-3.01$$
Step-by-Step Solution
Verified Answer
The solution to the equation is \( x = -41.16 \).
1Step 1: Isolate the Variable
First, we need to isolate the term with the variable. Start by subtracting 2.87 from both sides of the equation:\[\frac{1}{7}x + 2.87 - 2.87 = -3.01 - 2.87\]This simplifies to:\[\frac{1}{7}x = -5.88\]
2Step 2: Remove the Fraction
To eliminate the fraction \( \frac{1}{7} \), multiply both sides by 7:\[7 \times \frac{1}{7}x = 7 \times (-5.88)\]This simplifies to:\[ x = -41.16\]
3Step 3: Verify the Solution
Substitute \( x = -41.16 \) back into the original equation to ensure it satisfies:\[\frac{1}{7}(-41.16) + 2.87 = -3.01\]Calculate \( \frac{1}{7}(-41.16) \) which is \(-5.88\) and check:\[-5.88 + 2.87 = -3.01\]Since both sides equal, our solution \( x = -41.16 \) is verified.
Key Concepts
Isolating the VariableRemoving FractionsVerifying Solutions
Isolating the Variable
In many math problems, especially when solving linear equations, the first goal is to isolate the variable. This means rearranging the equation until the variable stands alone on one side of the equation sign. Let's consider the equation: \( \frac{1}{7} x + 2.87 = -3.01 \).
Start by focusing on the side of the equation that contains the variable. We have \( \frac{1}{7} x + 2.87 \). To remove the 2.87, perform the reverse operation by subtracting it from both sides. Why subtract? Because doing the opposite cancels the term with 2.87:
\[ \frac{1}{7} x + 2.87 - 2.87 = -3.01 - 2.87 \]
This leaves us with \( \frac{1}{7} x = -5.88 \), thus isolating \( x \) in its fraction form on one side.
Start by focusing on the side of the equation that contains the variable. We have \( \frac{1}{7} x + 2.87 \). To remove the 2.87, perform the reverse operation by subtracting it from both sides. Why subtract? Because doing the opposite cancels the term with 2.87:
\[ \frac{1}{7} x + 2.87 - 2.87 = -3.01 - 2.87 \]
This leaves us with \( \frac{1}{7} x = -5.88 \), thus isolating \( x \) in its fraction form on one side.
Removing Fractions
Fractions can make equations appear more complex, but there is a straightforward approach to eliminate them. Once you have isolated the variable, like in \( \frac{1}{7} x = -5.88 \), the next goal is to simplify the equation by getting rid of the fraction.
Here, \( \frac{1}{7} \) implies that \( x \) is divided by 7. To remove this division, multiply the whole equation by 7. By doing this, you balance out the fraction:
And there you have it, \( x \) is now free of fractions, appearing as a whole number in the equation.
Here, \( \frac{1}{7} \) implies that \( x \) is divided by 7. To remove this division, multiply the whole equation by 7. By doing this, you balance out the fraction:
- Multiply both sides by 7: \( 7 \times \frac{1}{7} x = 7 \times (-5.88) \)
- This simplifies to: \( x = -41.16 \)
And there you have it, \( x \) is now free of fractions, appearing as a whole number in the equation.
Verifying Solutions
The final and vital step in solving equations is checking that your solution is correct. Verifying your solution assures that you didn't make any mistakes during your calculations.
Begin by substituting your value of \( x \) back into the original equation to see if it holds true. For example, with \( x = -41.16 \):
Since both sides of the original equation equal \(-3.01\), this confirms that the solution \( x = -41.16 \) is indeed correct. Verifying in this way ensures confidence in your math technique and the accuracy of your results.
Begin by substituting your value of \( x \) back into the original equation to see if it holds true. For example, with \( x = -41.16 \):
- Plug \( x \) back in: \( \frac{1}{7}(-41.16) + 2.87 \)
- Calculate \( \frac{1}{7}(-41.16) \), which results in \(-5.88\)
- Add 2.87: \(-5.88 + 2.87 = -3.01\)
Since both sides of the original equation equal \(-3.01\), this confirms that the solution \( x = -41.16 \) is indeed correct. Verifying in this way ensures confidence in your math technique and the accuracy of your results.
Other exercises in this chapter
Problem 16
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