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Chapter 2: Problem 4

In \(3-20,\) solve each equation and check. $$ \frac{3}{4} x=14-x $$

Short Answer

Expert verified
The solution to the equation is \(x = 8\), and it checks out when substituted back into the original equation.

Step by step solution

01

Move all terms to one side

To start solving the equation \(\frac{3}{4}x = 14 - x\), move all terms involving \(x\) to one side to facilitate combining them. Add \(x\) to both sides of the equation:\[\frac{3}{4}x + x = 14\]
02

Combine like terms

Now, combine the \(x\) terms on the left side of the equation. Remember that adding \(1x\) is the same as adding \(\frac{4}{4}x\):\[\frac{3}{4}x + \frac{4}{4}x = 14\]\[\frac{7}{4}x = 14\]
03

Solve for x

To solve for \(x\), multiply both sides of the equation by the reciprocal of \(\frac{7}{4}\), which is \(\frac{4}{7}\):\[x = 14 \times \frac{4}{7}\]Calculate the right side:\[x = 14 \cdot \frac{4}{7} = 2 \cdot 4 = 8\]So, \(x = 8\).
04

Check the solution

Substitute \(x = 8\) back into the original equation to verify the solution:\[\frac{3}{4}(8) = 14 - 8\]Calculate both sides:\[\frac{3}{4}(8) = 6\]\[14 - 8 = 6\]Both sides equal 6, confirming that \(x = 8\) is the correct solution.

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Key Concepts

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

Combining Like Terms
When tackling linear equations, combining like terms is a fundamental aspect. It means bringing similar mathematical elements together to simplify expressions. In our example, the equation we start with is \( \frac{3}{4}x = 14 - x \). To solve, we first need all terms that involve \( x \) on one side for easier combining.
By adding \( x \) to both sides, we end up with \( \frac{3}{4}x + x = 14 \). Now, it's important to remember that \( x \) can be rewritten with a common denominator as \( \frac{4}{4}x \). This way, both terms involving \( x \) take the same form.
  • Adding \( \frac{3}{4}x \) and \( \frac{4}{4}x \) gives us \( \frac{7}{4}x \), making calculations straightforward.
  • This process of combining helps reduce the complexity and solve for \( x \) more efficiently.
It clarifies the equation as you work, keeping steps organized and accessible.
Reciprocal in Algebra
The reciprocal plays a crucial role when solving equations. In algebra, the reciprocal of a fraction \( \frac{a}{b} \) is \( \frac{b}{a} \). It is essential when you want to isolate a variable.
In our equation, after combining like terms, we have \( \frac{7}{4}x = 14 \). To solve for \( x \), we need to eliminate the fraction attached to it.
  • To do this, we multiply both sides by the reciprocal of the fraction \( \frac{7}{4} \), which is \( \frac{4}{7} \).
  • This action cancels out the fraction so we can easily solve for \( x \).
So, multiplying gives us \( x = 14 \times \frac{4}{7} \). Performing the calculation, \( x = 8 \) emerges as the solution. Employing reciprocals simplifies the equation, allowing direct access to the value of \( x \) without extra hurdles.
Checking Solutions in Algebra
After finding a solution to an equation, it is crucial to check its accuracy. Verification means substituting the solution back into the original equation and ensuring both sides equal.
For our equation \( \frac{3}{4}x = 14 - x \), we obtained \( x = 8 \). By substituting \( 8 \) back into the equation, we confirm its correctness.
  • On substituting, the left side becomes \( \frac{3}{4}(8) = 6 \).
  • The right side calculates as \( 14 - 8 = 6 \).
Both sides equate to 6, confirming that \( x = 8 \) is indeed the correct solution. Checking is a key step in algebra, ensuring no mistakes were made during calculation. By following this process, confidence in the obtained solution is gained, reinforcing the learning and solving experience.

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Most popular questions from this chapter

Matthew said that \(\frac{a}{b}+\frac{c}{d}=\frac{a d+b c}{b d}\) when \(b \neq 0, d \neq 0 .\) Do you agree with Matthew? Justify your answer.

In \(3-20\) , perform the indicated additions or subtractions and write the result in simplest form. In each case, list any values of the variables for which the fractions are not defined. $$ \frac{1}{x}+\frac{1}{x-2}-\frac{2}{x^{2}-2 x} $$

In \(3-20,\) solve each equation and check. $$ \frac{2 x}{3}+1=\frac{3 x}{4} $$

In \(13-24,\) divide and express each quotient in simplest form. In each case, list any values of the variables for which the fractions are not defined. $$ \frac{a^{2}}{8 a} \div \frac{3 a}{4} $$

In \(3-14,\) solve and check each inequality. $$ \frac{x}{x+5}-\frac{1}{x+5}>4 $$
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