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Chapter 2: Problem 7

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

Short Answer

Expert verified
The solution is \(x = 12\). Verifying this, both sides of the original equation equal 9.

Step by step solution

01

Eliminate Fractions

To eliminate fractions, multiply the entire equation by the least common multiple of the denominators (3 and 4) which is 12. This gives: \[ 12 \times \left( \frac{2x}{3} + 1 \right) = 12 \times \frac{3x}{4} \] Simplifying both sides, we have: \[ 8x + 12 = 9x \]
02

Move Variable Terms to One Side

Now, subtract \(8x\) from both sides of the equation to isolate the variable terms on one side. This gives: \[ 8x + 12 - 8x = 9x - 8x \] Resulting in: \[ 12 = x \]
03

Check the Solution

Substitute \(x = 12\) back into the original equation to verify the solution: \[ \frac{2(12)}{3} + 1 = \frac{3(12)}{4} \] This simplifies to \(8 + 1 = 9\) on the left side and \(9 = 9\) on the right side, 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.

Eliminating Fractions
Fractions can sometimes make equations seem more complicated than they actually are. One effective strategy to simplify an equation is to eliminate the fractions. This can be done by finding a common multiple of all the denominators involved. By multiplying the entire equation by this common multiple, the fractions are removed, making the equation easier to manage.

Let's look at the equation from our example, \( \frac{2x}{3} + 1 = \frac{3x}{4} \). The denominators are 3 and 4. When we identify their least common multiple, we can use it to tidy up the equation.
Least Common Multiple
The Least Common Multiple (LCM) of two numbers is the smallest number that is a multiple of both. It helps in eliminating fractions by providing a number that can be used to cancel out all denominators.
  • First, identify the denominators: 3 and 4.
  • The multiples of 3 are: 3, 6, 9, 12, ...
  • The multiples of 4 are: 4, 8, 12, 16, ...
12 is the smallest number that appears in both lists, making it the LCM.

We can use 12 to remove the fractions in an equation by multiplying each term, like so:

\[ 12 \left( \frac{2x}{3} + 1 \right) = 12 \left( \frac{3x}{4} \right) \]

This step simplifies our equation significantly.
Checking Solutions
After solving an equation, it's important to check if your solution is correct. You do this by substituting your answer back into the original equation and verifying whether both sides of the equation are equal.

In our case, when we found \(x = 12\), we substitute it back into the original equation:

\[ \frac{2(12)}{3} + 1 = \frac{3(12)}{4} \]

Simplifying both sides gives:
  • Left side: \( 8 + 1 = 9 \)
  • Right side: \( 9 = 9 \)
Since both sides are equal, it confirms that our solution \(x = 12\) is indeed correct.
Isolating the Variable
In order to find the solution to a linear equation, we need to isolate the variable, which means getting the variable by itself on one side of the equation.

After eliminating fractions and simplifying the equation from our example, we had:

\[ 8x + 12 = 9x \]

To isolate \(x\), subtract \(8x\) from both sides:

\[ 8x + 12 - 8x = 9x - 8x \]

This results in \( 12 = x \), meaning that we have successfully isolated the variable. From here, it's simple to verify the solution, as shown in the previous section.

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Most popular questions from this chapter

In \(3-20,\) solve each equation and check. $$ 0.4 x+8=0.5 x $$

In \(25-30,\) perform the indicated operations and write the result in simplest form. In each case, list any values of the variables for which the fractions are not defined. $$ (3 b)^{2} \div \frac{3 b}{b+2} \cdot \frac{2 b+4}{b} $$

What is the solution set of \(\frac{|x|}{x}<0 ?\) Justify your answer.

In \(3-20,\) solve each equation and check. $$ \frac{7}{2 x-3}=\frac{4}{x} $$

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. $$ (2 x+7) \div \frac{1}{2 x^{2}+5 x-7} $$
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