Problem 64 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve equation. Use words or set notation to identify equations that have no solution, or equations that are true for all real numbers. \(2+3(2 x-7)=9-4(3 x+1)\)

Step-by-Step Solution

Answer

The solution to the equation is \(x = 1.11\).

Step 1: Distribute Terms

Start by distributing 3 into \(2x - 7\) on the left side and -4 into \(3x + 1\) on the right side of the equation. The equation becomes \(2 + 6x - 21 = 9 - 12x - 4\).

Step 2: Simplify Both Sides

Combine like terms on both sides of the equation. The equation simplifies to \(-15 + 6x = 5 - 12x\).

Step 3: Rearrange Terms to Solve for 'x'

Rearrange the equation to group x terms on one side and constants on the other. This results in \(6x + 12x = 5 + 15\). That simplifies to \(18x = 20\).

Step 4: Solve for 'x'

Finally, solve for x by dividing both sides of the equation by 18. The solution is \(x = 20/18 = 1.11\)

Key Concepts

Distributive Property

The distributive property is an important tool in algebra when dealing with expressions that have parentheses. This property allows us to multiply a single term by each term inside the parentheses individually.
For example, when we encounter an expression like \(3(2x - 7)\), the distributive property tells us to share the 3 by multiplying it with both \(2x\) and \(-7\). So, \(3 \times 2x\) becomes \(6x\) and \(3 \times (-7)\) becomes \(-21\). This makes the expression \(6x - 21\).
Similarly, apply the distributive property on the right side of the equation \(-4(3x + 1)\). This results in \(-12x - 4\), as \(-4\) is multiplied by both \(3x\) and \(1\).

• Remember: Every term within the parentheses gets multiplied by the term outside.
• Always check your signs (positive or negative) when applying the distributive property.

Combining Like Terms

Combining like terms is a straightforward yet essential step when simplifying algebraic expressions. It involves adding or subtracting terms that have the same variable part. This simplifies the expression and makes it easier to solve.
In our given equation, once we have used the distributive property, we obtain: \(2 + 6x - 21\) and \(9 - 12x - 4\). Before moving forward, we need to combine like terms on either side of the equation.

• On the left side, combine the constant terms: \(2\) and \(-21\), which simplifies to \(-19\). Put it together with the variable term to get \(-19 + 6x\).
• On the right side, combine the constants \(9\) and \(-4\) to get \(5\). Thus, it simplifies to \(5 - 12x\).

Combining the terms makes the equation simpler and sets it up for the next steps in solving.

Rearranging Equations

Rearranging equations is a critical part of algebra, especially when we want to isolate a particular variable. The goal is to have all like terms on the same side, so the structure is clean and straightforward for solving.
From our simplified equation \(-19 + 6x = 5 - 12x\), we need to form an equation where all \(x\)-terms are on one side and the constants on the other. We do this by adding \(12x\) to both sides and adding \(19\) to the constants.

• Add \(12x\) to \(6x\) to group all \(x\)-terms: \(6x + 12x = 18x\).
• Address the constant terms by taking \(19\) to the right side so it joins with \(5\), resulting in \(5 + 19 = 24\).

Now, the equation looks like \(18x = 24\), which is easier to solve for \(x\).

Division by a Constant

Division by a constant is often the final step when solving linear equations. In this stage, we simplify the equation by isolating the variable.
From our rearranged equation \(18x = 24\), we need \(x\) alone on one side. To achieve this, divide every term by 18.
By following these steps:

• The left side becomes \(\frac{18x}{18} = x\).
• The right side results in \(\frac{24}{18}\), simplifying to a fractional or decimal form that concludes the solution.

This action results in \(x = \frac{4}{3}\) (or approximately \(1.33\) when expressed as a decimal). Understanding division by a constant is crucial because it's a basic arithmetic operation that helps isolate the variable, leading us to our solution.