Problem 74
Question
Solve each linear or quadratic equation \(2(3 x+1)+15=-7\)
Step-by-Step Solution
Verified Answer
The solution is \(x = -4\).
1Step 1: Expand the Equation
First, distribute the 2 across the parentheses. This means multiplying 2 by each term inside the parentheses. So, calculate \(2 \times (3x + 1)\). This simplifies to \(6x + 2\). The equation now becomes \(6x + 2 + 15 = -7\).
2Step 2: Combine Like Terms
Next, combine the constant terms on the left side of the equation. Add 2 and 15 to get \(17\). Now the equation looks like \(6x + 17 = -7\).
3Step 3: Isolate the Variable Term
Subtract 17 from both sides to isolate the term with the variable on one side. This gives \(6x = -7 - 17\). Simplifying the right side, you have \(6x = -24\).
4Step 4: Solve for the Variable
Finally, divide both sides of the equation by 6 to solve for \(x\). So, \(x = \frac{-24}{6}\). Simplifying the fraction, you find \(x = -4\).
Key Concepts
Distributive PropertyCombining Like TermsIsolating the Variable
Distributive Property
The distributive property is a fundamental concept in algebra that is used to simplify expressions and equations. When you encounter an expression like \( a(b + c) \), the distributive property allows you to "distribute" the \( a \) across the terms inside the parentheses. This means you apply \( a \) to both \( b \) and \( c \) individually.
For example, in the equation \( 2(3x + 1) + 15 = -7 \), we distribute 2 to both \( 3x \) and 1:
By using the distributive property, you open up the expression and make it easier to work with the individual terms, preparing the equation for further simplification.
For example, in the equation \( 2(3x + 1) + 15 = -7 \), we distribute 2 to both \( 3x \) and 1:
- Multiply 2 by \( 3x \) to get \( 6x \).
- Multiply 2 by 1 to get 2.
By using the distributive property, you open up the expression and make it easier to work with the individual terms, preparing the equation for further simplification.
Combining Like Terms
Combining like terms is an essential step in simplifying algebraic expressions. It involves adding or subtracting terms that have the same variable raised to the same power. This simplification allows you to work efficiently with equations.
In our example equation, \( 6x + 2 + 15 = -7 \), you look for terms that can be combined:
Combining like terms helps condense equations, making them neater and often easier to solve by reducing the clutter on one side of the equation.
In our example equation, \( 6x + 2 + 15 = -7 \), you look for terms that can be combined:
- The constants 2 and 15 are like terms, as they both do not have a variable component.
Combining like terms helps condense equations, making them neater and often easier to solve by reducing the clutter on one side of the equation.
Isolating the Variable
Isolating the variable is a crucial step in solving linear equations. The goal is to get the variable alone on one side of the equation, which then allows you to determine its value.
With the equation now \( 6x + 17 = -7 \), we need to isolate \( x \). This is done by removing the constant term on the same side as the variable. Subtract 17 from both sides:
Next, solve for \( x \) by dividing both sides by 6 to get \( x = \frac{-24}{6} \). Simplifying this gives \( x = -4 \).
Isolating the variable is the step that ultimately allows you to find what value the variable represents in the equation, thus solving it.
With the equation now \( 6x + 17 = -7 \), we need to isolate \( x \). This is done by removing the constant term on the same side as the variable. Subtract 17 from both sides:
- Left side: \( 6x + 17 - 17 \) simplifies to \( 6x \).
- Right side: \( -7 - 17 \) simplifies to \( -24 \).
Next, solve for \( x \) by dividing both sides by 6 to get \( x = \frac{-24}{6} \). Simplifying this gives \( x = -4 \).
Isolating the variable is the step that ultimately allows you to find what value the variable represents in the equation, thus solving it.
Other exercises in this chapter
Problem 74
In your own words, describe how to add or subtract two rational expressions with the same denominator.
View solution Problem 74
Perform each indicated operation. $$ \frac{8}{15} \div \frac{5}{8} $$
View solution Problem 75
Explain why the \(L C D\) of the rational expressions \(\frac{7}{x+1}\) and \(\frac{9 x}{(x+1)^{2}}\) is \((x+1)^{2}\) and not \((x+1)^{3}\).
View solution Problem 75
Which of the following are incorrect and why? $$ \frac{5 a-15}{5} \text { simplifies to } a-3 $$
View solution