Problem 13
Question
\(5-22=\) Solve the equation. $$ |4 x+7|=9 $$
Step-by-Step Solution
Verified Answer
The solutions are \(x = \frac{1}{2}\) and \(x = -4\).
1Step 1: Understanding Absolute Values
The equation involves an absolute value, which means that we must consider both the positive and negative scenarios that satisfy the equation. So, we write: 1. \(|4x + 7| = 9\), which translates to two separate equations: - \(4x + 7 = 9\) - \(4x + 7 = -9\).
2Step 2: Solving the First Equation
Solve the equation \(4x + 7 = 9\). Subtract 7 from both sides to get:\[ 4x = 9 - 7 \]Simplify the right side:\[ 4x = 2 \]Now, divide both sides by 4:\[ x = \frac{2}{4} \]Simplify the fraction:\[ x = \frac{1}{2} \].
3Step 3: Solving the Second Equation
Now, solve the equation \(4x + 7 = -9\). Subtract 7 from both sides:\[ 4x = -9 - 7 \]Simplify the right side:\[ 4x = -16 \]Now, divide both sides by 4:\[ x = \frac{-16}{4} \]Simplify the fraction:\[ x = -4 \].
4Step 4: Writing the Solution Set
Since the absolute value equation has two potential solutions, our final answer includes both solutions from the steps above:\[ x = \frac{1}{2} \quad \text{or} \quad x = -4 \].
Key Concepts
Understanding Absolute ValueSolving Linear EquationsDetermining the Solution Set
Understanding Absolute Value
Absolute value is a fundamental concept in mathematics, particularly when dealing with equations. The absolute value of a number refers to its distance from zero on the number line, regardless of its direction.
Therefore, the absolute value of any real number is always non-negative. For example, the absolute value of both -3 and 3 is 3, because both are 3 units away from zero.
When we solve equations that include absolute values, like \(|4x + 7| = 9\), we must consider two scenarios. One scenario is the expression inside the absolute value being equal to the positive result, and the other scenario is it being equal to the negative result.
Therefore, the absolute value of any real number is always non-negative. For example, the absolute value of both -3 and 3 is 3, because both are 3 units away from zero.
When we solve equations that include absolute values, like \(|4x + 7| = 9\), we must consider two scenarios. One scenario is the expression inside the absolute value being equal to the positive result, and the other scenario is it being equal to the negative result.
- This dual scenario approach accounts for the two possible distances from zero that could satisfy the absolute value equation.
- Thus, the equation splits into two separate linear equations.
Solving Linear Equations
Linear equations are equations of the first degree, which means they involve variables raised only to the power of one. In the context of an absolute value equation, once the equation is split into two linear equations, these equations can be solved independently.
For instance, based on our original problem \(|4x + 7| = 9\), it splits into two linear equations:
In our example:
For instance, based on our original problem \(|4x + 7| = 9\), it splits into two linear equations:
- \(4x + 7 = 9\)
- \(4x + 7 = -9\)
In our example:
- For \(4x + 7 = 9\): Subtracting 7 from both sides and dividing by 4 gives \(x = \frac{1}{2}\).
- For \(4x + 7 = -9\): Subtracting 7 from both sides and dividing by 4 gives \(x = -4\).
Determining the Solution Set
The solution set of an equation contains all the possible values that satisfy it. After working through an absolute value equation, it often results in two possible solutions, due to the nature of absolute values producing two possible cases.
For our equation \(|4x + 7| = 9\), the solution set is derived from solving both linear equations formed from considering the positive and negative cases – namely, \(4x + 7 = 9\) and \(4x + 7 = -9\).
For our equation \(|4x + 7| = 9\), the solution set is derived from solving both linear equations formed from considering the positive and negative cases – namely, \(4x + 7 = 9\) and \(4x + 7 = -9\).
- The first equation provides the solution \(x = \frac{1}{2}\).
- The second equation yields the solution \(x = -4\).
Other exercises in this chapter
Problem 12
Determine whether the given value is a solution of the equation. \(x^{2}-b x+\frac{1}{4} b^{2}=0\) (a) \(x=\frac{b}{2} \quad\) (b) \(x=\frac{1}{b}\)
View solution Problem 13
Find the real and imaginary parts of the complex number. $$ \sqrt{3}+\sqrt{-4} $$
View solution Problem 13
Solve the linear inequality. Express the solution using interval notation and graph the solution set. $$ 2 x-5>3 $$
View solution Problem 13
Solve the equation by factoring. $$ 6 x^{2}+5 x=4 $$
View solution