Problem 9
Question
\(5-22=\) Solve the equation. $$ |x-3|=2 $$
Step-by-Step Solution
Verified Answer
The solutions are x = 5 and x = 1.
1Step 1: Understanding Absolute Value
The equation is \(|x-3|=2\). An absolute value equation \(|a|=b\) indicates that the distance of \(a\) from 0 is \(|b|\). This means \(a=b\) or \(a=-b\).
2Step 2: Set up Potential Equations
Given \(|x-3|=2\), we understand this as two potential equations: \(x-3=2\) and \(x-3=-2\). This is because \(x-3\) can either be 2 units to the right of zero or 2 units to the left.
3Step 3: Solve the First Equation
Start with solving \(x-3=2\). Add 3 to both sides to isolate \(x\): \(x=2+3\), which simplifies to \(x=5\).
4Step 4: Solve the Second Equation
Now solve \(x-3=-2\). Similarly, add 3 to both sides: \(x=-2+3\), which simplifies to \(x=1\).
5Step 5: Combine Solutions
The solutions to the original absolute value equation \(|x-3|=2\) are \(x=5\) and \(x=1\). These are the points that satisfy the equation.
Key Concepts
Solving EquationsDistance from ZeroIsolating Variables
Solving Equations
Solving equations is a fundamental part of algebra. It involves finding the value of a variable that makes an equation true. To solve an equation, we often start by simplifying both sides, combining like terms, or using techniques like factoring when necessary. In the case of absolute value equations, which involve a number's distance from zero, using absolute value properties is crucial. For example, if the equation is \(|x-3|=2\), it means that the expression \(|x-3|\) is exactly 2 units away from zero on the number line.
To solve such equations, recognize that if \(|a|=b\), then \(a\) can be equal to either \(b\) or \(-b\). This dual possibility is because absolute value measures how far a number is from zero, regardless of direction.
When solving absolute value equations, you'll set up two separate equations to consider both scenarios represented by \(b\) and \(-b\). From there, resolve these equations as you would any linear equation, determining the precise values for the variable.
To solve such equations, recognize that if \(|a|=b\), then \(a\) can be equal to either \(b\) or \(-b\). This dual possibility is because absolute value measures how far a number is from zero, regardless of direction.
When solving absolute value equations, you'll set up two separate equations to consider both scenarios represented by \(b\) and \(-b\). From there, resolve these equations as you would any linear equation, determining the precise values for the variable.
Distance from Zero
The absolute value of a number fundamentally represents its numerical distance from zero. It symbolizes magnitude without some influence of direction. This is why the absolute value of a positive number is the number itself, and for a negative, it is the positive counterpart of that number. In the mathematical language, the absolute value of \(a\) is written as \(|a|\).
In the equation \(|x-3|=2\), the expression \(x-3\) must be 2 units away from zero on a number line. This could mean two possible positions: 2 units either to the right or to the left of zero. Thus, absolute value questions often resolve into two potential solutions.
Remember, the concept of distance from zero helps in visualizing why we break down absolute value problems into two separate linear equations. Each solution satisfies the condition of being at the specified distance.
In the equation \(|x-3|=2\), the expression \(x-3\) must be 2 units away from zero on a number line. This could mean two possible positions: 2 units either to the right or to the left of zero. Thus, absolute value questions often resolve into two potential solutions.
Remember, the concept of distance from zero helps in visualizing why we break down absolute value problems into two separate linear equations. Each solution satisfies the condition of being at the specified distance.
Isolating Variables
In algebra, isolating the variable means rearranging an equation to have the variable on one side and everything else on the other. This process helps pinpoint the exact value of the variable, effectively solving the equation.
For the absolute value equation \(|x-3|=2\), after splitting it into two linear equations, you proceed to isolate \(x\) in each. This involves simple steps of "undoing" operations around the variable. For \(x-3=2\), you add 3 to both sides to isolate \(x\), resulting in \(x=5\). Similarly, for \(x-3=-2\), adding 3 both sides results in \(x=1\).
These operations
For the absolute value equation \(|x-3|=2\), after splitting it into two linear equations, you proceed to isolate \(x\) in each. This involves simple steps of "undoing" operations around the variable. For \(x-3=2\), you add 3 to both sides to isolate \(x\), resulting in \(x=5\). Similarly, for \(x-3=-2\), adding 3 both sides results in \(x=1\).
These operations
- Keep the equation balanced by performing equal actions on both sides.
- Lead to finding exact numerical solutions for \(x\).
Other exercises in this chapter
Problem 8
Determine whether the given value is a solution of the equation. \(\frac{1}{x}-\frac{1}{x-4}=1\) (a) \(x=2 \quad\) (b) \(x=4\)
View solution Problem 9
Find the real and imaginary parts of the complex number. $$ 3 $$
View solution Problem 9
Let \(S=\left\\{-2,-1,0, \frac{1}{2}, 1, \sqrt{2}, 2,4\right\\} .\) Determine which elements of \(S\) satisfy the inequality. $$ \frac{1}{x} \leq \frac{1}{2} $$
View solution Problem 9
Solve the equation by factoring. $$ 3 x^{2}-5 x-2=0 $$
View solution