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\(5-22=\) Solve the equation. $$ |x-3|=2 $$

Short Answer

Expert verified
The solutions are x = 5 and x = 1.

Step by step solution

01

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 \(a=-b\).
02

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.
03

Solve the First Equation

Start with solving \(x-3=2\). Add 3 to both sides to isolate \(x\): \(x=2+3\), which simplifies to \(x=5\).
04

Solve the Second Equation

Now solve \(x-3=-2\). Similarly, add 3 to both sides: \(x=-2+3\), which simplifies to \(x=1\).
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.
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.
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
  • Keep the equation balanced by performing equal actions on both sides.
  • Lead to finding exact numerical solutions for \(x\).
Hence, the isolated variable represents the specific value estimated to satisfy the given absolute value equation.

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Most popular questions from this chapter

What's Wrong Here? It is tempting to try to solve an in- equality as if it were an equation. For instance, we might try to solve \(1<3 / x\) by multiplying both sides by \(x\) , to get \(x<3,\) so the solution would be \((-\infty, 3) .\) But that's wrong; for example, \(x=-1\) lies in this interval but does not satisfy the original in- equality. Explain why this method doesn't work (think about the sign of \(x\) ). Then solve the inequality correctly.

Evaluate the expression and write the result in the form a bi. $$ i^{1002} $$

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Falling Ball Using calculus, it can be shown that if a ball is thrown upward with an initial velocity of 16 \(\mathrm{ft} / \mathrm{s}\) from the top of a building 128 \(\mathrm{ft}\) high, then its height \(h\) above the ground \(t\) seconds later will be $$ h=128+16 t-16 t^{2} $$ During what time interval will the ball be at least 32 \(\mathrm{ft}\) above the ground?

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