/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Rewrite each expression without ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Rewrite each expression without using absolute value notation. $$|x-3|+|x-4| \text { given that } x<3$$

Short Answer

Expert verified
-2x + 7

Step by step solution

01

Understand the Absolute Values

The given expression involves absolute values: \(|x-3|+|x-4|\). Absolute values simplify to linear expressions depending on whether the expression inside the absolute value is positive or negative.
02

Consider the Condition x

Given that \(x < 3\), both \(x-3\) and \(x-4\) are negative. Hence, the absolute values can be rewritten using the property \(|a| = -a\) if \(a < 0\).
03

Remove the Absolute Value for |x-3|

Since \(x<3\), the expression \(x-3\) is negative. Thus, \(|x-3| = -(x-3)\). Simplifying this gives \(|x-3| = -x + 3\).
04

Remove the Absolute Value for |x-4|

Similarly, since \(x<3\), \(x-4\) is negative. Thus, \(|x-4| = -(x-4)\). Simplifying this gives \(|x-4| = -x + 4\).
05

Combine the Simplified Expressions

Add the simplified expressions from steps 3 and 4: \(-x + 3 + (-x + 4)\) simplifies to \(-2x + 7\).

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Key Concepts

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

Absolute Value Properties
Absolute value is a number's distance from zero on the number line, without considering direction. This means it is always non-negative. For example, the absolute value of both \(5\) and \(-5\) is \(5\). The mathematical sentence for absolute value is denoted by vertical bars, like \(|a|\).
Consider \(|x-3|\) in our exercise. Here, the concept used is that if \(a < 0\), then \(|a| = -a\). This is because when a number inside the absolute value is negative, its absolute form is its negative counterpart, which makes it positive.
This property allows us to remove the absolute value signs by using conditions given in the problem, like \(x < 3\) for the expression \(|x-3| + |x-4|\). Because both \(x-3\) and \(x-4\) are negative in this range, we apply the property to rewrite them without absolute value brackets.
Linear Expressions
Linear expressions are algebraic expressions that involve constant terms and terms with variables raised only to the first power. They are written in the form \(ax + b\), where \(a\) and \(b\) are constants.
In the context of our problem, \(|x-3|\) becomes \(-x+3\) and \(|x-4|\) becomes \(-x+4\) when simplified. These are both linear expressions because they follow the form \(ax + b\).
Understanding linear expressions is key in algebra because they form the basis for solving equations and inequalities. They represent lines on a graph, and in this case, they are used to evaluate values under specific conditions, such as when \(x < 3\). Remember, a linear expression changes predictably, and understanding that change is vital for solving many beginner algebra problems.
Simplifying Expressions
Simplifying expressions means making them as concise and clear as possible. It involves combining like terms and removing any unnecessary components, such as absolute value signs in this exercise.
For example, when we look at \(-x+3\) and \(-x+4\), we see that both have \(-x\) as a common term. By combining these like terms, we simplify the expression.
So, instead of dealing with two separate linear expressions, these can be combined to give \(-2x+7\). Remember: simplification does not change the value the expression represents; it only makes it easier to work with and understand. This step is crucial for solving more complex equations efficiently.
Inequalities
Inequalities express a relationship of one expression being greater than or less than another. They are symbolized by signs such as \(<\), \(>\), \(\leq\), and \(\geq\). Unlike equations, inequalities do not always represent definite values but rather a range of possible values.
In this exercise, we deal with the inequality \(x < 3\). This tells us the range of values \(x\) can take. It indicates that any number less than \(3\) satisfies the given condition, which impacts how we simplify the absolute value expressions.
Understanding how inequalities operate is critical for finding solutions that must fit within specific boundaries, like those found in real-world scenarios or complex math problems. Recognizing when and how to apply them helps solve problems involving conditions, like those given in many algebraic expressions and constraints.

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Most popular questions from this chapter

Graph the equation after determining the \(x\) - and \(y\) -intercepts and whether the graph possesses any of the three types of symmetry described on page 58 $$y=1 / x^{2}$$

Show that the slope of the line passing through the two points \(\left(a, a^{2}\right)\) and \(\left(x, x^{2}\right)\) is \(x+a .\) Hint: You'll need to use difference of squares factoring from intermediate algebra. If you need a review, see Appendix B.4.

(a) Sketch the line \(y=\frac{1}{2} x-5\) and the point \(P(1,3) .\) Follow parts (b)-(d) to calculate the perpendicular distance from point \(P(1,3)\) to the line. (b) Find an equation of the line that passes through \(P(1,3)\) and is perpendicular to the line \(y=\frac{1}{2} x-5\) (c) Find the coordinates of the point where these two lines intersect. Hint: From intermediate algebra, to find where two lines \(y=m x+b\) and \(y=M x+B\) intersect, set the expressions \(m x+b\) and \(M x+B\) equal to each other, and solve for \(x\) (d) Use the distance formula to find the perpendicular distance from \(P(1,3)\) to the line \(y=\frac{1}{2} x-5\)

Graph the equation after determining the \(x\) - and \(y\) -intercepts and whether the graph possesses any of the three types of symmetry described on page 58 $$y=-1 / x$$

You \(\%\) need to recall the following definitions and results from elementary geometry. In a triangle, a line segment drawn from a vertex to the midpoint of the opposite side is called a median. The three medians of a triangle are concurrent; that is, they intersect in a single point. This point of intersection is called the centroid of the triangle. A line segment drawn from a vertex perpendicular to the opposite side is an altitude. The three altitudes of a triangle are concurrent; the point where the altitudes intersect is the orthocenter of the triangle. This exercise illustrates the fact that the altitudes of a triangle are concurrent. Again, we'll be using \(\triangle A B C\) with vertices \(A(-4,0), B(2,0),\) and \(C(0,6) .\) Note that one of the altitudes of this triangle is just the portion of the \(y\) -axis extending from \(y=0\) to \(y=6 ;\) thus, you won't need to graph this altitude; it will already be in the picture. (a) Using paper and pencil, find the equations for the three altitudes. (Actually, you are finding equations for the lines that coincide with the altitude segments.) (b) Use a graphing utility to draw \(\triangle A B C\) along with the three altitude lines that you determined in part (a). Note that the altitudes appear to intersect in a single point. Use the graphing utility to estimate the coordinates of this point. (c) Using simultaneous equations (from intermediate algebra), find the exact coordinates of the orthocenter. Are your estimates in part (b) close to these values?

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