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Chapter 1: Problem 50

Solve each equation for \(x\) in terms of the other letters. $$a x+b=b x-a$$

Short Answer

Expert verified
\(x = \frac{-a - b}{a-b} \)

Step by step solution

01

Isolate Terms with 'x' on One Side

Start by getting all terms with the variable \(x\) on one side of the equation. Move \(b x\) to the left side by subtracting \(b x\) from both sides.\[ a x - b x + b = -a \]
02

Combine Like Terms

Combine the terms with \(x\) on the left side of the equation.\[ (a-b)x + b = -a \]
03

Isolate the \(x\) Variable

Subtract \(b\) from both sides to further isolate \(x\).\[ (a-b)x = -a - b \]
04

Solve for \(x\)

To completely isolate \(x\), divide both sides of the equation by \((a-b)\), assuming \(a eq b\).\[ x = \frac{-a - b}{a-b} \]

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Key Concepts

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

Solving Equations
When solving equations, the main goal is to find the value of the variable that makes the equation true. This process involves several steps, including moving terms from one side of the equation to the other and transforming the equation until the variable is isolated.
For example, in the equation given, move terms around so like terms are grouped together, making it easier to solve for the variable.
  • Start by deciding which variable you want to isolate (usually all terms with the variable on one side of the equation).
  • Then, perform arithmetic operations such as addition, subtraction, multiplication, or division symmetrically to maintain equality.
  • Ensure you follow the correct order of operations to avoid mistakes.
Once the equation is simplified and rearranged, you can find the variable's value that balances the equation.
Variable Isolation
Variable isolation is essential for finding the solution of an equation. It refers to getting the variable you are solving for alone on one side of the equation. This is especially important in equations where more than one term contains the variable of interest.
In the example provided, achieving variable isolation required moving all terms involving 'x' to one side and the constants to the other. This was done by:
  • Subtracting terms to cancel out constants or other variable terms from the side where you are isolating the main variable.
  • Simplifying complex expressions by performing arithmetic operations.
Thus, when isolating 'x,' ensure the end step results in only 'x' equals something, as neat and simple as possible.
Combining Like Terms
Combining like terms refers to simplifying an expression by adding or subtracting coefficients that have the same variable component.
In the equation originally given, there are two terms involving 'x,' specifically 'ax' and 'bx' on different sides. To combine these terms effectively:
  • Move them to the same side of the equation.
  • Add or subtract the coefficients (numbers in front of the variable) where applicable.
This makes the equation a lot easier to solve, as it reduces the number of separate terms. In this exercise, by combining like terms, we simplified _ ax - bx_ to _(a-b)x_ to help isolate the variable efficiently.

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Most popular questions from this chapter

In each of parts (a) through (d), first solve the equation for \(y\) so that you can enter it in your graphing utility. Then use the graphing utility to graph the equation in an appropriate viewing rectangle. In each case, the graph is a line. Given that the \(x\) - and \(y\) -intercepts are (in every case here) integers, read their values off the screen and write them down for easy reference when you get to part (e). (a) \(\frac{x}{2}+\frac{y}{3}=1\) (c) \(\frac{x}{6}+\frac{y}{5}=1\) (b) \(\frac{x}{-2}+\frac{y}{-3}=1\) (d) \(\frac{x}{-6}+\frac{y}{-5}=1\) (e) On the basis of your results in parts (a) through (d), describe, in general, the graph of the equation \(\frac{x}{a}+\frac{y}{b}=1,\) where \(a\) and \(b\) are nonzero constants.

The set of real numbers satisfying the given inequality is one or more intervals on the number line. Show the interval(s) on a number line. $$|x|>1$$

Is the graph of the line \(y=0\) the \(x\) -axis or the \(y\) -axis?

(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 \(y=\frac{1}{2} x-5\)

Rewrite each statement using absolute value notation, as in Example 5. The number \(y\) is less than one unit from the number \(t\).
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