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Chapter 6: Problem 74

In Exercises 74-75, solve each proportion for \(x\). \(\frac{x+a}{a}=\frac{b+c}{c}\)

Short Answer

Expert verified
The value of \(x\) in the given proportion is \( \frac{ab}{c} \)

Step by step solution

01

Perform Cross Multiplication

Start by doing a cross multiplication. It means multiplying the denominator of the left ratio with the numerator of the right ratio and vice versa. So we get: \( (x+a) * c = a * (b+c) \)
02

Distribute and Simplify

Distribute \(c\) into \((x+a)\) and \(a\) into \((b+c)\) in order to eliminate the parentheses: \( cx+ac = ab+ac \)
03

Isolate the Variable x

The next step is to isolate \(x\) . We can subtract ac from both sides of the equation to get \(x\) alone on the left side. Doing so, we get: \( cx = ab \)
04

Solve for x

Finally, to solve for \(x\), we can divide both sides of the equation by \(c\), resulting in the solution: \( x = \frac{ab}{c} \)

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Key Concepts

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

Understanding Proportions
A proportion is an equation that shows two ratios are equivalent. In our problem, we have two fractions, \(\frac{x+a}{a}\) and \(\frac{b+c}{c}\), which are set equal to each other. The key idea is that if the two ratios are equal, they have the same value when simplified. This allows us to make predictions about unknown values, like \(x\), by creating equations that leverage the relationships between these quantities.
  • Composed of two equal ratios.
  • Often solved using cross multiplication.
  • Useful to find unknowns when direct calculation isn't possible.
By setting the cross products equal, we maintain the equality of the original proportions, allowing us to solve for unknowns.
Solving for X
To solve for \(x\) in a proportion like \(\frac{x+a}{a}=\frac{b+c}{c}\), the first step is to perform cross multiplication. This involves multiplying \((x+a)\) by \(c\) and \(a\) by \((b+c)\). The result is the equation: \((x+a) \cdot c = a \cdot (b+c)\).
Next, distribute the constants \(c\) and \(a\) over the terms in the parentheses:
  • \(c(x+a)\) becomes \(cx + ac\).
  • \(a(b+c)\) becomes \(ab + ac\).
This expanded equation, \(cx + ac = ab + ac\), is now ready for simplification. By subtracting \(ac\) from both sides, we isolate \(x\) and simplify the equation to \(cx = ab\). The final step is dividing by \(c\), giving us \(x = \frac{ab}{c}\). This process effectively isolates \(x\) by undoing the operations applied to it.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying equations to solve for unknowns. In solving our proportion, we engaged in several key manipulations:
  • Cross Multiplication: \((x+a)\cdot c\) and \(a\cdot (b+c)\) set up the equation.
  • Distribution: Expanded the equation to get rid of parentheses.
  • Subtraction: Removed like terms on both sides to isolate the variable.
  • Division: Solved for \(x\) by dividing by \(c\).
Each step simplifies the equation and moves us closer to isolating \(x\). By using these techniques, complex equations become simpler and more manageable, enabling us to solve for unknown quantities efficiently.

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Most popular questions from this chapter

Solve the equations in Exercises 53-72 using the quadratic formula. \(x^{2}-x=14\)

Solve the quadratic equations in Exercises 37-52 by factoring. \(x^{2}-2 x-15=0\)

The formula for converting Fahrenheit temperature, \(F\), to Celsius temperature, \(C\), is $$ C=\frac{5}{9}(F-32) \text {. } $$ If Celsius temperature ranges from \(15^{\circ}\) to \(35^{\circ}\), inclusive, what is the range for the Fahrenheit temperature?

Solve the equations in Exercises 53-72 using the quadratic formula. \(x^{2}+8 x+15=0\)

Solve the equations in Exercises 53-72 using the quadratic formula. \(3 x^{2}=5 x-1\)
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