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Chapter 10: Problem 27

Solve for \(x\) and \(y\) $$\frac{x}{5}=\frac{20}{x}=y$$

Short Answer

Expert verified
Solutions are \((10, 2)\) and \((-10, -2)\).

Step by step solution

01

Understand the Problem

We have a system of equations: \( \frac{x}{5} = y \) and \( \frac{20}{x} = y \). We need to find the values for \(x\) and \(y\).
02

Set Equations Equal

The equation \( \frac{x}{5} = y \) and \( \frac{20}{x} = y \) both equal \( y \). Thus, we can set these expressions equal to each other: \( \frac{x}{5} = \frac{20}{x} \).
03

Cross-Multiply

By cross-multiplying the equation \( \frac{x}{5} = \frac{20}{x} \), we get \( x^2 = 100 \).
04

Solve for x

To solve \( x^2 = 100 \), take the square root of both sides. Therefore, \( x = 10 \) or \( x = -10 \).
05

Substitute to Find y

Substitute \( x = 10 \) into \( \frac{x}{5} = y \): \( y = \frac{10}{5} = 2 \). Similarly, substitute \( x = -10 \): \( y = \frac{-10}{5} = -2 \).
06

Write the Solutions

The possible solutions are \((x, y) = (10, 2)\) and \((x, y) = (-10, -2)\).

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Key Concepts

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

Simultaneous Equations
Simultaneous equations refer to a set of equations that contain multiple variables.These equations are meant to be solved together in order to find a solution that satisfies both (or all) equations at the same time.
In our exercise, we are given two equations:
  • \( \frac{x}{5} = y \)
  • \( \frac{20}{x} = y \)
The goal is to find values of \(x\) and \(y\) that satisfy both equations. This type of problem requires identifying a common solution, sharing variables, and ensuring the same conditions apply to both equations. Solving simultaneous equations often involves substituting one equation into another, or setting the expressions equal to each other, as demonstrated here.
Cross-Multiplication
Cross-multiplication is a technique used to eliminate fractions by multiplying across the equals sign diagonally. It's especially helpful in solving equations that have fractions.
In this exercise, we utilize cross-multiplication with our expression \( \frac{x}{5} = \frac{20}{x} \) to simplify it. The steps include:
  • Multiply the numerator of the first fraction by the denominator of the second: \( x \times x \).
  • Then multiply the denominator of the first fraction by the numerator of the second: \( 5 \times 20 \).
  • Set the two products equal to each other, giving us \( x^2 = 100 \).
This process allows us to transform the equation into a simpler form that is easier to solve and removes any fractional components.
Square Roots
To solve the equation \( x^2 = 100 \), taking square roots of both sides helps determine the actual values of \( x \).
This involves finding a number which, when squared, provides 100.
  • The square root of 100 is 10.
  • We must also consider the negative square root, thus \( x = -10 \), because squaring both 10 and -10 returns the positive value 100.
This understanding comes from the fact that squaring any real number results in a non-negative number, indicating two potential solutions in this context, one positive and one negative. This type of solution is common when dealing with quadratic equations.
System of Equations
A system of equations is a set of two or more equations that you need to solve together. A solution to this system must satisfy all equations simultaneously.
In this exercise, we deal with two expressions for \( y \) and deduce by setting both expressions equal to each other:
  • \( \frac{x}{5} = y \)
  • \( \frac{20}{x} = y \)
After simplifying to \( x^2 = 100 \) and finding possible values for \( x \), we back-substitute these values into either equation to find corresponding \( y \) values:
  • When \( x = 10 \), \( y = 2 \).
  • When \( x = -10 \), \( y = -2 \).
Thus, the solutions are two pairs of \( (x, y) \) that solve this system, demonstrating how a system can have multiple valid solutions.

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Most popular questions from this chapter

The following exercises are about \(\triangle \mathrm{ABC}\), in which \(\mathrm{AB}=8 \mathrm{cm}, \mathrm{AC}=10 \mathrm{cm},\) and \(\mathrm{BC}=$$6 \mathrm{cm}\). Use the Angle Bisector Theorem to find AD and DB. Then, measure \(\overline{A D}\) and \(\overline{D B}\) on your drawing to check your answers.

Find the geometric mean between each of the following pairs of numbers. Example: 12 and 15 $$ \begin{aligned} \text { Solution: } \frac{12}{x} &=\frac{x}{15} \\ x^{2} &=180 \\ x &=\sqrt{180} \text { (because } x \text { is positive }) \\ &=\sqrt{36} \cdot 5 \\ &=6 \sqrt{5} \end{aligned} $$ 8 and 10

Solve for \(x\) and \(y\) $$\frac{x-1}{x}=\frac{2}{7}=\frac{y}{y+1}$$

Two ratios that are famous in the history of the number \(\pi\) are \(\frac{22}{7}\) and \(\frac{355}{113}\) Express \(\frac{22}{7}\) in decimal form to the nearest hundredth.

(IMAGE CAN'T COPY). In \(\triangle \mathrm{BUD}\) and \(\triangle \mathrm{PES}, \angle \mathrm{B}=\angle \mathrm{P}, \frac{\mathrm{BU}}{\mathrm{PE}}=\frac{\mathrm{BD}}{\mathrm{PS}}\) \(\overline{\mathrm{UA}} \perp \overline{\mathrm{BD}},\) and \(\overline{\mathrm{ET}} \perp \overline{\mathrm{PS}}\) Why is \(\triangle \mathrm{BUD} \sim \triangle \mathrm{PES} ?\)
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