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Chapter 6: Problem 22

Solve. If no solution exists, state this. $$ \frac{x}{5}=\frac{20}{x} $$

Short Answer

Expert verified
The solutions are \( x = 10 \) and \( x = -10 \).

Step by step solution

01

Cross Multiply

To eliminate the fractions, cross multiply the equation \ \( \frac{x}{5} = \frac{20}{x} \). This gives us \( x^2 = 100 \).
02

Solve for x

Solve the equation \( x^2 = 100 \) by taking the square root of both sides. This results in two solutions: \( x = 10 \) and \( x = -10 \).
03

Verify Solutions

Substitute each solution back into the original equation to ensure they are valid. \ For \( x = 10 \): \( \frac{10}{5} = 2 \) and \( \frac{20}{10} = 2 \). Both sides are equal. \ For \( x = -10 \): \( \frac{-10}{5} = -2 \) and \( \frac{20}{-10} = -2 \). Both sides are equal.

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Key Concepts

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

Cross Multiplication
Cross multiplication is a method used to solve equations where two ratios (fractions) are set equal to each other. It essentially eliminates the fractions by multiplying the numerator of one ratio by the denominator of the other ratio. This process converts the rational equation into a simpler algebraic equation that can be solved more easily.

In our example, the original equation is \(\frac{x}{5} = \frac{20}{x}\). To eliminate the fractions, we cross multiply:

\(x \cdot x = 5 \cdot 20\)

This results in: \(x^2 = 100\)

With the fractions out of the way, we now have a straightforward quadratic equation to solve.
Quadratic Equations
A quadratic equation is a second-order polynomial equation in a single variable \(x\) with the general form \(ax^2 + bx + c = 0\). In our case, the equation we obtained from cross multiplication is \(x^2 = 100\).

To solve for \(x\), we need to isolate \(x\). We do this by taking the square root of both sides of the equation:

\(\sqrt{x^2} = \sqrt{100}\)

This simplifies to:

\(x = \pm 10\)

Hence, we have two solutions: \(x = 10\) and \(x = -10\). Quadratic equations often have two solutions due to the \(\pm\) when taking the square root.
Verifying Solutions
After solving an equation, it's crucial to verify the solutions by substituting them back into the original equation. This ensures that the solutions are correct and satisfy the initial condition.

Let's substitute \(x = 10\) and \(x = -10\) back into the original equation \(\frac{x}{5} = \frac{20}{x}\).
  • For \(x = 10\):
    \(\frac{10}{5} = 2\) and \(\frac{20}{10} = 2\). Both sides are equal, so \(x = 10\) is a valid solution.
  • For \(x = -10\):
    \(\frac{-10}{5} = -2\) and \(\frac{20}{-10} = -2\). Both sides are equal, so \(x = -10\) is also a valid solution.
Verification helps ensure that there are no extraneous solutions, providing confidence in the correctness of the answers.

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Most popular questions from this chapter

Simplify. $$ \frac{a^{3}-2 a^{2}+2 a-4}{a^{3}-2 a^{2}-3 a+6} $$

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Find an equation of variation in which: \(y\) varies directly as \(x\) and inversely as \(w\) and the square of \(z,\) and \(y=4.5\) when \(x=15, w=5,\) and \(z=2\).
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