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Chapter 7: Problem 48

For exercises 43-58, (a) solve. (b) check. $$ \frac{4}{a+6}=\frac{9}{a-4} $$

Short Answer

Expert verified
The solution is \( a = -14 \) and it satisfies the original equation.

Step by step solution

01

Cross-multiply

Cross-multiply to eliminate the fractions. This means multiplying the numerator of each fraction by the denominator of the other fraction: \[ 4(a-4) = 9(a+6) \]
02

Distribute

Distribute the numbers inside the parentheses: \[ 4a - 16 = 9a + 54 \]
03

Move all terms involving the variable to one side

Subtract 4a from both sides to combine like terms: \[ -16 = 5a + 54 \]
04

Isolate the variable

Subtract 54 from both sides: \[ -16 - 54 = 5a \] This simplifies to: \[ -70 = 5a \]
05

Solve for the variable

Divide both sides by 5: \[ a = -14 \]
06

Check

Substitute \( a = -14 \) back into the original equation to verify: \[ \frac{4}{-14 + 6} = \frac{9}{-14 - 4} \] This simplifies to: \[ \frac{4}{-8} = \frac{9}{-18} \] Both sides simplify to: \[ -\frac{1}{2} = -\frac{1}{2} \] The solution is verified.

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Key Concepts

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

Cross-Multiplication
When solving rational equations, a key technique used is cross-multiplication. This helps eliminate the fractions, making it easier to solve for the variable. Cross-multiplication involves multiplying the numerator of one fraction by the denominator of the other fraction. For example, in the equation \[ \frac{4}{a+6}=\frac{9}{a-4} \], you would multiply 4 by \(a-4\) and 9 by \(a+6\).
This results in the equation: \[ 4(a-4) = 9(a+6) \].
Now, we no longer have fractions, and we can proceed to solve the equation.
Cross-multiplication works because it produces an equivalent equation without changing the relationship between the sides. Just remember that this method only works when each side of the equation is a single fraction.
Distributive Property
The distributive property is essential when dealing with equations that have variables and constants inside parentheses.
It means you need to distribute (or multiply) the number outside the parentheses by each term inside the parentheses. For example, if you have \[ 4(a-4) \], distribute the 4:
\[ 4a - 16 \].
Similarly, for the other side of our example equation \[ 9(a+6) \], distribute the 9:
\[ 9a + 54 \].
Using the distributive property simplifies the equation and gets rid of parentheses, making it easier to combine like terms and solve for the variable.
Solving for Variables
Once you've used cross-multiplication and the distributive property, your goal is to isolate the variable. This means getting the variable on one side of the equation and constants on the other.
Here鈥檚 how it works:
  • First, combine like terms involving the variable by moving them to one side. For example, subtract \(4a\) from both sides of the equation \(4a - 16 = 9a + 54\), resulting in \[ -16 = 5a + 54 \].
  • Next, isolate the variable by performing inverse operations. Subtract 54 from both sides: \[ -16 - 54 = 5a \], which simplifies to \[ -70 = 5a \].
    Finally, divide both sides by 5 to solve for \(a\):
    \[ a = -14 \].
    Don鈥檛 forget to check your solution by substituting it back into the original equation to ensure both sides are equal.
    This final step verifies your solution and confirms its correctness.

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Most popular questions from this chapter

The relationship of the radius of a circle, \(x\), and the circumference of the circle, \(y\), is a direct variation. The radius of a circle is \(10 \mathrm{~cm}\), and the circumference is \(62.8 \mathrm{~cm}\). a. Find the constant of proportionality, \(k\). b. Write an equation that represents this relationship. c. Find the circumference of a circle with a radius of \(20 \mathrm{~cm}\).

The relationship of \(x\) and \(y\) is an inverse variation. When \(x=2, y=10\). a. Find the constant of proportionality, \(k\). b. Write an equation that represents this inverse variation. c. Find \(y\) when \(x=5\).

For exercises 49-52, the formula \(C=\frac{P_{m} P_{i}}{T F}\) describes the cost of insurance, \(C\). Is the relationship of the given variables a direct variation or an inverse variation? $$ C, T, \text { and } F \text { are constant; the relationship of } P_{i} \text { and } P_{m} $$

For a fixed length of household copper wire, the relationship of the cross- sectional area, \(x\), and the resistance, \(y\), is an inverse variation. When the cross-sectional area is \(3.14 \times 10^{-6} \mathrm{~m}^{2}\), the resistance is \(5.4 \times 10^{-3} \mathrm{ohm}\). a. Find the constant of proportionality, \(k\). Use scientific notation. Include the units of measurement. b. Write an equation that represents this relationship. c. Find the resistance when the cross-sectional area is \(2.05 \times 10^{-6} \mathrm{~m}^{2}\). Round the mantissa to the nearest tenth.

For exercises 43-58, (a) solve. (b) check. $$ \frac{z+2}{4}=\frac{z-8}{12} $$
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