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Chapter 6: Problem 24

Solve each equation, and check the solutions. $$ \frac{2 z+1}{5}=\frac{7 z+5}{15} $$

Short Answer

Expert verified
The solution is \( z = -2 \).

Step by step solution

01

Cross-Multiply to Eliminate the Fractions

To eliminate the fractions, cross-multiply the terms on both sides of the equation: \( \frac{2z + 1}{5} = \frac{7z + 5}{15} \) becomes \( (2z + 1) * 15 = (7z + 5) * 5 \).
02

Simplify the Equation

Distribute the numbers outside the parentheses: \( 15(2z + 1) = 5(7z + 5) \) becomes \( 30z + 15 = 35z + 25 \).
03

Move All Terms with z to One Side

Subtract 30z from both sides to get the z terms on one side: \( 30z + 15 - 30z = 35z + 25 - 30z \) simplifies to \( 15 = 5z + 25 \).
04

Solve for z

Subtract 25 from both sides to isolate the term with z: \( 15 - 25 = 5z \) simplifies to \( -10 = 5z \). Finally, divide both sides by 5: \( \frac{-10}{5} = \frac{5z}{5} \), resulting in \( z = -2 \).
05

Check the Solution

Substitute \( z = -2 \) back into the original equation to verify: \( \frac{2(-2) + 1}{5} = \frac{7(-2) + 5}{15} \). Simplify both sides: \( \frac{-3}{5} = \frac{-9}{15} \), and since \( \frac{-9}{15} = \frac{-3}{5} \), the solution satisfies the original equation.

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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 handy method used to eliminate the fractions in a rational equation. It's like creating a bridge to simpler arithmetic. When you have an equation like \( \frac{a}{b} = \frac{c}{d} \), cross-multiplication involves multiplying the numerator of one fraction by the denominator of the other. So, this becomes \( a * d = b * c \).

Let鈥檚 look at our example: \( \frac{2z + 1}{5} = \frac{7z + 5}{15} \). Cross-multiplying gets rid of the fractions:
  • Multiply the numerator of the first fraction by the denominator of the second fraction: \( (2z + 1) * 15 \)
  • Multiply the numerator of the second fraction by the denominator of the first fraction: \( (7z + 5) * 5 \)
This simplifies the equation to \( (2z + 1) * 15 = (7z + 5) * 5 \). Now, we can work without fractions and solve for the variable!
Simplifying Equations
Simplifying equations helps make complex problems more manageable. Once you鈥檝e cross-multiplied in our problem, you鈥檒l distribute the numbers outside the parentheses. Start with:
\( 15(2z + 1) = 5(7z + 5) \) which becomes:
  • Distribute 15 inside: \( 15 * 2z + 15 * 1 = 30z + 15 \)
  • Distribute 5 inside: \( 5 * 7z + 5 * 5 = 35z + 25 \)
Now, we have a more straightforward equation: \( 30z + 15 = 35z + 25 \).

Next, we move all terms containing \( z \) to one side and constants to the other. Subtract 30z from both sides:
  • \( 30z + 15 - 30z = 35z + 25 - 30z \)
Which simplifies to:
\( 15 = 5z + 25 \)
Then, isolate \( z \) by subtracting 25:
  • \( 15 - 25 = 5z \)
This simplifies further to:
\( -10 = 5z \). Finally, divide both sides by 5 to find \( z \):
  • \( \frac{-10}{5} = \frac{5z}{5} \)
Resulting in: \( z = -2 \).
Checking Solutions
Checking your solutions is a crucial step in solving equations. It ensures you haven鈥檛 made a mistake. Once we find \( z = -2 \), we substitute it back into the original equation to verify:
\( \frac{2(-2) + 1}{5} = \frac{7(-2) + 5}{15} \). Simplify both sides:
  • Left side: \( \frac{-4 + 1}{5} = \frac{-3}{5} \)
  • Right side: \( \frac{-14 + 5}{15} = \frac{-9}{15} \)
We can see \( \frac{-9}{15} = \frac{-3}{5} \), confirming both sides are equal, so the solution \( z = -2 \) is correct.


Recap: Always plug your solution back into the original equation to check its validity. It鈥檚 the best way to catch any errors and ensure your solution is correct.

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