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Chapter 6: Problem 30

Solve each equation, and check your 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

Eliminate the Fraction

To eliminate the fraction, multiply both sides of the equation by 15 (the common denominator). This gives: \[15 \times \frac{2z + 1}{5} = 15 \times \frac{7z + 5}{15}\]
02

Simplify the Equation

Simplify both sides to remove the fractions. We get: \[3(2z + 1) = 7z + 5\]
03

Distribute and Combine Like Terms

Distribute the 3 on the left side: \[6z + 3 = 7z + 5\]Next, move all terms involving z to one side of the equation and constant terms to the other side. Subtract 6z from both sides: \[3 = z + 5\]Then subtract 5 from both sides: \[-2 = z\]
04

Check the Solution

Substitute z = -2 back into the original equation to verify the solution. Original equation: \[\frac{2(-2) + 1}{5} = \frac{7(-2) + 5}{15}\]This simplifies to: \[\frac{-4 + 1}{5} = \frac{-14 + 5}{15}\]\[\frac{-3}{5} = \frac{-9}{15}\]\[-0.6 = -0.6\]The left-hand side equals the right-hand side, confirming that z = -2 is the correct solution.

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Key Concepts

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

Eliminating Fractions in Linear Equations
When solving linear equations, dealing with fractions can be tricky. The first step is to eliminate the fractions. This simplifies the equation and makes it easier to solve. To eliminate fractions:
  • Find a common denominator for all fractions involved
  • Multiply every term in the equation by that common denominator
For example, in the equation \(\frac{2 z+1}{5} = \frac{7 z+5}{15}\), the common denominator is 15. Multiplying both sides by 15 removes the fractions: \[15 \times \frac{2z+1}{5} = 15 \times \frac{7z+5}{15} \] Simplify if necessary, and then you have an equation without fractions!
Distributing in Equations
Once fractions are eliminated in an equation, the next step often involves distribution. Distribution helps in spreading out terms inside parentheses to make the equation simpler.
Consider the equation: \(\frac{2z+1}{5} = \frac{7z+5}{15}\)
After eliminating fractions, we have \[3(2z+1) = 7z + 5 \] Distribute the 3 on the left-hand side: \[3 \times 2z + 3 \times 1 = 6z + 3 \] This results in: \[6z + 3 = 7z + 5 \] Now the equation is without fractions or parentheses, making it much simpler to solve.
Checking Solutions in Algebra
After solving an equation, it's crucial to verify that your solution is correct. This step ensures that no mistakes were made. Here鈥檚 how to check your solution:
  • Substitute the solution back into the original equation
  • Simplify both sides and ensure they are equal
For example, we found the solution \z = -2\. Substituting this back into the original equation: \(\frac{2(-2) + 1}{5} = \frac{7(-2) + 5}{15}\) simplifies to \(\frac{-3}{5} = \frac{-9}{15}\) which checks out because \(-0.6 = -0.6\). If both sides of the equation are equal, the solution is correct.

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Most popular questions from this chapter

Divide. Write each answer in lowest terms. $$ \frac{35 x^{8}}{7 x^{9}} \div \frac{5 x^{5}}{9 x^{6}} $$

Divide. Write each answer in lowest terms. $$ \frac{(x-3)^{2}}{6 x} \div \frac{x-3}{x^{2}} $$

Multiply or divide as indicated. Write each answer in lowest terms. $$ \frac{7}{12} \div \frac{15}{4} $$

Multiply. Write each answer in lowest terms. $$ \frac{(x-y)^{2}}{2} \cdot \frac{24}{3(x-y)} $$

Write each rational expression in lowest terms. $$ \frac{5-x}{5+x} $$
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