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Chapter 2: Problem 25

For exercises 11-46, (a) solve. (b) check. $$ 0=6 z $$

Short Answer

Expert verified
The solution is \( z = 0 \).

Step by step solution

01

Title - Isolate the variable

To solve the equation, isolate the variable by dividing both sides of the equation by 6. This will give: \[ z = \frac{0}{6} \]
02

Title - Simplify the fraction

Simplify the fraction by performing the division: \[ z = 0 \]
03

Title - Check the solution

Substitute the found value back into the original equation to verify: \[ 0 = 6 \times 0 \] Simplify the right side: \[ 0 = 0 \] Since both sides of the equation are equal, the solution is correct.

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Key Concepts

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

Isolating Variables
In the given problem, our goal was to isolate the variable, which in this case is 'z'. To isolate a variable means to get the variable by itself on one side of the equation. This is a crucial step in solving linear equations.
To isolate 'z', we divide both sides of the equation by 6:
  • We start with the equation: \[0 = 6z\]
  • Divide both sides by 6: \[z = \frac{0}{6}\]
This gives us 'z' on one side of the equation. We now have a simpler expression that only includes the variable 'z' and a number.
Simplifying Fractions
Once we have the variable isolated, the next step is to simplify any fractions that appear. Simplifying fractions means performing the division to reduce the fraction to its simplest form.
In our example, we have the fraction \[\frac{0}{6}\]. When we perform the division:
  • We calculate \t\(\frac{0}{6} = 0\)
Hence, we find that 'z' equals to 0. Simplifying fractions is an important step because it gives us the cleanest, simplest form of our solution, making it easier to understand and verify.
Checking Solutions
Finally, it is very important to check our solution to ensure its correctness. To do this, we substitute the solved value of the variable back into the original equation and verify both sides are equal.
In this case, our solution for 'z' was 0. To check, we plug 0 back into the original equation \[0 = 6z\].
  • Substituting 'z' with 0, we get: \[0 = 6 \times 0\]
  • Simplifying, we see that \[0 = 0\]
Both sides of the equation are equal, confirming our solution is correct. This is a vital last step as it ensures there are no errors in our calculations and that the solution accurately satisfies the equation.

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Most popular questions from this chapter

a. Solve \(D=\frac{M}{V}\) for \(V\). b. The density \(D\) of gasoline is \(\frac{0.737 \mathrm{~g}}{1 \mathrm{~mL}}\), and the mass \(M\) is \(50 \mathrm{~g}\). Find the volume, \(V\). Round to the nearest tenth.

For exercises 53-62, (a) clear the fractions or decimals and solve. (b) check the direction of the inequality sign. $$ \frac{1}{4} h+3 \leq \frac{2}{3} h+2 $$

Each of the three sides of an equilateral triangle is the same length. If \(P\) is the perimeter of the triangle and \(L\) is the length of a side, write a formula in \(P\) and \(L\) for the perimeter of an equilateral triangle and solve this formula for \(L\).

For exercises 37-52, (a) solve. (b) use a number line graph to represent the solution. (c) check the direction of the inequality sign. $$ 7 x+3<52 $$

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