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Chapter 6: Problem 20

Solve and check each equation. \(25-6 x=-83\)

Short Answer

Expert verified
The solution for the equation \(25-6x = -83\) is \(x = 18\)

Step by step solution

01

Isolate the Term Containing the Variable

In this exercise, subtract \(25\) from both sides to isolate the term \(-6x\). This results in \(-6x = -83 - 25\). Simplify the equation to find \(-6x = -108\).
02

Solve for the variable

To obtain \(x\), we have to remove \(-6\). We do this by dividing both sides of the equation by \(-6\). This leads to \(x = -108 / -6\). After performing the division, we find \(x = 18\).
03

Check the Solution

Substitute \(x\) with \(18\) in the original equation \(25-6x = -83\). This becomes \(25 - 6 * 18 = -83\). After simplification, we get \(-83 = -83\). Since both sides of the equation are equal, our answer is correct.

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Key Concepts

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

Equation Solving Steps
When you're tackling algebraic equations, following a structured set of steps can make the process straightforward and manageable. In our example, we started with the equation:
  • \(25 - 6x = -83\)
The first step is to isolate the term with the variable. We subtract \(25\) from both sides to move closer towards solving for \(x\). This leads us to:
  • \(-6x = -83 - 25\)
Simplifying the equation, we find:
  • \(-6x = -108\)
The final step is to solve for \(x\), which we achieve by dividing both sides by \(-6\). Following these systematic steps helps eliminate errors and simplifies the problem-solving process.
Variable Isolation
Variable isolation means getting the variable by itself on one side of the equation. Here, our variable is \(x\). Initially, the variable is embedded within the term \(-6x\). To untangle \(x\), we focus on eliminating any additions or subtractions on its side of the equation. We removed \(25\) from the equation by subtracting it from both sides:
  • \(-6x = -108\)
Now \(x\) is part of a simple equation. The next move is to divide by \(-6\), thereby reversing the multiplication. This operation leaves \(x\) standing alone:
  • \(x = 18\)
By isolating the variable, we streamline our efforts to solve for it directly.
Checking Solutions
After calculating the potential solution to an equation, it's crucial to check if it's correct. We do this by substituting the found value back into the original equation. In our case, we discovered that \(x = 18\). We substitute this value back into:
  • \(25 - 6x = -83\)
Replacing \(x\) yields:
  • \(25 - 6 \times 18 = -83\)
Simplifying the left side confirms that both sides balance:
  • \(-83 = -83\)
This verification method acts as a check against missteps, ensuring the reliability of our solution.

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Most popular questions from this chapter

Use FOIL to find the products in Exercises 1-8. \((x+3)(x+5)\)

Solve the equations in Exercises 53-72 using the quadratic formula. \(4 x^{2}=12 x-9\)

Factor the trinomials or state that the trinomial is prime. Check your factorization using FOIL multiplication. \(6 x^{2}-17 x+12\)

Solve the quadratic equations in Exercises 37-52 by factoring. \(2 x^{2}=7 x+4\)

Explain how to multiply two binomials using the FOIL method. Give an example with your explanation.
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