Question

Verify that the given value is a solution of the given equation. $$ 0.01 x-1=0, x=100 $$

Short answer

Question: Verify that x = 100 is a solution to the given equation, 0.01x - 1 = 0. Answer: Yes, x = 100 is a solution to the equation, as substituting this value results in a true statement: 0 = 0.

Step-by-step solution

Substitute the given value of x into the equation

We are given the equation, 0.01x - 1 = 0, and the value of x = 100. We substitute x = 100 into the equation:\[0.01(100) - 1 = 0\]

Simplify the equation

Now that we have the value of x in the equation, we can simplify it:\[1 - 1 = 0\]

Check for equality

After simplification, we have the equation 0 = 0, which is a true statement. Since both sides of the equation are equal after substituting the value x = 100, we can conclude that x = 100 is indeed a solution to the given equation, 0.01x - 1 = 0.

Key concepts

Equation Solving

Solving an equation means finding the value of the variable that makes the equation true. An equation is like a balance scale. Both sides of the equation must have equal value. When solving, our goal is to determine the value(s) of the variable that accomplish this balance. In the example provided, the task is to check if a specific number, 100 in this instance, when used in place of the variable \(x\), satisfies the equation \(0.01x - 1 = 0\). By replacing \(x\) with 100 and verifying both sides of the equation are equal, we can effectively "solve" the equation. Whenever we solve equations, we often:

• Substitute possible values where necessary.
• Perform operations to isolate the variable.
• Check that the solution satisfies the original equation.

The main takeaway is ensuring both sides maintain equality, securing the balance of the equation. Finding solutions often requires methodically applying mathematical operations to simplify challenges.

Substitution Method

The substitution method is a powerful tool in algebra that involves replacing a variable with a known or given value. This method is especially useful when you have equations in which you are asked to verify specific values as potential solutions. For example, if the equation we are given is \(0.01x - 1 = 0\) and the value of \(x\) provided is 100, we solve by substituting 100 for \(x\):1. Begin with the original equation: \(0.01x - 1 = 0\).2. Substitute \(x = 100\) into the equation to get \(0.01(100) - 1 = 0\).By choosing and inserting a defined value of \(x\), you directly test whether this value solves the equation. The ease of the substitution method lies in its straightforwardness, transforming the variable expression into simple arithmetic. Whenever a specific number makes the equation true, it confirms we've found a valid solution.

Simplification

Simplification in algebra involves reducing an equation to its simplest form. This often means making the equation easier to interpret, solve, or verify. The simplification process involves combining like terms, performing arithmetic operations, or expressing a formula in a more basic way. Let's say we've substituted \(x = 100\) into the given equation, \(0.01(100) - 1 = 0\). Simplification is the next step:1. Calculate the multiplication: \(0.01 \times 100 = 1\).2. Apply arithmetic operations: \(1 - 1 = 0\).The result, \(0 = 0\), indicates both sides of the equation align perfectly, affirming the calculated solution.Simplification not only helps verify solutions, but also helps in understanding the relationship between components in the equation. When all is reduced to its simplest form, learners can focus on core relationships and ensure clarity in problem-solving. The appeal of simplification lies in transforming a complex task into more manageable pieces.