Problem 65 - Intermediate Algebra

Question

Solve each equation. Check each result. See Example 8. $$ 0.04(12)+0.01 t-0.02(12+t)=0 $$

Step-by-Step Solution

Answer

The solution to the equation is \(t = 24\).

Step 1: Distribute and Simplify

First, distribute the terms across the parentheses in the equation. This means multiplying 0.04 by 12 and 0.02 by both 12 and \(t\). This gives us:\[0.04 imes 12 + 0.01 t - 0.02 imes 12 - 0.02t = 0\]Which simplifies to:\[0.48 + 0.01t - 0.24 - 0.02t = 0\]

Step 2: Combine Like Terms

Next, combine the like terms. This means adding or subtracting the coefficients of \(t\) and the constants:\[0.48 - 0.24 + 0.01t - 0.02t = 0\]This simplifies to:\[0.24 - 0.01t = 0\]

Step 3: Solve for t

Isolate \(t\) by moving the constant to the other side of the equation:\[0.01t = 0.24\]Then, divide both sides by 0.01:\[t = \frac{0.24}{0.01}\]Which gives:\[t = 24\]

Step 4: Check the Solution

Substitute \(t = 24\) back into the original equation to ensure it satisfies the equation:\[0.04(12) + 0.01(24) - 0.02(12 + 24) = 0\]Calculate each term:\[0.48 + 0.24 - (0.02 imes 36) = 0\]\[0.48 + 0.24 - 0.72 = 0\]Which simplifies to:\[0 = 0\]The solution \(t = 24\) is confirmed to be correct.

Key Concepts

Combining Like Terms

When working with linear equations, combining like terms is an essential process. It involves simplifying expressions by grouping similar terms together. Like terms are terms that contain the same variable raised to the same power. For instance, in an equation like \(0.01t - 0.02t\), both terms involve the variable \(t\) raised to the first power. To combine these, you simply add or subtract their coefficients (the numbers in front of the variables). In this example, you subtract \(0.02\) from \(0.01\):

• \(0.01t - 0.02t = -0.01t\)

On the other hand, constant terms like \(0.48\) and \(0.24\), which do not involve variables, are also combined separately:

• \(0.48 - 0.24 = 0.24\)

By systematically combining like terms, the equation gets simpler and easier to solve. This combined term approach paves the way to isolating the variable later in the process.

Distribution in Algebra

Distribution is a very useful tool in algebra, particularly when handling equations with parentheses. This technique involves distributing, or multiplying, a term outside the parentheses by each term inside. In the problem at hand, we distribute in these steps:- Distribute \(0.04\) across \(12\) to get \(0.48\).- Distribute \(-0.02\) over each term inside the parentheses \((12 + t)\), resulting in the two terms \(-0.24 - 0.02t\).Here’s how the distribution looks step-by-step:

• \(0.04 \times 12 = 0.48\)
• \(-0.02 \times 12 = -0.24\)
• \(-0.02 \times t = -0.02t\)

Distribution simplifies expressions, making it feasible to combine like terms and eventually solve the equation for the variable.

Checking Solutions

After finding a solution to a linear equation, it’s crucial to verify that the solution is correct. This step is called "checking the solution" and ensures that there are no errors.To check the solution, substitute the found value back into the original equation and simplify:Given that we calculated \(t = 24\), let's substitute this back into the initial equation:

• \(0.04(12) + 0.01(24) - 0.02(12 + 24)\)
• Simplifies to: \(0.48 + 0.24 - 0.72\)
• Which results in: \(0\)

If the left-hand side equals the right-hand side of the original equation (which is 0 in this case), the solution is verified as correct. Thus, substituting back and performing the arithmetic confirms that \(t = 24\) satisfies the equation, providing confidence in the solution's accuracy.