Problem 26 - Prealgebra

Question

Solve each equation using the methods shown in this section. $$10 a+3=4(a-1)+1$$

Step-by-Step Solution

Answer

The solution is \(a = -1\).

Step 1: Expand and Simplify

Start by expanding the right-hand side of the equation. This will involve distributing the term outside the parentheses across the terms inside the parentheses. The equation is \(4(a-1)+1\). Expanding gives us: \(4a - 4 + 1\). Simplify further: \(4a - 3\). The equation now reads: \(10a + 3 = 4a - 3\).

Step 2: Move Variables to One Side

Next, we need to get all terms with \(a\) on one side of the equation. To do this, subtract \(4a\) from both sides: \(10a + 3 - 4a = 4a - 3 - 4a\). This simplifies to \(6a + 3 = -3\).

Step 3: Isolate the Variable

Now, isolate \(a\) by getting rid of the constant term on the left. Subtract 3 from both sides: \(6a + 3 - 3 = -3 - 3\). This simplifies to \(6a = -6\).

Step 4: Solve for the Variable

Finally, solve for \(a\) by dividing both sides by 6: \(\frac{6a}{6} = \frac{-6}{6}\). This results in \(a = -1\).

Key Concepts

Equation Solving

When we talk about solving linear equations, we're discussing the process of finding the value of the variable that makes the equation true. In a linear equation, variables are to the first power and typically found on one or both sides of an equation.
To solve an equation, follow these principles:

• Keep the equation balanced by performing the same operations on both sides.
• Use operations that "undo" each other, such as addition and subtraction or multiplication and division.
• Simplify the equation as much as possible to make it easier to work with.

For instance, in the original exercise, the goal was to find the correct value of the variable, denoted as \(a\), that satisfies both sides of the equation. As we break down the process, remember that maintaining balance in the equation is essential to getting the right answer.

Variable Isolation

Isolating the variable is a critical step in solving equations. To isolate the variable, you need to get your variable of interest on one side of the equation and the constants on the other. This involves several key steps:

• Move all terms containing the variable to one side of the equation. This might mean adding or subtracting terms from both sides.
• Eliminate any constants on the side of the variable by using operations like addition, subtraction, multiplication, or division.

In our solution process, after simplifying the initial equation, it became necessary to shift terms around so that all occurrences of \(a\) were on the same side. Specifically, after distributing terms, we subtracted \(4a\) from both sides to get \(6a + 3 = -3\). Following this, we subtracted 3 from each side to further isolate \(6a\). Finally, the division of both sides by 6 gave us \(a = -1\).
Each of these steps helps narrow down toward isolating the variable, which is crucial in finding the solution for \(a\).

Distributive Property

The distributive property is a key concept often used when solving linear equations. It states that multiplying a number by a sum is the same as multiplying that number by each addend and then adding the products. This property can be expressed as: \(a(b + c) = ab + ac\).
In the given problem, the distributive property needed to be applied to the expression \(4(a-1)\) on the right side of the equation.

• The multiplier (4) was distributed to both terms inside the parentheses: \(a\) and \(-1\).
• This resulted in the expression being expanded to \(4a - 4\).

Applying the distributive property is fundamental because it transforms complex equations into simpler, more manageable forms, allowing us to proceed with solving the equation by isolating variables. Understanding how to distribute multiplication over addition or subtraction prepares us for handling more complicated equations in algebra.