Problem 29 - Algebra 1: Concepts and Skills

Question

CHECKING SOLUTIONS OF EQUATIONS. Check to see if the given value of the variable is or is not a solution of the equation. $$ 6 d-5=31 ; d=6 $$

Step-by-Step Solution

Answer

Yes, the given value of d (which is 6) is a solution to the equation.

Step 1: Understand the equation

The given equation is \( 6d - 5 = 31 \). In this equation, d is the variable, 6 and -5 are the coefficients of d, and 31 is the constant.

Step 2: Substitute d with 6

Substitute d in the equation with the given value which is 6. The equation then becomes \(6*6 - 5 = 31\).

Step 3: Solve the equation

After substituting 6 for d, the left side of the equation then becomes \( 6*6-5 = 36-5 \) which simplifies to 31.

Step 4: Compare both sides of the equation

Now, both sides of the equation are compared. If they are equal, then the given value of d (which is 6) is a solution to the equation. Here, both sides of the equation are equal (31 = 31), so d = 6 is indeed a solution to the equation.

Key Concepts

Equations

An equation is a mathematical statement that asserts the equality of two expressions. In simple terms, it's like a statement saying that two different expressions yield the same result when calculated. Equations are usually made up of variables, coefficients, and constants.
For instance, take the equation from our exercise:

• The variable is \( d \).
• The coefficients are 6 and -5, which interact with the variable.
• The constant on the right side is 31.

Understanding how to read and interpret each part is crucial. Equations are at the heart of algebra, and learning to work with them is like learning a new language. Once you become familiar, solving them becomes a straightforward process. Equations can demonstrate relationships between quantities, allowing us to solve for unknown variables.

Substitution Method

The substitution method is a technique used to solve equations. You replace a variable with a given value to check if it satisfies the equation. It's a handy tool when working with equations, and it's straightforward – just plug in the value and follow through with the arithmetic.
Applied to our exercise, we have the equation \(6d - 5 = 31\). We were given that \(d = 6\). By substituting 6 for \(d\), we transform the equation into \(6 \times 6 - 5 = 31\).

• First, calculate \(6 \times 6\) which equals 36.
• Then, subtract 5 from 36, leaving 31.

This method simplifies solving equations under specific conditions, making it less intimidating and more approachable for students.

Variable Value Assessment

Variable value assessment involves determining if a chosen value truly solves an equation. It is essential because it confirms whether a supposed answer makes an equation true.
After substituting \(d = 6\) into \(6d - 5 = 31\), we found both sides of the equation was equal, resulting in 31 = 31.

• This demonstrates that \(d = 6\) is a valid solution.
• If the two sides did not match, then \(d = 6\) would not be a solution, prompting a reevaluation.

Always check after substituting back the solution into the original equation. Ensuring you confirm the validity ensures accuracy in your math work, and it's a crucial part of problem-solving.