Problem 38 - Elementary Algebra

Question

For the following problems, solve each conditional equation. If the equation is not conditional, identify it as an identity or a contradiction. $$ 16(y-1)+11=-85 $$

Step-by-Step Solution

Answer

Answer: The value of y in the given equation is -5.

Step 1: Distribute the term outside the parenthesis

Multiply 16 with each term inside the parenthesis: \((y-1)\). $$ 16(y-1)+11=-85 \Rightarrow 16y - 16 + 11 = -85 $$

Step 2: Combine like terms

Combine the constants on the left side of the equation. $$ 16y - 16 + 11 = -85 \Rightarrow 16y - 5 = -85 $$

Step 3: Add 5 to both sides

In order to isolate the term with the variable (16y) on the left side, we can add 5 to both sides of the equation. $$ 16y - 5 + 5 = -85 + 5 \Rightarrow 16y = -80 $$

Step 4: Divide both sides by 16

To solve for the variable (y), we need to get rid of the coefficient (16). To do this, we will divide both sides of the equation by 16. $$ \frac{16y}{16} = \frac{-80}{16} \Rightarrow y = -5 $$ Since we have a unique solution for y, the equation is conditional.

Key Concepts

Conditional Equations

Conditional equations are equations that hold true only under certain conditions or for specific values. When we solve such equations, we aim to find the values of the variables that satisfy the equation. In our example, the equation is:

• \(16(y-1) + 11 = -85\)

If substituting any value for \(y\) satisfies this equation, then it is a conditional equation. In this case, we found that \(y = -5\) is the only solution that makes the equation valid. If there were no solutions, it would be called a contradiction. If all values of \(y\) solved the equation, we'd call it an identity. But here, we notice \(y = -5\) is our unique solution, marking this equation as conditional.

Distribution in Algebra

Distribution is a crucial algebraic tool used to simplify and solve equations by eliminating parentheses. The distributive property states that for any numbers \(a\), \(b\), and \(c\), we have:

• \(a(b + c) = ab + ac\)

This rule helps us "distribute" a single term across terms inside a parenthesis. In the equation \(16(y-1) + 11 = -85\), we distribute \(16\) to each term inside the parentheses:

• \(16 imes y - 16 imes 1 = 16y - 16\)

By expanding the parentheses, it becomes easier to manipulate and solve the equation. This step transforms the left side to \(16y - 16 + 11\), setting up subsequent steps for simplification.

Combining Like Terms

Combining like terms is another essential skill in solving algebraic equations, allowing us to simplify expressions. Like terms have the same variables and exponents. In our equation \(16y - 16 + 11 = -85\), the like terms are the constant numbers on the left side:

• \(-16\) and \(+11\)

We add these constants together to simplify the expression:

• \(-16 + 11 = -5\)

Now, the equation stands at \(16y - 5 = -85\). By reducing terms whenever possible, equations become much simpler and shorter, leading us closer to a solution.

Isolating Variables

Isolating variables is the ultimate goal when solving equations because it allows us to find the exact value of the unknown. Our steps to isolate \(y\) began from the equation \(16y - 5 = -85\). Here's how:

• Add \(5\) to both sides to get rid of \(-5\): \(16y - 5 + 5 = -85 + 5\) which simplifies to \(16y = -80\).
• Divide both sides by \(16\) to isolate \(y\): \( \frac{16y}{16} = \frac{-80}{16} \).
• This results in \(y = -5\).

Through these actions, first by adding or subtracting constants and then adjusting by division or multiplication, we enhance clarity in revealing the variable's value hidden within the equation.