Problem 71 - Elementary Algebra Within Reach

Question

Solve the equation and check your solution. (Some of the equations have no solution.) $$5 z-2=2(3 z-4)$$

Step-by-Step Solution

Answer

The solution to the given equation is \(z = 6\).

Step 1: Expand the right-hand side

Expand the right-hand side of the equation to eliminate the parenthesis. Using the distributive property of multiplication over addition gives \(2 * 3z - 2 * 4\), which simplifies to \(6z - 8\). So the equation becomes: \(5z - 2 = 6z - 8\).

Step 2: Gather like terms together

Next, we need to get all terms containing z on one side of the equation and numerical terms on the other side. By subtracting \(6z\) from both sides, we get \(-z - 2 = -8\). Then by adding 2 to both sides, we get \(-z = -6\).

Step 3: Solve for z

Solving for z implies getting z alone on one side of the equation. This can be done by multiplying both sides of the equation by -1. Thus, we get \(z = 6\).

Step 4: Verify the solution

To verify the solution, substitute z = 6 back into the original equation to check if the left-hand side equals the right-hand side. The original equation becomes \(5*6 - 2 = 2*(3*6 - 4)\), which simplifies to \(28 = 28\). Since both sides are equal, our solution z = 6 is the correct solution.

Key Concepts

Distributive Property

The distributive property is a key concept in algebra that helps simplify equations by removing parentheses. In the equation given, we have an expression on the right side: \(2(3z - 4)\). The distributive property tells us that we can multiply each term inside the parentheses by the number outside. This means we compute \(2 \times 3z\) and \(2 \times -4\). After simplifying, we get \(6z - 8\). Now our equation looks simpler: \(5z - 2 = 6z - 8\).

This property streamlines the process of solving linear equations and ensures each term is accounted for correctly. Remember, it works as a pattern:

• Multiply each term inside the bracket by the outside term.
• Sum or subtract the resulting products to simplify your expression.

Variable Isolation

Variable isolation involves getting the variable (in this case, \(z\)) all by itself on one side of the equation. The goal here is to solve for \(z\), which means determining its value. First, we need to move all terms involving \(z\) to one side and constant terms to the other. We began with the equation \(5z - 2 = 6z - 8\).

Start by subtracting \(6z\) from both sides, which gives \(-z - 2 = -8\). This process of balancing ensures we maintain equality. Then, you add 2 to both sides to collect like terms, resulting in \(-z = -6\).

To isolate \(z\), multiply each side by \(-1\) to change \(-z\) to \(z\). Thus, \(z = 6\). Variable isolation is essentially a step-by-step extraction of the variable, cleaning up the equation as you go.

Checking Solutions

Once you've found a potential solution, it’s important to verify that it works. This step reassures you that your solution is correct. To check our solution of \(z = 6\), substitute it back into the original equation: \(5z - 2 = 2(3z - 4)\).

Replace \(z\) with 6: The equation becomes \(5 \times 6 - 2 = 2 \times (3 \times 6 - 4)\). Simplify both sides, which calculates to \(30 - 2 = 2(18 - 4)\), simplifying further to \(28 = 28\).

Both sides equal, confirming the solution is accurate. This step ensures that no errors were made in calculation and that the found value of \(z\) satisfies the original condition of the equation. Always check your solutions to confirm your answers.

No Solution Cases

Sometimes, you might encounter equations that don't have solutions at all. These cases occur when you simplify an equation and end up with a statement that is not true, like \(0 = 7\). This signals that no single value of \(z\) will satisfy the equation.

In such cases,

• You might notice terms cancel each other out completely.
• End up with a contradiction during simplification.

These types of equations are known as inconsistent equations. They typically arise in scenarios where logical constraints of the problem definition are violated. It's important to recognize when equations have no solutions.

While the original problem was not a "no solution" case, being able to identify them is a useful skill when you start solving different types of equations.