Problem 54 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve equation and check your proposed solution in. \(1.4(z-5)-0.2=0.5(6 z-8)\)

Step-by-Step Solution

Answer

The solution to the equation is z = -7.

Step 1: Expand Both Sides

Expand both sides of the equation, which gives us \(1.4z - 7 - 0.2 = 3z - 4\). Simplifying further we get \(1.4z - 7.2 = 3z - 4\).

Step 2: Transfer like terms

Rearrange the equation by transferring the z terms to one side and the constant terms to the other side which gives \(1.4z - 3z = 4 + 7.2\). Simplifying further we get \(-1.6z = 11.2\).

Step 3: Solve for z

To find z, divide both sides of the equation by -1.6, which gives \(z = 11.2 ÷ -1.6\). This gives us \(z= -7\).

Step 4: Verify the solution

Substitute the value of z into the original equation as follows: 1.4(-7-5)-0.2=0.5(6*-7-8) which simplifies to -16.8 -0.2 = -24. Solving for the left hand side gives us -17 which equals to -17 on the right hand side. Since both sides are equal, we can confirm that z=-7 is the correct solution.

Key Concepts

Equation Solving

To solve equations successfully, it's essential to follow a series of steps to simplify and unravel the unknowns. In every algebraic equation, there are two sides that represent equal values. The goal is to manipulate these sides in such a way that you can solve for the unknown variable. This often involves operations such as addition, subtraction, multiplication, and division to make the equation simpler and isolate the variable.

• Start by expanding both sides, as seen here: from \(1.4(z-5)-0.2\) to \(1.4z - 7.2\). This step simplifies the equation and gives you a clearer path to follow.
• Continue by combining like terms, ensuring each side has terms that can be worked with easily. This involves recognizing similar terms, such as constants and variable terms, to combine them appropriately.

Remember, patience and precision in these steps ensure no mistakes are made, leading to an accurate solution.

Variable Isolation

Variable isolation is a central theme in algebra that refers to the process of rearranging an equation so that the unknown variable appears by itself on one side of the equation. This is crucial for finding the solution to the equation. In our exercise, this meant focusing on getting \(z\) by itself.

• First, isolate the variable on one side of the equation. In this example, terms with \(z\) need to be on one side, which was re-arranged to \(1.4z - 3z\).
• Next, ensure that all other constants are moved to the opposite side, resulting in an equation like \(-1.6z = 11.2\).

Finally, division comes into play to solve for \(z\), ensuring that \(z\) is by itself. Careful calculations help maintain accuracy and lead you to the solution of the variable.

Verification of Solutions

After solving for the variable, it's vital to verify the solution to ensure its correctness. Verification involves plugging the found value back into the original equation to check consistency.

• Take the solution \(z = -7\) and substitute it back into the original equation \(1.4(z-5)-0.2=0.5(6z-8)\).
• Simplify both sides independently to establish whether the equation holds true. For example, both sides should equal \(-17\) once simplified with \(z = -7\).

By making both sides equal, we confirm that our solution is correct. This step is crucial as it also teaches checking work, reducing the chances of errors going unnoticed.