Problem 12 - Algebra 1: Concepts and Skills

Question

Tell which equation you would use to isolate a variable. Explain. $$ \begin{array}{c} {5 c+3 d=11} \\ {5 c-d=5} \end{array} $$

Step-by-Step Solution

Answer

To isolate a variable in the given system of equations, subtract the second equation from the first. The resulting equation, 4d = 6, can be solved for 'd', which equals 1.5.

Step 1: Rearrange the equations

Set the equations so that the like terms will line up vertically. From the original problem, we already have two equations in the correct form: \[\begin{array}{c}{5c + 3d = 11} \{5c - d = 5}\end{array}\]

Step 2: Subtract the equations

Subtract the second equation from the first one. This means subtracting each part of the second equation from the corresponding part of the first:\( (5c + 3d) - (5c - d) = 11 - 5\)This simplifies to:\(5c + 3d - 5c + d = 11 - 5\)which further simplifies to:\(4d = 6\)

Step 3: Solve for the isolated variable

Now that the equation is simplified, we can solve for 'd' by dividing each side of the equation by 4 \(4d /4 = 6/4\)which simplifies to:\(d = 1.5\)

Key Concepts

Isolation of Variables

When solving systems of equations, one of the foundational techniques is the isolation of variables. This technique involves rearranging an equation to focus on one particular variable. By doing this, you create a simple pathway to find the value of that variable.

Here's why you might isolate a variable:

• To simplify a system of equations by reducing it to just one equation with one unknown.
• To make it easier to substitute one variable's value into another equation, simplifying the computational process.
• To gain better insight into the relationship between different variables.

In our example, by subtracting the second equation from the first, the terms involving 'c' cancel out. This leaves us with a straightforward equation in 'd', allowing us to isolate 'd' to find its value. Once isolated, the calculation becomes straightforward – you perform simple arithmetic operations like division to solve for the isolated variable.

Linear Equations

Linear equations are fundamental in algebra and represent equations of the first degree. This means the highest power of the variable in the equation is one. In a simple form, they can often be expressed as: \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants, and \(x\) and \(y\) are variables.

The system of equations we've worked with is a classic example. We have two linear equations:

• \(5c + 3d = 11\)
• \(5c - d = 5\)

Both equations have variables raised only to the power of one, ensuring they are linear. Each equation forms a straight line when graphed on a coordinate plane. The solution to the system, like any linear system, is the point where these lines intersect. In practical problems, solving linear equations helps in understanding relationships and interdependencies between different variables.

Solving Equations

Solving equations involves finding the value of unknown variables that satisfy the given relations or conditions. In systems of equations, this process can require various methods, such as substitution or elimination. In our specific exercise, we use the elimination method: - We subtract one equation from the other, eliminating 'c'. - We are left with an equation with only 'd', which makes it straightforward to solve.

The process:

• Subtract the equations: \((5c + 3d) - (5c - d) = 11 - 5\)
• Simplify to get \(4d = 6\).
• Divide both sides by 4 to solve for \(d\), resulting in \(d = 1.5\).

Each operation simplifies the problem further. The goal is to progress step by step until you isolate and find the value of each variable involved. Understanding solving techniques enhances problem-solving skills and offers a structured way to approach various mathematical challenges.