Problem 2 - Algebra 2

Question

Solve each system of equations by using substitution. \(4 c+2 d=10\) \(c+3 d=10\)

Step-by-Step Solution

Answer

The solution is \(c = 1\) and \(d = 3\).

Step 1: Solve for One Variable

Let's solve the second equation for the variable \(c\). The equation is \(c + 3d = 10\). To isolate \(c\), subtract \(3d\) from both sides:\[c = 10 - 3d\].

Step 2: Substitute into the First Equation

Now, substitute \(c = 10 - 3d\) into the first equation \(4c + 2d = 10\):Replace \(c\) with \(10 - 3d\):\[4(10 - 3d) + 2d = 10\].

Step 3: Simplify the Equation

Expand the equation from the previous step:\[40 - 12d + 2d = 10\]Combine like terms: \[40 - 10d = 10\].

Step 4: Solve for \(d\)

Let's isolate \(d\) by moving \(40\) to the right side. Subtract \(40\) from both sides:\[-10d = 10 - 40\]This simplifies to:\[-10d = -30\].Divide both sides by \(-10\) to solve for \(d\):\[d = 3\].

Step 5: Substitute \(d\) Back to Find \(c\)

Use \(d = 3\) to find \(c\) using the equation from Step 1. Substitute \(d = 3\) into \(c = 10 - 3d\):\[c = 10 - 3(3)\]Simplify:\[c = 10 - 9 = 1\].

Step 6: Verify the Solution

Check the values \(c = 1\) and \(d = 3\) in the original equations:1. First equation: \(4(1) + 2(3) = 4 + 6 = 10\), which is correct.2. Second equation: \(1 + 3(3) = 1 + 9 = 10\), which is also correct.Both equations are satisfied, which verifies our solution.

Key Concepts

Substitution Method

The substitution method is a popular technique for solving systems of equations. It involves solving one of the equations for one variable, and then substituting this expression into the other equation. This way, you convert the system of equations into a single equation in one variable, which becomes easier to solve. In our exercise:

• We first solve the second equation for the variable \(c\): \(c = 10 - 3d\).
• This expression is then substituted into the first equation to find the value of \(d\).

By replacing \(c\) with the expression we derived, we created an equation in terms of \(d\), making it simpler to isolate the variables and find their values. This process highlights the essence of substitution — reducing the complexity of a simultaneous equation by focusing on just one variable at a time.
It's a powerful tool because it systematically reduces variables and simplifies the equation. This method is especially handy when one equation is already solved for one variable, or can be easily manipulated to get one variable on its own.

Solving Equations

Once you've used substitution to work the system down to a single equation, the next step is solving it. This part can vary in complexity depending on the equation's structure, but the fundamental principles of solving equations stay consistent. First, we simplify the expression through cautious manipulation of algebra to isolate the variable. For instance:

• We substituted \(c = 10 - 3d\) into the first equation and expanded: \(4(10 - 3d) + 2d = 10\).
• Then, simplifying the equation led us to \(-10d = -30\).
• To solve for \(d\), we divided both sides by \(-10\), giving \(d = 3\).

Each step requires thorough checking to ensure no arithmetic mistake has been made, preserving mathematical accuracy. After solving for one variable, substituting back to find the other completes the solution process. This checks and ensures the solution's integrity by consistently verifying with the original equations.

Algebraic Manipulation

Algebraic manipulation is crucial when working with systems of equations. This involves rearranging terms, combining like terms, and simplifying expressions, all paramount in accurately solving equations. In our example, manipulation came into play in multiple steps:

• Isolating \(c\) in the equation \(c + 3d = 10\) by subtracting \(3d\) from both sides.
• Substituting the expression for \(c\) into another equation then expanding and combining like terms \(: 40 - 12d + 2d = 10\).
• Simplifying further by moving constants to one side \(-10d = -30\), and eventually isolating \(d\).

This manipulation often requires working with equations over multiple steps, maintaining balance across the equation. It's like a mathematical juggling act: keeping the equation balanced while moving and adjusting parts. These techniques ensure a systematic approach to reaching a solution. Each move must be justified by preserving the equation's properties, avoiding disruption of the equality we aim to solve.