Problem 12 - Algebra 1: Concepts and Skills

Question

Use linear combinations to solve the linear system. Then check your solution. \(3 v-2 w=1\) \(2 v+2 w=4\)

Step-by-Step Solution

Answer

The solution to the system of equations is \(v = 1/3\) and \(w = 5/3\).

Step 1: Rewrite the second equation to allow for an easy combination

The second equation can be rewritten as \(2v + 2w = 4\), dividing through by 2 gives us \(v + w = 2\) which is easier to use.

Step 2: Subtract the equations to isolate a variable

Subtract the second equation from the first to isolate one of the variables. Doing so gives us \((3v - 2w) - (v + w) = 1 - 2\), which simplifies to \(2v - w = -1\). Then, the new equation can be rearranged to solve for w: \(w = 2v + 1\).

Step 3: Substitute w into the second equation and solve for the first variable

Put \(w = 2v + 1\) into the second equation \(v + w = 2\), we get \(v + (2v + 1) = 2\), it simplifies to \(3v + 1 = 2\) and then to \(3v = 1\), so \(v = 1/3\).

Step 4: Substitute v back into w equation to get w

Substitute \(v = 1/3\) back into the solution for w, we get \(w = 2(1/3) + 1 = 2/3 + 1 = 5/3\).

Step 5: Check the solution

The final step is to substitute the found values for \(v\) and \(w\) back into the original system of equations to make sure that they satisfy both equations. Substituting \(v = 1/3\) and \(w = 5/3\) into the original equations gives \(3(1/3) - 2(5/3) = 1\) and \(2(1/3) + 2(5/3) = 4\), which are both true, verifying that the solution found is indeed correct.

Key Concepts

Linear Combinations

Linear combinations involve manipulating equations in a way that allows us to solve for one or more unknowns. When dealing with a system of linear equations, the goal is often to either eliminate one of the variables or make the coefficients more manageable. This technique helps to focus on one variable at a time.

Here's what you might do:

• Add or subtract the equations from each other
• Multiply one or both equations by a constant to align coefficients, making cancellation easier
• Combine the equations to reduce the number of variables

In our exercise, one equation was transformed to make the system easier to work with, enabling a cleaner path to a solution through combining the equations.

Substitution Method

The substitution method is another powerful tool used to solve systems of equations. The core idea is to express one of the variables in terms of the other variable(s). This is particularly useful when one of the equations is simpler or when a variable is easily isolated.

To use substitution, follow these steps:

• Isolate a variable in one of the equations, if not already done
• Substitute this expression into the other equation(s)
• Solve the resulting single-variable equation

In the exercise, after determining a simpler form of one equation, substitution allowed us to solve for 'v' and 'w' easily. By finding 'w' in terms of 'v' and substituting back into a simplified equation, we obtained specific values for both variables.

Algebraic Manipulation

Algebraic manipulation refers to the series of steps we take to rearrange and simplify equations or expressions to solve them more easily. This involves using operations like addition, subtraction, multiplication, division, and factoring, all guided by algebra's rules.

Key algebraic manipulation techniques include:

• Rearranging terms to isolate variables
• Multiplying or dividing entire equations to simplify (e.g., eliminating fractions)
• Combining like terms to reduce complexity

In our exercise, algebraic manipulation was crucial to simplify the equations. By properly simplifying and rearranging, the complexity of solving the system was greatly reduced, making it possible to derive the needed values for 'v' and 'w.'