Problem 13 - Algebra 1

Question

Use linear combinations to solve the system of linear equations. $$\begin{aligned} &3 v-2 w=1\\\ &2 v+2 w=4 \end{aligned}$$

Step-by-Step Solution

Answer

The solutions for the system of linear equations are \(v = 1\) and \(w = 1\)

Step 1: Multiply the equations

To eliminate one of the variables from the system, multiply the first equation by 2 and the second equation by 3 such that the coefficients of \(w\) in both equations cancel out when subtracted: \\[ 2*(3v - 2w) = 2*1\\ 3*(2v + 2w) = 3*4 ]

Step 2: Subtract one equation from the other

You should get the following:\(6v - 4w = 2\) and \(6v + 6w = 12\). Now subtract the first equation from the second which will cancel out the \(w\) term: \\[ (6v + 6w) - (6v - 4w) = 12 - 2 ]

Step 3: Solve for \(w\)

This should result in the equation \(10w = 10\). Solve this equation for \(w\) by dividing both sides by 10. You find that \(w = 1\)

Step 4: Substitute \(w\) into one of the original equations

Now that you have the value of \(w\), you can substitute it into either of the original equations to solve for \(v\). Let's substitute into the first original equation: \\[ 3v - 2*1 = 1 ]

Step 5: Solve for \(v\)

This results in the equation \(3v = 3\). Solve this equation for \(v\) by dividing both sides by 3. You find that \(v=1\)

Key Concepts

Understanding Systems of Linear Equations

A system of linear equations consists of two or more equations that share the same set of variables. Here, we are focusing on a system of two equations involving the variables \(v\) and \(w\): - \(3v - 2w = 1\)- \(2v + 2w = 4\)These equations are considered linear because the variables have a power of one and are not multiplied together. Each equation represents a straight line when graphed in a coordinate plane.
The solution to a system of linear equations is the point where the lines intersect. In practical terms, it’s the set of values for the variables that makes both equations true at the same time. Finding this solution can tell you important information about the relationships between different quantities.
In our exercise, we solve this system using the method of linear combinations, which involves manipulating the equations to eliminate one variable, making it easier to find the values of the others.

Methods of Solving Linear Equations

Solving linear equations means finding the values of the variables that satisfy all equations in the system. Several methods can be used including substitution, elimination, and graphing. Here, we focus on using "linear combinations" or the method of elimination. This method involves combining the equations to cancel out one of the variables.

• **Substitution**: Solve one equation for one variable and substitute into the other equation.
• **Elimination by Linear Combinations**: Modify equations so that adding or subtracting them will eliminate one variable.
• **Graphing**: Plot both equations on a graph and find where they intersect.

Linear combinations are effective because they keep the system balanced while reducing the number of variables, ultimately allowing us to solve for the remaining ones. Using this method, find the result for one variable first and then backtrack to find the other.

Variable Elimination: Making Equations Simpler

Variable elimination is a strategic move to simplify a system of equations. It allows solving for one variable by manipulating the equations and removing the other. Here's the process used in our exercise:Start by identifying coefficients of the variable you want to eliminate—in this case, \(w\). We multiply the first equation by 2 and the second by 3 to align the \(w\) coefficients:

• First equation becomes: \(6v - 4w = 2\)
• Second equation becomes: \(6v + 6w = 12\)

Next, subtract the first equation from the second:\(6v + 6w - (6v - 4w) = 12 - 2\), which simplifies to \(10w = 10\). Solving this gives \(w = 1\).
Finally, substitute \(w = 1\) back into one of the original equations to solve for \(v\). This leads to finding \(v = 1\). Through elimination, we efficiently find both variables without having to deal with complex algebraic manipulations.