Problem 16 - Algebra 1: Concepts and Skills

Question

Use linear combinations to solve the linear system. Then check your solution. \(v-w=-5\) \(v+2 w=4\)

Step-by-Step Solution

Answer

The solutions for the system of equations are \(v = -2\) and \(w = 3\).

Step 1: Rearrange the first equation

We can rearrange the first equation to isolate \(v\). Doing so gives us: \(v = w - 5\). This form of the equation is very useful because we can substitute \(v\) from this equation into the second equation to find the value of \(w\).

Step 2: Substitute \(v\) into the second equation

Substitute the value of \(v\) from Step 1 into the second equation: \(w - 5 + 2w = 4\), which simplifies to: \(3w - 5 = 4\). This equation now only has one variable, \(w\), and can be solved as a simple linear equation.

Step 3: Solve for \(w\)

Add 5 to both sides of the equation: \(3w = 9\). Then, divide by 3 to isolate \(w\), yielding: \(w = 3\).

Step 4: Substitute \(w\) back into the first equation

Substitute the value of \(w\) (3) into the first equation: \(v - 3 = -5\), which then can be rearranged to: \(v = -5 + 3\).

Step 5: Solve for \(v\)

Perform the operation on the RHS, yielding the solution: \(v = -2\).

Step 6: Check your solutions

Substitute the values found for \(v\) and \(w\) into both original equations to verify the solutions. The left hand side (LHS) of both equations should equal the right hand side (RHS) when the values of \(v\) and \(w\) are substituted into the equations. We have: \(v\) (which is -2) - \(w\) (which is 3) is indeed equal to -5 and \(v\) (which is -2) + 2 * \(w\) (which is 3) is indeed equal to 4. Thus, the solutions for \(v\) and \(w\) check out in both equations and are therefore correct.

Key Concepts

Linear Equations

Linear equations are mathematical expressions that represent straight lines when graphed on a coordinate plane. They typically take the form of \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) and \( y \) are variables. In a system of linear equations, two or more linear equations are considered together. The solution to such a system is the set of values for the variables that make all equations true simultaneously. For the example from the exercise, we have:

• \( v - w = -5 \)
• \( v + 2w = 4 \)

Here, \( v \) and \( w \) are the variables we need to solve for. The goal is to find one specific solution for both equations at the same time.

Substitution Method

The substitution method is a popular technique to solve systems of linear equations. It involves solving one equation for one variable and then substituting that expression into the other equation. This creates a single equation with one variable that can be solved more easily.In our exercise, we first rearranged the first equation \( v - w = -5 \) to isolate \( v \), giving us \( v = w - 5 \). By doing this, we express one variable in terms of another. From here, we substitute this expression into the second equation \( v + 2w = 4 \), resulting in \( w - 5 + 2w = 4 \), which simplifies to \( 3w - 5 = 4 \). The substitution method simplifies the process by eliminating one variable early on, making it easier to reach the solution.

Solution Verification

Checking your solution is an important step when solving systems of equations. It confirms that the values found satisfy all of the original equations.To verify, substitute the obtained values of \( v \) and \( w \) back into the original equations:

• For \( v - w = -5 \), substituting \( v = -2 \) and \( w = 3 \) gives: \( -2 - 3 = -5 \), which holds true.
• For \( v + 2w = 4 \), substituting \( v = -2 \) and \( w = 3 \) results in \( -2 + 2(3) = 4 \), which also holds true.

If both equations are verified as true, the solution is confirmed as correct. This adds a level of assurance and accuracy to your work.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to make them more manageable. This skill is crucial in solving systems of linear equations.During the exercise, algebraic manipulation was used to:

• Rearrange \( v - w = -5 \) to get \( v = w - 5 \), making substitution possible.
• Simplify \( w - 5 + 2w = 4 \) to \( 3w - 5 = 4 \), so we can solve for \( w \).

Algebraic manipulation makes it possible to simplify complex equations and solve for unknown variables. It often involves applying basic operations such as addition, subtraction, multiplication, and division to both sides of an equation. Mastery of these skills is essential for tackling more complicated algebraic problems.