Problem 57 - Algebra 2

Question

Solve each system of equations. $$ \begin{array}{l}{x+y=5} \\ {x+y+z=4} \\ {2 x-y+2 z=-1}\end{array} $$

Step-by-Step Solution

Answer

The solution is \( x = 2 \), \( y = 3 \), and \( z = -1 \).

Step 1: Label the Equations

First, we need to label each equation in the system for easier reference during our solution process: 1. Equation (1): \( x + y = 5 \)2. Equation (2): \( x + y + z = 4 \)3. Equation (3): \( 2x - y + 2z = -1 \)

Step 2: Solve the First Two Equations for z

Subtract Equation (1) from Equation (2) to eliminate \( x + y \):\( (x + y + z) - (x + y) = 4 - 5 \)This simplifies to:\( z = -1 \)

Step 3: Substitute z in Equation (2)

Replace \( z = -1 \) in Equation (2):\( x + y - 1 = 4 \)Rearrange to solve for \( x + y \):\( x + y = 5 \)This confirms that the equation \( x + y = 5 \) holds as it repeats from Equation (1).

Step 4: Substitute z in Equation (3)

Substitute \( z = -1 \) into Equation (3):\( 2x - y + 2(-1) = -1 \)Simplify to:\( 2x - y - 2 = -1 \)Rearranging gives:\( 2x - y = 1 \)

Step 5: Solve for x and y Using Modified Equations

Now, we solve the simplified equations together:1. \( x + y = 5 \) 2. \( 2x - y = 1 \)Add these two equations to eliminate \( y \):\( (x + y) + (2x - y) = 5 + 1 \)Resulting in:\( 3x = 6 \)Solve for \( x \):\( x = 2 \)

Step 6: Solve for y Using x Value

Using the value \( x = 2 \), substitute back into \( x + y = 5 \):\( 2 + y = 5 \)So, solving for \( y \):\( y = 5 - 2 = 3 \)

Step 7: Summarize the Solution

The solution to the system of equations is:\( x = 2 \)\( y = 3 \)\( z = -1 \)

Key Concepts

Linear Equations

Linear equations are a fundamental part of algebra that help us understand relationships between variables.They are called "linear" because they represent straight lines when plotted on a graph.Each equation is usually in the form of: \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) and \( y \) are variables.
Linear equations can have one, two, or more variables, but they are always of the first degree, meaning no variable is raised to a power higher than one.
The beauty of linear equations is that they can be easily solved using various methods such as graphing, substitution, or elimination.This makes them a versatile tool in solving a wide range of mathematical problems, from simple ones to complex systems of equations.

Substitution Method

The substitution method is a straightforward technique for solving systems of equations.It involves solving one of the equations for one variable and then substituting that expression into the other equation.This effectively reduces the system to a single equation with one variable, which you can then solve more easily.
For example, if you have two equations:

• \( x + y = 5 \)
• \( 2x - y = 1 \)

You can solve the first equation for \( y \), getting \( y = 5 - x \), and substitute this expression for \( y \) in the second equation.
This substitution allows you to solve the second equation for \( x \) directly.Once \( x \) is found, you can back-substitute to find \( y \).In the given exercise, once \( z = -1 \) was found using this method, it was substituted back into the equations to find values for \( x \) and \( y \).The substitution method excels in cases where one equation can be easily solved for a variable, providing quick results with minimal calculations.

Elimination Method

The elimination method is another popular technique to solve systems of linear equations.This method involves adding or subtracting equations from each other to eliminate one of the variables, which simplifies the problem to a single variable equation.
Consider the system:

• \( x + y = 5 \)
• \( 2x - y = 1 \)

To eliminate \( y \), you can add these two equations together to form: \( (x + y) + (2x - y) = 5 + 1 \), which simplifies to \( 3x = 6 \).Solving this gives \( x = 2 \), and you can substitute back to find \( y \).
This method is especially useful when the system of equations is already well set for elimination, where adding or subtracting variables directly cancels them out.In the original exercise, after finding \( z = -1 \), elimination was used to solve the modified equations effectively.The elimination method is powerful as it often provides a clear path to the solution with relatively simple arithmetic.