Problem 55 - Algebra and Trigonometry

Question

Solve the system of equations: \(\left\\{\begin{array}{c}x-y-z=0 \\ 2 x+y+3 z=-1 \\ 4 x+2 y-z=12\end{array}\right.\)

Step-by-Step Solution

Answer

x = 1, y = 3, z = -2

Step 1: Write Down the Equations

The system of equations is given as:1) \( x - y - z = 0 \)2) \( 2x + y + 3z = -1 \)3) \( 4x + 2y - z = 12 \)

Step 2: Add Equations 1 and 3

Add equation 1 and equation 3 to eliminate \(y\) and \(z\)\(x - y - z + 4x + 2y - z = 0 + 12\)This simplifies to:\(5x + y - 2z = 12 \)Now we have a new equation: 4) \(5x + y - 2z = 12\)

Step 3: Solve for One Variable

From equation 1, we can express \( y \) in terms of \( x \) and \( z \):\( y = x - z \)

Step 4: Substitute into Other Equations

Substitute \( y = x - z \) into equations 2 and 4:2 becomes: \( 2x + (x - z) + 3z = -1 \)4 becomes: \( 5x + (x - z) - 2z = 12 \)Simplify:Equation 2: \( 3x + 2z = -1 \)Equation 4: \( 6x - 3z = 12 \)

Step 5: Solve the Simplified System

Now, solve these two equations:1) \( 3x + 2z = -1 \)2) \( 6x - 3z = 12 \)Multiply the first equation by 3:\( 9x + 6z = -3 \)Add this to equation 2:\( 6x - 3z + 9x + 6z = 12 - 3 \)\( 15x + 3z = 9 \)Solve for \( x \). Substituting \( x = 1 \) back:\( 3 \times 1 + 2z = -1\)\( 3 + 2z = -1\)\( 2z = -4\)\( z = -2 \)

Step 6: Substitute Back to Find y

Using \( x = 1 \) and \( z = -2 \), substitute back into our expression for \( y \):\( y = x - z = 1 - (-2) = 3 \)Thus, we have found \( x = 1, y = 3, z = -2 \)

Key Concepts

Linear Equations

Linear equations are equations that graph as straight lines when plotted on a coordinate plane. They are of the form \(ax + by + cz + \ldots = d\), where \(a, b, c,\) and \(d\) are constants and \(x, y, z\) are variables. In our exercise, we work with three linear equations:

• \( x - y - z = 0 \)
• \( 2x + y + 3z = -1 \)
• \( 4x + 2y - z = 12 \)

All these equations involve combining and solving for the variables \(x, y,\) and \(z\). By understanding the components and structure of linear equations, it becomes easier to use different methods to solve them.

Variables

Variables are symbols used to represent unknown values in equations. In our system of equations, \(x, y,\) and \(z\) are the variables we need to solve for. Knowing the values of these variables will satisfy all the given equations simultaneously. Understanding how to manipulate variables and isolate them is key to solving systems of equations. In the solution steps provided, you will see that we progressively isolate and solve for each variable by using algebraic techniques such as substitution and elimination.

Substitution Method

The substitution method is an algebraic technique used to solve systems of equations. It involves solving one of the equations for one variable and then substituting that expression into the other equations. Here's how we used substitution in the exercise:

• Solved equation 1 for \(y\) in terms of \(x\) and \(z\): \( y = x - z \)
• Substituted \( y = x - z \) into the other equations to simplify them.

This gave us new, simpler equations that were easier to solve. Substitution is particularly useful when one equation is already solved for a variable or can be easily rearranged to solve for a variable.

Elimination Method

The elimination method involves adding or subtracting equations to eliminate one of the variables. Here, we used elimination to simplify the equations:

• We added equation 1 to equation 3 to eliminate \(y\) and \(z\): \( x - y - z + 4x + 2y - z = 0 + 12 \)

This gave us a new equation: \(5x + y - 2z = 12\). By continuing with the elimination method, we paired equations in a way that allowed us to simplify and solve for each variable. This technique is particularly effective when the coefficients of one variable in two equations are opposites or can easily be made opposites through multiplication.