Problem 27 - Precalculus: Functions and Graphs

Question

Use the method of substitution to solve the system. $$\left\\{\begin{array}{rr} x+2 y-z= & -1 \\ 2 x-y+z= & 9 \\ x+3 y+3 z= & 6 \end{array}\right.$$

Step-by-Step Solution

Answer

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

Step 1: Express x from the first equation

Start by isolating \(x\) in the first equation: \(x + 2y - z = -1\). Move all other terms to the right side: \(x = -1 - 2y + z\). This expression will be used to substitute \(x\) in the other equations.

Step 2: Substitute x in the second equation

Replace \(x\) in the second equation, \(2x - y + z = 9\), with the expression found in step 1: \(2(-1 - 2y + z) - y + z = 9\). Simplify the equation: \(-2 - 4y + 2z - y + z = 9\), which simplifies to \(-5y + 3z = 11\).

Step 3: Substitute x in the third equation

Replace \(x\) in the third equation, \(x + 3y + 3z = 6\), with \(-1 - 2y + z\): \((-1 - 2y + z) + 3y + 3z = 6\). This simplifies to \(-1 + y + 4z = 6\), so \(y + 4z = 7\).

Step 4: Solve the system of two equations

Now we solve the simpler system: \(-5y + 3z = 11\) and \(y + 4z = 7\). Use substitution and solve for \(y\) in terms of \(z\) from the second equation: \(y = 7 - 4z\). Substitute \(y\) in the first equation: \(-5(7 - 4z) + 3z = 11\). Simplify: \(-35 + 20z + 3z = 11\), hence \(23z = 46\). Solving this gives: \(z = 2\).

Step 5: Find y using z = 2

Substitute \(z = 2\) back into \(y = 7 - 4z\): \(y = 7 - 4(2)\), which simplifies to \(y = -1\).

Step 6: Find x using y = -1 and z = 2

Substitute \(y = -1\) and \(z = 2\) into the expression for \(x\) from Step 1: \(x = -1 - 2(-1) + 2\), which simplifies to \(x = 3\).

Step 7: Confirm the solution

Substitute \(x = 3\), \(y = -1\), and \(z = 2\) back into the original equations to verify correctness. All original equations should hold true, confirming that \(x = 3\), \(y = -1\), and \(z = 2\) is the correct solution.

Key Concepts

Substitution Method

The substitution method is a popular technique for solving systems of linear equations. It's particularly useful when one of the equations in the system can be easily solved for one of the variables. Here's how it works:

• First, solve one of the equations for one variable in terms of the other variables. This step involves basic algebraic manipulation.
• Next, substitute this expression into the other equations. This reduces the number of equations and variables, as the system becomes smaller.
• Continue the process step by step until you solve for all variables.

In our original exercise, we started by expressing \(x\) in terms of \(y\) and \(z\) from the first equation. By doing this, we could replace \(x\) in the remaining equations, allowing us to focus on fewer variables in each step. This method simplifies the problem into a more manageable form and is a neat way to step through complex systems methodically.
It is essential to ensure your algebraic manipulations are precise at every stage to avoid errors that could impact the final solution.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations and their representations through matrices and vector spaces. When dealing with systems of linear equations, such as the one in our exercise, linear algebra provides tools to find and describe their solutions.

• Systems of linear equations can have zero, one, or infinitely many solutions. Understanding the structure of the matrix representation helps in determining the nature of the solutions.
• Matrix operations can elegantly solve these systems, particularly when applied using techniques like Gaussian elimination or matrix inversion.
• Linear equations are equations of the first degree, which means all variables appear to the first power and are neither multiplied together nor appear as a function other than a constant multiplier.

Though our exercise focused more on substitution, understanding the broader context of linear algebra helps to appreciate why certain methods work well and how they relate to each other in this fascinating field.

Solution Verification

Verifying the solution of a system of equations is crucial. You need to ensure that your solution satisfies all the original equations, which confirms correctness. Here’s how you can verify:

• After finding the potential solution, substitute the values back into the original equations. Verify each equation's left side equals the right side.
• This check involves simple substitution and basic arithmetic, where you plug the variable values into each equation.
• If all original equations hold true with the found values, you can confidently conclude that the solution is correct.

For example, in our exercise, we substituted \(x = 3\), \(y = -1\), and \(z = 2\) back into all the original equations. Each equation was satisfied, confirming that this set of values is indeed the correct solution. Always perform this crucial verification step to ensure accuracy in your results.

Solving Equations

Solving equations is a foundational skill in algebra that involves finding values for variables that make equations true. Here’s a simple breakdown of effective techniques:

• Identify the type of equation: linear, quadratic, polynomial, etc. Each type has specific methods of solution.
• Use algebraic manipulation: Move terms from one side of the equation to the other by adding, subtracting, multiplying, or dividing.
• Look for opportunities to simplify: Combine like terms or use factoring where applicable.
• For systems of equations, use methods like substitution, elimination, or matrix operations.

In our context, solving equations involved isolating one variable at a time and substituting back into other equations. This approach breaks down complex problems into simpler parts, allowing for a step-by-step resolution. Remember, practice in solving equations leads to proficiency.