Problem 25 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the system of linear equations and check any solution algebraically. $$\left\\{\begin{array}{cc} 4 x+y-3 z= & 11 \\ 2 x-3 y+2 z= & 9 \\ x+y+z= & -3 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution to the given system of linear equations is \(x = -28\), \(y = 12\), and \(z = 13\).

Step 1: Simplify the equations

Before starting with the elimination, we will simplify the equations: Equation 1: \(4x + y - 3z = 11\), Equation 2: \(2x - 3y + 2z = 9\), and Equation 3: \(x + y + z = -3\).

Step 2: Process of Elimination

We can start by multiplying Equation 2 by 2 and Equation 3 by 4. This will give us coefficients for x in Equations 2 and 3 that can be eliminated with Equation 1: New Equation 2: \(4x - 6y + 4z = 18\) and New Equation 3: \(4x + 4y + 4z = -12\). Now, we subtract Equation 2 from Equation 1 and subtract Equation 3 from Equation 1. This will give us two new equations: Equation 1 - Equation 2: \(7y - 7z = -7\) (which simplifies to \(y - z = -1\)) and Equation 1 - Equation 3: \(-3y + z = 23\). Now we subtract the new Equation 2 from the new Equation 3 to find \(2y = 24\), which further simplifies to \(y = 12\).

Step 3: Solve for Remaining Variables

Next, we'll substitute \(y = 12\) into original Equation 3 to find values of x and z. After substituting y, Equation 3 becomes: \(x + 12 + z = -3\), which simplifies to \(x + z = -15\). And substituting y into the new Equation 2 from Step 2 (\(y - z = -1\)) will give us \(12 - z = -1\), which simplifies to \(z = 13\). Lastly, substitute z into \(x + z = -15\) to find \(x = -15 - 13 = -28\).

Step 4: Verify the Solution

We finally check these solutions by substituting x, y, z back into the original equations to verify that they hold true. After substitution, all original equations hold true, which verifies our solution.

Key Concepts

Systems of Linear Equations

A system of linear equations is a collection of two or more linear equations that share common variables. In such systems, a solution is a set of values for the variables that makes all the equations true simultaneously.
For our exercise, the system consists of three equations with variables \(x\), \(y\), and \(z\):

• Equation 1: \(4x + y - 3z = 11\)
• Equation 2: \(2x - 3y + 2z = 9 \)
• Equation 3: \(x + y + z = -3\)

The objective is to find one set of \((x, y, z)\) that satisfies all three equations at the same time. Handling three variables requires strategies that can be efficiently managed through methods like elimination or substitution.

Elimination Method

The elimination method involves eliminating one variable at a time by combining equations in such a way that one variable is removed. This simplifies the system, making it easier to solve for the remaining variables.
When using elimination, you'll often multiply equations by constants to align coefficients so they can be either added or subtracted to remove a variable.

In our problem, equations were adjusted by multiplying and subtracting:

• Multiplied Equation 2 by 2: \(4x - 6y + 4z = 18\)
• Multiplied Equation 3 by 4: \(4x + 4y + 4z = -12\)

Subtraction led to new simplified equations:

• \(y - z = -1\)
• \(-3y + z = 23\)
• From these, solving for one variable became possible with relative ease.

Algebraic Verification

Once a potential solution is found, it's important to verify it by substituting the solution back into the original equations. This step ensures the solution is indeed correct for the entire system, catching any possible errors from prior calculations.
Algebraic verification not only confirms the solution's accuracy but also reinforces conceptual understanding by revisiting the relationships between the variables.

• Substitute \(x=-28\), \(y=12\), and \(z=13\) back into each of the original equations.
• Check if each equation holds true with these values:
• Equation 1 gives \(4(-28) + 12 - 3(13) = 11\)
• Equation 2 gives \(2(-28) - 3(12) + 2(13) = 9\)
• Equation 3 gives \((-28)+12+13=-3\)

If all equations are satisfied, the solution is indeed correct.

Substitution Method

While this system was solved mainly through elimination, understanding the substitution method is valuable for solving systems when elimination seems cumbersome. Substitution involves solving one equation for one variable and then replacing that variable in another equation.
This method is handy when one of the equations is easily solvable for one variable, making it suitable for substituting into other equations to gradually narrow down the values of all variables.

• Choose an equation to solve for one variable, like Equation 3: \(x + y + z = -3\).
• From Equation 3, solve for a variable, such as \(x = -3 - y - z\).
• Substitute \(x\) into the other equations, replacing \(x\) wherever it appears.
• This reduces the complexity, breaking down the process until each variable is solved.

Substitution is often used when starting the solution or when equations are naturally aligned for this approach.