Problem 50 - Precalculus: Mathematics for Calculus

Question

Solving a Linear System Solve the system of linear equations. $$\left\\{\begin{array}{rr} 2 x-3 y+5 z= & 14 \\ 4 x-y-2 z= & -17 \\ -x-y+z= & 3 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \((-1, 3, 5)\).

Step 1: Choose a Method

We will use the substitution method to solve the given system of linear equations. This involves solving one of the equations for one of the variables, and then substituting that expression into the other equations.

Step 2: Solve for a Variable

Let's start with the third equation \(-x-y+z=3\). Solve for\( z \):\[ z = x + y + 3 \].

Step 3: Substitute into Another Equation

Substitute \( z = x + y + 3 \) into the first equation:\(2x - 3y + 5(x + y + 3) = 14\). Simplify:\[ 2x - 3y + 5x + 5y + 15 = 14 \]\[ 7x + 2y = -1 \].

Step 4: Substitute into the Remaining Equation

Substitute \( z = x + y + 3 \) into the second equation:\(4x - y - 2(x + y + 3) = -17 \). Simplify:\[ 4x - y - 2x - 2y - 6 = -17 \]\[ 2x - 3y = -11 \].

Step 5: Solve the New System of Equations

Now, solve the system of equations \( 7x + 2y = -1 \) and \( 2x - 3y = -11 \). We multiply the second equation by 2 to eliminate \( y \):\( 4x - 6y = -22 \).Multiply the first equation by 3:\( 21x + 6y = -3 \).

Step 6: Add Equations to Eliminate a Variable

Add the two new equations:\[ 21x + 6y + 4x - 6y = -3 - 22 \]\[ 25x = -25 \]Solve for\( x \):\[ x = -1 \].

Step 7: Substitute Back to Find y

Substitute \( x = -1 \) back into one of the equations, say \( 7x + 2y = -1 \):\( 7(-1) + 2y = -1 \)\[ -7 + 2y = -1 \]\[ 2y = 6 \]\[ y = 3 \].

Step 8: Solve for z

Use\( x = -1 \) and \( y = 3 \) in \( z = x + y + 3 \).\( z = -1 + 3 + 3 \)\[ z = 5 \].

Step 9: Write the Solution

The solution to the system of equations is \((x, y, z) = (-1, 3, 5)\).

Key Concepts

Substitution Method

The substitution method is a powerful technique used to solve systems of linear equations. It involves isolating one of the variables in one of the equations and then substituting this expression into the other equations.
This step-by-step pattern simplifies the complexity of working with multiple variables, reducing the system to fewer variables.

• Start by selecting one equation and solve it for one variable.
• Replace the chosen variable in the other equations with the expression from the first equation.
• Solve the new system, which has one less variable.
• Repeat the process if necessary until all variables are found.

In this exercise, we solved for \( z \) from the third equation and substituted this expression into the other two equations to find the values of \( x \) and \( y \). Finally, these values were back-substituted to solve for \( z \). This method is particularly useful for systems where one equation can be easily manipulated to isolate a variable.

Linear Equations

Linear equations are equations where each term is either a constant or the product of a constant and a single variable. They form straight lines when graphed.
When dealing with multiple variables, the equations describe lines in a multidimensional space.
Understanding the structure and behavior of these equations is vital for solving them. Consider a typical linear equation:\[ ax + by + cz = d \] Here, \( a, b, \) and \( c \) are coefficients, while \( x, y, \) and \( z \) are variables.
The constant \( d \) represents the equation's specific outcome or result.

• Linear equations do not have squares, cubes, or any forms more complex than a first-degree term.
• They allow for prediction and understanding of relationships between variables when combined.

In the given exercise, each equation is linear and forms part of a greater system. Recognizing this helps break down the problem into simpler parts and facilitates the use of methods like substitution.

System of Equations

A system of equations consists of multiple equations that share variables. Solving a system means finding the set of variable values that satisfy all equations simultaneously.
There are several methods to solve these systems, and the choice may depend on the specific system and the solver's preference.

• The system in this problem includes three linear equations.
• It demonstrates how each equation represents a different constraint.

Systems can be dependent (having infinitely many solutions), inconsistent (having no solution), or independent (having a single solution). The exercise you solved was an independent system.
By carefully applying operations like substitution, combining equations, and reducing variables, you find the single unique solution: \( (x, y, z) = (-1, 3, 5) \). Understanding systems of equations is crucial, as it applies to real-world contexts such as economics, engineering, and science where multiple conditions must be met simultaneously.