Problem 15 - Algebra 2

Question

Solve each system of equations. \(3 x+y+z=4\) \(2 x+2 y+3 z=3\) \(x+3 y+2 z=5\)

Step-by-Step Solution

Answer

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

Step 1: Write the System

We start with the given system of equations: \[ \begin{align*}3x + y + z &= 4 \quad (1) \2x + 2y + 3z &= 3 \quad (2) \x + 3y + 2z &= 5 \quad (3)\end{align*} \]

Step 2: Eliminate z from Equations (1) and (2)

Multiply equation (1) by 3 to eliminate \(z\) when combined with equation (2):\[ 9x + 3y + 3z = 12 \quad (4) \]Subtract equation (2) from equation (4):\[ (9x + 3y + 3z) - (2x + 2y + 3z) = 12 - 3 \]\[ 7x + y = 9 \quad (5) \]

Step 3: Eliminate z from Equations (2) and (3)

Multiply equation (3) by 3 to eliminate \(z\) when combined with equation (2):\[ 3x + 9y + 6z = 15 \quad (6) \]Subtract equation (2) from equation (6):\[ (3x + 9y + 6z) - (2x + 2y + 3z) = 15 - 3 \]\[ x + 7y + 3z = 12 \quad (7) \]

Step 4: Solve for y from Equations (5) and (7)

Using equation (5):\[ 7x + y = 9 \quad \Rightarrow \quad y = 9 - 7x \quad (8) \]

Step 5: Substitute y into Equation (3)

Substitute \(y = 9 - 7x\) from equation (8) into equation (3):\[ x + 3(9 - 7x) + 2z = 5 \]\[ x + 27 - 21x + 2z = 5 \]\[ -20x + 27 + 2z = 5 \]\[ -20x + 2z = -22 \]\[ 2z = 20x - 22 \]\[ z = 10x - 11 \quad (9) \]

Step 6: Substitute z into Equation (2)

Put \(z = 10x - 11\) into equation (2):\[ 2x + 2y + 3(10x - 11) = 3 \]\[ 2x + 2(9 - 7x) + 30x - 33 = 3 \]\[ 2x + 18 - 14x + 30x - 33 = 3 \]\[ 18x - 15 = 3 \]\[ 18x = 18 \]\[ x = 1 \]

Step 7: Find y and z Values

Substitute \(x = 1\) back into equations for \(y\) and \(z\):From equation (8), \(y = 9 - 7 \times 1 = 2\).From equation (9), \(z = 10 \times 1 - 11 = -1\).

Step 8: Write the Solution

The solution to the system of equations is \(x = 1\), \(y = 2\), and \(z = -1\).

Key Concepts

Linear Equations

Linear equations are fundamental tools in mathematics. They involve variables raised to the first power and have a standard form of \( ax + by + cz = d \). In these equations:

• \( x, y, \) and \( z \) are variables.
• \( a, b, c \) are coefficients, representing the numbers multiplied with the variables.
• \( d \) is the constant term on the opposite side of the equation.

These equations are called "linear" because their graphs are straight lines. When you have more than one linear equation, it forms a system of equations, which can have one solution, no solution, or infinite solutions. The goal is to find the values of the variables that satisfy all equations simultaneously.

Substitution Method

The substitution method is a technique for solving systems of equations, particularly useful when one of the equations is simple and easily solvable. The core idea is:

• Solve one equation for one variable in terms of the others.
• Substitute the expression into the other equations.

This isolates variables one by one, simplifying the problem. For example, if you have an equation like \( y = 9 - 7x \), substitute this into another equation containing \( y \). This reduces the number of variables in that equation, making it easier to solve step-by-step until all variables are determined. This method is reliable when one equation is already solved for a variable or can be easily manipulated to do so.

Elimination Method

The elimination method is another strategy for solving systems of equations. Instead of focusing on single equations, this method involves combining equations to "eliminate" one variable. You follow these steps:

• Multiply equations, if needed, so the coefficients of one variable are equal.
• Add or subtract the equations to cancel out that variable.

The goal is to reduce the system to fewer variables and equations. For instance, you might multiply an equation by a number to match coefficients of \( z \) in two different equations, and then subtract one from the other to eliminate \( z \). This way, you work towards equations with only two variables, making them simpler to solve. Repeat the process as needed until you have a solvable equation for one variable.

Solution of Equations

Finding the solution to a system of equations means determining the values of the variables that satisfy all equations simultaneously. In linear systems, solutions represent the point of intersection of the lines or planes described by the equations.

• A unique solution implies exactly one set of values for the variables.
• No solution indicates the equations represent parallel lines or planes that never intersect.
• Infinite solutions occur when the equations represent the same line or plane, overlapping completely.

In our exercise, the solution was found using substitution and elimination, revealing that \( x = 1, y = 2, \) and \( z = -1 \). It's important to check these values in all original equations to ensure they satisfy each one, confirming the correctness of the solution.