Question

Solve each system of equations using any method you wish. \(\left\\{\begin{aligned} 2 x-3 y+z &=4 \\\\-3 x+2 y-z &=-3 \\\\-5 y+z &=6 \end{aligned}\right.\)

Short answer

The solution set is \( ( x, -1 - x, 1 - 5x ) \).

Step-by-step solution

- Label the Equations

Label each equation for reference.Equation 1: \(2x - 3y + z = 4\)Equation 2: \(-3x + 2y - z = -3\)Equation 3: \(-5y + z = 6\)

- Solve Equation 3 for z

Isolate variable \(z\) in Equation 3 to use in substitution.\( -5y + z = 6 \)Add \(5y\) to both sides:\(z = 6 + 5y\)

- Substitute z into Equation 1 and Equation 2

Replace \(z\) in Equations 1 and 2 with \(6 + 5y\).For Equation 1:\( 2x - 3y + (6 + 5y) = 4 \)Simplify:\( 2x + 2y + 6 = 4 \)Subtract 6 from both sides:\( 2x + 2y = -2 \)Divide by 2:\( x + y = -1 \) \(Equation \ 4\)For Equation 2:\( -3x + 2y - (6 + 5y) = -3\)Simplify:\( -3x + 2y - 6 - 5y = -3 \)Combine like terms:\( -3x - 3y - 6 = -3 \)Add 6 to both sides:\( -3x - 3y = 3 \)Divide by -3:\( x + y = -1 \) \(Equation \ 5\)

- Verify Consistency

Compare Equations 4 and 5:\( x + y = -1 \) is consistent.

- Solve for a Variable Using Equation 4

Solve for \(x\) or \(y\) using \(x + y = -1\).For \(y\):\( y = -1 - x \)

- Substitute into Expression for z

Use the value of \(y\) in \(z = 6 + 5y\) to find \(z\).\(z = 6 + 5(-1 - x)\)Simplify:\(z = 6 - 5 - 5x\)\(z = 1 - 5x\)

- Express Solution Set

Solutions will be expressed in terms of \(x\):\(x = x,\) \( y = -1 - x,\) \(z = 1 - 5x\).Solution Set: \( ( x, -1 - x, 1 - 5x ) \)

Key concepts

linear equations

Understanding linear equations is crucial for solving systems of equations. A linear equation is a mathematical expression that represents a straight line when plotted on a graph. It typically has the form \[ ax + by + cz = d \] where \(a\), \(b\), \(c\), and \(d\) are constants, and \(x\), \(y\), and \(z\) are variables. These equations are 'linear' because the variables are only raised to the power of one. The solution to a linear equation in two variables \(x\) and \(y\) is a pair \((x, y)\) that makes the equation true. In three variables, the solution is a triplet \((x, y, z)\) that satisfies the equation. Systems of linear equations are sets of linear equations that we solve together to find values for all variables involved.

substitution method

The substitution method is an effective technique for solving systems of linear equations. Here's how it works:

• First, solve one of the equations for one of the variables in terms of the others. This is often the simplest form to rearrange.
• Next, substitute this expression into the other equations. This substitution will replace the variable with its equivalent expression.
• Repeat the substitution until you have an equation with a single variable. Solve this equation to find the value of that variable.
• Back-substitute this value into the expression you found initially to find the other variables.

In our problem, after isolating \(z\) in Equation 3: \[ z = 6 + 5y \], we substituted this expression into Equations 1 and 2, turning them into equations with only \(x\) and \(y\). By solving these, we ultimately found the solution in terms of \(x\).

solution set

A solution set encompasses all possible solutions that satisfy the given system of equations. In this context, the solution set for the system is represented in parametric form, meaning the solutions are expressed in terms of a parameter, typically one of the variables. For the given exercise, we found that:

• \(x = x\)
• \(y = -1 - x\)
• \(z = 1 - 5x\)

Here, \(x\) can be any real number, and for every \(x\) value chosen, there corresponds a \(y\) and \(z\) that satisfies all three original equations. We call this the 'parameter' or 'free variable' solution to the system. Always verify the solutions by substituting back into the original equations to ensure they work for all given conditions.