Problem 16 - Algebra A Combined Function

Question

Solve each system. $$ \left\\{\begin{aligned} 5 x+y+3 z &=1 \\ x-y+3 z &=-7 \\ -x+y &=1 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{7}{6} \), \( y = \frac{13}{6} \), and \( z = -2 \).

Step 1: Examine the System of Equations

We have three equations with three unknowns: 1) \( 5x + y + 3z = 1 \) 2) \( x - y + 3z = -7 \) 3) \( -x + y = 1 \). Our goal is to find the values of \( x \), \( y \), and \( z \) that satisfy all three equations simultaneously.

Step 2: Solve for y in Equation 3

From equation 3, we have \( -x + y = 1 \).Rearrange this equation to solve for \( y \): \( y = x + 1 \).

Step 3: Substitute y from Step 2 into Equation 2

Substitute \( y = x + 1 \) into equation 2: \( x - (x + 1) + 3z = -7 \).Simplify the equation: \( -1 + 3z = -7 \) or \( 3z = -6 \).

Step 4: Solve for z

From the equation \( 3z = -6 \), divide both sides by 3 to solve for \( z \): \( z = -2 \).

Step 5: Substitute z into Equation 3

Substitute \( z = -2 \) back into the expression for \( y \): Since \( y = x + 1 \), we need \( y \) for other equations, so we substitute \( z \) next into a different equation.

Step 6: Substitute y from Step 2 and z from Step 4 into Equation 1

Substitute \( y = x + 1 \) and \( z = -2 \) into equation 1:\( 5x + (x + 1) + 3(-2) = 1 \).Simplify and solve for \( x \): \( 6x - 6 = 1 \) or \( 6x = 7 \).

Step 7: Solve for x

Divide both sides by 6 to solve for \( x \): \( x = \frac{7}{6} \).

Step 8: Find y using x from Step 7

Now that we have \( x = \frac{7}{6} \) and we previously had \( y = x + 1 \), substitute to find \( y \):\( y = \frac{7}{6} + \frac{6}{6} = \frac{13}{6} \).

Step 9: Summary of the Solution

We have found the solutions are:\( x = \frac{7}{6} \), \( y = \frac{13}{6} \), and \( z = -2 \). These values satisfy all three equations of the system.

Key Concepts

Linear Equations

Linear equations are mathematical expressions that represent straight lines when plotted graphically. They typically take the form \( ax + by + cz = d \), where \( a \), \( b \), and \( c \) are coefficients, and \( d \) is a constant term.
In linear equations, the variables \( x \), \( y \), and \( z \) are raised to the power of 1, making them easy to manipulate and solve.Characteristics of linear equations:

• They graph as straight lines on a coordinate plane.
• Solve for one variable in terms of others to simplify the system.
• Can have one solution, no solution, or infinitely many solutions depending on the system.

When a system of linear equations is given, like in the original exercise, the goal is to find specific values for each variable that work with all the provided equations. Solving systems of linear equations usually involves methods like substitution or elimination.

Substitution Method

The substitution method is a powerful technique for solving systems of equations. It involves expressing one variable in terms of the others and substituting it into another equation. This reduces the number of variables and simplifies the equations.Consider the basic steps of the substitution method:

• Solve one of the equations for one of the variables.
• Substitute this expression into another equation.
• Simplify the resulting equation to solve for another variable.
• Back-substitute the found values into the original equations to find the remaining variables.

In the original problem, we saw this method applied when \( y \) was expressed as \( x + 1 \) from the third equation. This expression was used in other equations to progressively find values for \( y \), \( x \), and \( z \). This method is straightforward and often crucial in systems where one equation is simpler or already isolated for a single variable.

Solution of Equations

The solution of equations in a system is the set of values that satisfies all given equations simultaneously. For a set of linear equations, the solution represents the point where all the lines intersect on a graph. In three dimensions, this can extend to where planes intersect.In the original exercise, once each variable \( x \), \( y \), and \( z \) was calculated using the substitution method, the final step was verifying these values in all the equations:

• \( x = \frac{7}{6} \)
• \( y = \frac{13}{6} \)
• \( z = -2 \)

These values satisfy each of the original equations, indicating they are correctly solved. To ensure accuracy, when you find a proposed solution, it's beneficial to plug back the values into the original equations to confirm that they all hold true. This not only confirms that the math is correct but also that the solution makes sense in the context of the problem. It's a vital step in the problem-solving process.