Problem 10 - Algebra 2

Question

Solve each system by substitution. Check your answers. $$ \left\\{\begin{aligned} x+2 y+3 z &=6 \\ y+2 z &=0 \\ z &=2 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution to the system is x = 8, y = -4, z = 2.

Step 1 - Substitute z in the second equation

The system of equations has already given us the value of z. Substitute z = 2 into the second equation, y + 2z = 0, to find the value of y.

Step 2 - Solve for y

Substitute z = 2 into the second equation: y + 2(2) = 0, which simplifies to y + 4 = 0. Solve for y by subtracting 4 from both sides of the equation to get y = -4.

Step 3 - Substitute y and z into the first equation

Now that we have the values for y and z, we can substitute both into the first equation, x + 2y + 3z = 6, to find the value of x.

Step 4 - Solve for x

Substitute y = -4 and z = 2 into the first equation: x + 2(-4) + 3(2) = 6, which simplifies to x - 8 + 6 = 6. Solve for x by adding 8 to both sides of the equation to get x = 8.

Step 5 - Check the solution

To check the solution (x = 8, y = -4, z = 2), substitute these values into all three original equations and verify if they satisfy each equation.

Key Concepts

Substitution Method

The substitution method is a fundamental technique in algebra used to find solutions to systems of linear equations. It involves solving one of the equations for one variable and then substituting the result into another equation.

For instance, take a look at the provided exercise with the system of three equations. Since the last equation directly gives us the value of variable z, it makes a perfect starting point. We use this known value to substitute into the other equations to find the values of the other unknowns, namely x and y.

This method is especially useful when one of the equations in the system already solves for one of the variables or can be easily manipulated to do so. By methodically replacing variables with their equivalent expressions from other equations, the substitution method can simplify a complex system into solvable single-variable equations.

Systems of Linear Equations

Systems of linear equations are collections of two or more linear equations involving the same set of variables. A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable.

In the context of our exercise, we have a system of three linear equations with three variables—x, y, and z. Solving the system means finding the values of these variables that satisfy all equations simultaneously.

Linear systems can have one solution, no solution, or infinitely many solutions. When we are able to find exact values for all variables, as in the provided exercise, we say the system has a unique solution. This result is often represented as a point in three-dimensional space, since our system involves three variables.

Algebraic Solutions

Algebraic solutions involve finding the exact values of variables that satisfy a set of equations. They differ from numerical or graphical solutions because they focus on using algebraic manipulations to arrive at the precise answers.

For example, in our problem, we obtained algebraic solutions by substituting the known value of z into the second equation, then using the result to find y, and finally, plugging these into the first equation to find x. This sequence of steps, which involved manipulating the equations, led us to the answers x = 8, y = -4, and z = 2.

Verifying our solutions is crucial to ensure accuracy. By substituting our found values back into the original equations, we confirm that they hold true, thus validating our algebraic solutions.