Problem 15 - College Algebra Essentials

Question

Solve each system. $$\left\\{\begin{array}{rr} x+y & =-4 \\ y-z & =1 \\ 2 x+y+3 z & =-21 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution to the system is \(x = -12\), \(y = 8\), and \(z = 7\)

Step 1: Substitution

From the first equation \(x+y = -4\), isolate \(x\), so \(x = -4 - y\). This expression of \(x\) can be used for substitution in the third equation.

Step 2: Substitute into third equation

Substitute \(x\) from step 1 into the third equation yielding a new equation which only includes \(y\) and \(z\): \(2(-4-y)+y+3z = -21\), which simplifies to \(-8 - 2y + y + 3z = -21\), and further simplifying yields: \(y - 3z = -13\). So now we have a pair of equations with the variables \(y\) and \(z\), namely: (1) \(y - z = 1\), and (2) \(y - 3z = -13\).

Step 3: Solve for \(y\) and \(z\)

Subtract equation (1) from equation (2) dismissing \(y\), yielding: \(y - z - (y - 3z) = 1 - (-13)\) which simplifies to \(2z = 14\), thus \(z = 7\). Substitute \(z = 7\) back into either equation (1) or (2) and solve for \(y\): \(y - 7 = 1\) , therefore, \(y = 8\)

Step 4: Solve for \(x\)

Substitute the value of \(y\) we found in step 3 back into our equation from step 1 to find \(x\): \(x = -4 - y = -4 - 8 = -12\), thus, \(x = -12\).

Key Concepts

Substitution Method

The substitution method is a critical technique used to solve systems of equations, particularly useful for linear equations. The main idea is to take one of the equations in the system, solve it for one variable, and then substitute that expression into the other equations. This turns a system that may initially have three unknowns into one that is more manageable.

In our exercise, we started with the equation \(x+y = -4\). By isolating \(x\), we express it as \(x = -4 - y\). This expression becomes a tool for substitution. It means wherever we see \(x\) in the other equations, we can replace it with \(-4 - y\) to reduce the system's complexity.

Using substitution effectively simplifies a system from three variables down to a simpler problem. The next steps involve solving each equation systematically and substituting back to find the values of the unknown variables.

Linear Equations

Linear equations form the backbone of many algebraic problems and are foundational in solving systems of equations. A linear equation usually appears in the form \(ax + by + cz = d\), where \(a\), \(b\), \(c\), and \(d\) are constants, and \(x\), \(y\), and \(z\) are variables.

In our exercise, we dealt with a system of linear equations:

• \(x + y = -4\)
• \(y - z = 1\)
• \(2x + y + 3z = -21\)

These equations are 'linear' because they consist solely of terms that are either constants or products of constants and a single variable. The graphic representation of such equations is usually a straight line in a two-dimensional space, and a plane in a three-dimensional space.

The essence of solving these equations lies in finding the values of the variables that satisfy all the equations simultaneously, which involves strategic manipulation and simplification using methods like substitution or elimination.

Algebraic Solutions

An algebraic solution involves solving equations using algebraic manipulations. This requires a series of logical steps to isolate variables and determine their values. In systems of equations, this process typically involves multiple steps and operations such as addition, subtraction, multiplication, and division.

In the problem at hand, after employing the substitution method, we proceeded with solving a pair of equations for two variables \(y\) and \(z\):

• \(y - z = 1\)
• \(y - 3z = -13\)

By subtracting these equations, \(2z = 14\) was obtained, giving us \(z = 7\). Substituting \(z = 7\) back into one of the original equations provides \(y = 8\).

Finally, using these values in the expression for \(x\), we find \(x = -12\). Through algebraic solutions, each variable's value was determined, demonstrating the power and systematic nature of algebra in solving equations.