Problem 159 - Elementary Algebra

Question

In the following exercises, solve the systems of equations by substitution. $$ \left\\{\begin{array}{l} -3 x-y=8 \\ 6 x+2 y=-16 \end{array}\right. $$

Step-by-Step Solution

Answer

Infinitely many solutions along the line \(y = -3x - 8\).

Step 1: Solve one equation for one variable

Choose the first equation \(-3x - y = 8\). Solve for \(y\) by isolating it on one side. Adding \(3x\) to both sides: \(-y = 3x + 8\) Then multiply both sides by \(-1\): \(y = -3x - 8\).

Step 2: Substitute the expression into the second equation

Take the expression found for \(y\) from Step 1 \(y = -3x - 8\) and substitute it into the second equation \(6x + 2y = -16\). Replace \(y\): \(6x + 2(-3x - 8) = -16\).

Step 3: Simplify and solve for \(x\)

Distribute the 2 on the left side of the equation: \(6x - 6x - 16 = -16\). Combine like terms: \(0x - 16 = -16\) This simplifies to \(-16 = -16\), which is a true statement indicating the system has infinitely many solutions along the line \(y = -3x - 8\).

Step 4: Verify the solution

For verification purposes, one can choose any value for \(x\) and solve for \(y\) using the simplified form \(y = -3x - 8\). Substituting any \(x\) into both original equations should result in valid equalities.

Key Concepts

Solving systems by substitution

Solving systems of equations by substitution is a handy method where you solve one of the equations for one variable, and then substitute that expression into the other equation. This eliminates one variable, allowing you to solve for the remaining one. In this method, the goal is to convert the system into a simpler form.
For example, consider the system:ewlineewlineewline[\begin{aligned}&-3x - y = 8 \&6x + 2y = -16ewlineewlineWe start by isolating \(y\) in the first equation:
Step 1: Add \(3x\) to both sides of $$-3x - y = 8.$$ This yields \(-y = 3x + 8\) Then, multiply by \(-1\) to get \(y = -3x - 8\).Next, substitute \(y = -3x - 8\) into the second equation \(6x + 2y = -16\). This will allow us to solve for \(x\).

Step-by-step algebra problems

When tackling algebra problems step-by-step, it’s crucial to perform each arithmetic operation carefully. Let’s continue our substitution example.
Step 2: Substitute \(y\) into the second equation:
Starting from \(6x + 2(-3x - 8) = -16\),we distribute the 2: \(6x - 6x - 16 = -16\). Combine like terms to simplify this to \(0x - 16 = -16\),which holds true as both sides are equal, indicating the system's consistency.Working in clear steps allows us to identify patterns and maintain accuracy, especially when variables cancel out or we encounter special cases like infinite solutions.

Consistent systems with infinite solutions

A consistent system with infinite solutions occurs when the two equations in the system represent the same line. This means every point on the line is a solution to the system.
Here's how it was identified in our example:
After substitution and simplification, our equation reduced to \(-16 = -16\). Since this holds true regardless of the values of \(x\) and \(y\), we know our solution set includes infinitely many points.
The general form of this solution for our example is \(y = -3x - 8\).To verify, you can substitute any \(x\) value into both original equations. For instance, if \(x = 0\), \(y = -8\). Both \(-3(0) - (-8) = 8\) and \(6(0) + 2(-8) = -16\) hold true.Understanding that consistent systems with infinite solutions mean the equations describe the same line helps reinforce the concept.