Systems of Linear Equations Study Notes

A system of linear equations involves two or more linear equations that share the same variables. In 8th grade, we typically focus on systems with two linear equations and two variables, usually \(x\) and \(y\).

What is a System of Linear Equations? 🤔

A linear equation is an equation whose graph is a straight line. When you have two or more such equations together, they form a system. For example:

\[ \begin{cases} y = 2x - 3 \\ x + y = 6 \end{cases} \]

This is a system of two linear equations with variables \(x\) and \(y\).

The Solution to a System 🎯

The solution to a system of two linear equations is an ordered pair \((x, y)\) that makes both equations true at the same time. Geometrically, this solution represents the point where the lines corresponding to each equation intersect.

For instance, if \((3, 3)\) is a solution to the system above:

• For \(y = 2x - 3\): \(3 = 2(3) - 3 \implies 3 = 6 - 3 \implies 3 = 3\) (True)
• For \(x + y = 6\): \(3 + 3 = 6 \implies 6 = 6\) (True)

Since \((3, 3)\) satisfies both equations, it is the solution to this system.

Methods for Solving Systems 🛠️

There are three primary methods to solve systems of linear equations:

• Graphing
• Substitution
• Elimination

Method 1: Graphing 📊

This method involves graphing both linear equations on the same coordinate plane. The point where the two lines intersect is the solution to the system.

Steps for Solving by Graphing:

1. Write each equation in slope-intercept form (\(y = mx + b\)) if not already. This makes graphing easier.
2. Graph the first line using its \(y\)-intercept and slope.
3. Graph the second line on the same coordinate plane.
4. Identify the point of intersection \((x, y)\). This is your solution.
5. Check your solution by plugging the \((x, y)\) values into both original equations.

Example:

Solve the system:

\[ \begin{cases} y = x + 1 \\ y = -2x + 4 \end{cases} \]

1. Both equations are already in \(y = mx + b\) form.
2. Graph \(y = x + 1\): \(y\)-intercept at \((0, 1)\), slope of \(1\) (up 1, right 1).
3. Graph \(y = -2x + 4\): \(y\)-intercept at \((0, 4)\), slope of \(-2\) (down 2, right 1).
4. The lines intersect at the point \((1, 2)\).
5. Check:
   • For \(y = x + 1\): \(2 = 1 + 1 \implies 2 = 2\) (True)
   • For \(y = -2x + 4\): \(2 = -2(1) + 4 \implies 2 = -2 + 4 \implies 2 = 2\) (True)

The solution is \((1, 2)\).

📌 Key Takeaway: Graphing is great for visualizing, but it can be difficult to find an exact solution if the intersection point involves fractions or decimals that are not easy to read from a graph.

Method 2: Substitution 🔄

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation.

Steps for Solving by Substitution:

1. Solve one of the equations for either \(x\) or \(y\). Choose the equation and variable that is easiest to isolate (e.g., a variable with a coefficient of \(1\) or \(-1\)).
2. Substitute the expression from Step 1 into the other equation. This will result in a single equation with only one variable.
3. Solve the new equation for the remaining variable.
4. Substitute the value found in Step 3 back into one of the original equations (or the equation from Step 1) to find the value of the other variable.
5. Write your solution as an ordered pair \((x, y)\).
6. Check your solution in both original equations.

Example:

Solve the system:

\[ \begin{cases} y = 2x - 3 \quad (\text{Equation 1}) \\ x + y = 6 \quad (\text{Equation 2}) \end{cases} \]

1. Equation 1 is already solved for \(y\): \(y = 2x - 3\).
2. Substitute \((2x - 3)\) for \(y\) in Equation 2: \[x + (2x - 3) = 6\]
3. Solve for \(x\): \[3x - 3 = 6\] \[3x = 9\] \[x = 3\]
4. Substitute \(x = 3\) into Equation 1 (or Equation 2): \[y = 2(3) - 3\] \[y = 6 - 3\] \[y = 3\]
5. The solution is \((3, 3)\).
6. Check (from earlier example): \((3, 3)\) satisfies both equations.

💡 Pro Tip: Substitution is very efficient when one variable is already isolated or can be easily isolated (coefficient of \(1\) or \(-1\)).

Method 3: Elimination (Addition/Subtraction) ➖➕

The elimination method involves adding or subtracting the equations to eliminate one of the variables. This often requires multiplying one or both equations by a constant first.

Steps for Solving by Elimination:

1. Align the equations so that like terms (e.g., \(x\) terms, \(y\) terms, constant terms) are in columns.
2. Decide which variable to eliminate. Make sure the coefficients of that variable are either opposites (e.g., \(3y\) and \(-3y\)) or the same (e.g., \(2x\) and \(2x\)).
   • If coefficients are opposites, add the equations.
   • If coefficients are the same, subtract one equation from the other.
   • If coefficients are neither, multiply one or both equations by a constant to create opposite or same coefficients.
3. Add or subtract the equations to eliminate one variable.
4. Solve the resulting single-variable equation.
5. Substitute the value found in Step 4 back into one of the original equations to find the value of the other variable.
6. Write your solution as an ordered pair \((x, y)\).
7. Check your solution in both original equations.

Example 1 (Opposite Coefficients):

Solve the system:

\[ \begin{cases} x + y = 5 \quad (\text{Equation 1}) \\ x - y = 1 \quad (\text{Equation 2}) \end{cases} \]

1. Terms are aligned.
2. The \(y\) terms have opposite coefficients (\(1y\) and \(-1y\)). Add the equations.
3. Add the equations: \[(x + y) + (x - y) = 5 + 1\] \[2x = 6\]
4. Solve for \(x\): \[x = 3\]
5. Substitute \(x = 3\) into Equation 1: \[3 + y = 5\] \[y = 2\]
6. The solution is \((3, 2)\).
7. Check:
   • For \(x + y = 5\): \(3 + 2 = 5 \implies 5 = 5\) (True)
   • For \(x - y = 1\): \(3 - 2 = 1 \implies 1 = 1\) (True)

Example 2 (Multiplying an Equation):

Solve the system:

\[ \begin{cases} 2x + y = 7 \quad (\text{Equation 1}) \\ x - 3y = 0 \quad (\text{Equation 2}) \end{cases} \]

1. Terms are aligned.
2. Let's eliminate \(x\). Multiply Equation 2 by \(2\) to get \(2x\): \[2(x - 3y) = 2(0)\] \[2x - 6y = 0 \quad (\text{New Equation 2})\]. Now we have: \[ \begin{cases} 2x + y = 7 \quad (\text{Equation 1}) \\ 2x - 6y = 0 \quad (\text{New Equation 2}) \end{cases} \] The \(x\) coefficients are the same. Subtract New Equation 2 from Equation 1.
3. Subtract the equations: \[(2x + y) - (2x - 6y) = 7 - 0\] \[2x + y - 2x + 6y = 7\] \[7y = 7\]
4. Solve for \(y\): \[y = 1\]
5. Substitute \(y = 1\) into Equation 2 (original): \[x - 3(1) = 0\] \[x - 3 = 0\] \[x = 3\]
6. The solution is \((3, 1)\).
7. Check:
   • For \(2x + y = 7\): \(2(3) + 1 = 7 \implies 6 + 1 = 7 \implies 7 = 7\) (True)
   • For \(x - 3y = 0\): \(3 - 3(1) = 0 \implies 3 - 3 = 0 \implies 0 = 0\) (True)

💡 Pro Tip: Elimination is very effective when all variables and constants are on one side of the equation (standard form \(Ax + By = C\)).

Types of Solutions 🔍

A system of two linear equations can have one of three possible outcomes:

One Solution (Intersecting Lines)

This is the most common case, where the two lines intersect at exactly one point. The system is called consistent and independent.

\[ \begin{cases} y = x + 1 \\ y = -2x + 4 \end{cases} \] Solution: \((1, 2)\)

No Solution (Parallel Lines)

If the two lines are parallel, they never intersect. This means there is no ordered pair \((x, y)\) that can satisfy both equations simultaneously. When solving algebraically, you will end up with a false statement (e.g., \(0 = 5\)). The system is called inconsistent.

\[ \begin{cases} y = 2x + 3 \\ y = 2x - 1 \end{cases} \] (Both lines have a slope of \(2\) but different \(y\)-intercepts, so they are parallel.)

Infinitely Many Solutions (Same Line)

If the two equations represent the exact same line, then every point on that line is a solution. There are infinitely many solutions. When solving algebraically, you will end up with a true statement (e.g., \(0 = 0\) or \(5 = 5\)). The system is called consistent and dependent.

\[ \begin{cases} y = x + 2 \\ 2y = 2x + 4 \end{cases} \] (The second equation is just the first equation multiplied by \(2\), so they are the same line.)

Summary of Types of Solutions

Type of Solution	Description (Graph)	Algebraic Result	System Classification
One Solution	Lines intersect at one point	A specific value for \(x\) and \(y\) (e.g., \(x=3, y=2\))	Consistent & Independent
No Solution	Parallel lines (never intersect)	False statement (e.g., \(0 = 5\))	Inconsistent
Infinitely Many Solutions	The same line (coincide)	True statement (e.g., \(0 = 0\))	Consistent & Dependent

Solving Word Problems with Systems 📝

Systems of linear equations are powerful tools for solving real-world problems. The key is to translate the problem into two equations.

Steps for Solving Word Problems:

1. Define Variables: Identify the two unknown quantities and assign a variable (e.g., \(x\) and \(y\)) to each.
2. Formulate Equations: Read the problem carefully to find two distinct relationships or conditions. Use these to write two linear equations.
3. Solve the System: Choose the most appropriate method (graphing, substitution, or elimination) to solve the system.
4. State the Answer: Write your solution in the context of the word problem, including units if applicable.
5. Check Your Answer: Make sure your solution makes sense in the original problem statement.

Example:

The sum of two numbers is 10. Their difference is 4. Find the two numbers.

1. Define Variables: Let \(x\) = the first number Let \(y\) = the second number
2. Formulate Equations: "The sum of two numbers is 10": \(x + y = 10\) "Their difference is 4": \(x - y = 4\)
3. Solve the System (using Elimination): \[ \begin{cases} x + y = 10 \\ x - y = 4 \end{cases} \] Add the two equations: \[(x + y) + (x - y) = 10 + 4\] \[2x = 14\] \[x = 7\] Substitute \(x = 7\) into the first equation: \[7 + y = 10\] \[y = 3\]
4. State the Answer: The two numbers are 7 and 3.
5. Check: Sum: \(7 + 3 = 10\) (True) Difference: \(7 - 3 = 4\) (True)