Chapter Notes: Systems of Equations

Table of Contents
1. What Is a System of Equations?
2. Solving Systems by Graphing
3. Solving Systems by Substitution
4. Solving Systems by Elimination
5. Special Cases
6. Choosing the Best Method
7. Applications of Systems of Equations
8. Summary
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Many real-world problems involve more than one unknown quantity and more than one condition or constraint. For example, you might know the total cost of buying apples and oranges, and you might also know how many pieces of fruit you bought altogether. A system of equations is a set of two or more equations that share the same variables. Solving a system means finding values for the variables that make all the equations true at the same time. In this chapter, you will learn three powerful methods for solving systems of two linear equations with two variables.

What Is a System of Equations?

A system of equations consists of two or more equations that involve the same variables. The solution to a system is an ordered pair (or set of values) that satisfies every equation in the system simultaneously.

For example, consider the system:

Here, \( x \) and \( y \) are the variables. A solution must make both equations true at the same time. If we try \( x = 6 \) and \( y = 4 \), we can check:

• First equation: \( 6 + 4 = 10 \) ✓
• Second equation: \( 6 - 4 = 2 \) ✓

Since both equations are satisfied, the ordered pair \( (6, 4) \) is the solution to the system.

Graphical Interpretation

Every linear equation in two variables represents a line on the coordinate plane. A system of two linear equations corresponds to two lines. The solution to the system is the point where the two lines intersect, because that point lies on both lines and therefore satisfies both equations.

There are three possible outcomes when graphing two lines:

• One solution: The lines intersect at exactly one point. The system is called consistent and independent.
• No solution: The lines are parallel and never intersect. The system is called inconsistent.
• Infinitely many solutions: The lines are identical (they coincide). The system is called consistent and dependent.

Solving Systems by Graphing

The graphing method involves graphing both equations on the same coordinate plane and identifying the point of intersection. This method gives a visual understanding of the solution.

Steps for Solving by Graphing

1. Rewrite each equation in slope-intercept form \( y = mx + b \) if necessary.
2. Graph the first equation using its slope and y-intercept.
3. Graph the second equation on the same set of axes.
4. Identify the point where the two lines intersect.
5. Check the solution by substituting the coordinates into both original equations.

Example:  Solve the system by graphing:
\( y = 2x + 1 \)
\( y = -x + 4 \)

What is the solution?

Solution:

First equation: \( y = 2x + 1 \) has slope 2 and y-intercept 1. Start at (0, 1) and use the slope to find another point: go up 2 and right 1 to reach (1, 3).

Second equation: \( y = -x + 4 \) has slope -1 and y-intercept 4. Start at (0, 4) and use the slope to find another point: go down 1 and right 1 to reach (1, 3).

Both lines pass through the point (1, 3), so this is the intersection point.

Check: First equation: \( 3 = 2(1) + 1 = 3 \) ✓
Second equation: \( 3 = -(1) + 4 = 3 \) ✓

The solution is (1, 3).

Advantages and Limitations

Graphing is useful for visualizing the relationship between two equations and understanding the concept of a solution. However, it has limitations. If the solution involves fractions or decimals, it may be difficult to read the exact coordinates from a graph. For precise answers, algebraic methods are more reliable.

Solving Systems by Substitution

The substitution method is an algebraic technique that involves solving one equation for one variable and then substituting that expression into the other equation. This reduces the system to a single equation with one variable.

Steps for Solving by Substitution

1. Solve one of the equations for one variable in terms of the other variable.
2. Substitute this expression into the other equation.
3. Solve the resulting equation for the remaining variable.
4. Substitute the value found back into the expression from Step 1 to find the other variable.
5. Check the solution in both original equations.

Example:  Solve the system using substitution:
\( y = 3x - 5 \)
\( 2x + y = 10 \)

What is the solution?

Solution:

The first equation is already solved for \( y \): \( y = 3x - 5 \).

Substitute \( 3x - 5 \) for \( y \) in the second equation:
\( 2x + (3x - 5) = 10 \)

Combine like terms:
\( 5x - 5 = 10 \)

Add 5 to both sides:
\( 5x = 15 \)

Divide by 5:
\( x = 3 \)

Now substitute \( x = 3 \) back into \( y = 3x - 5 \):
\( y = 3(3) - 5 = 9 - 5 = 4 \)

Check in both equations:
First: \( 4 = 3(3) - 5 = 4 \) ✓
Second: \( 2(3) + 4 = 6 + 4 = 10 \) ✓

The solution is (3, 4).

Example:  Solve the system using substitution:
\( x + 2y = 7 \)
\( 3x - y = 5 \)

What is the solution?

Solution:

Solve the first equation for \( x \):
\( x = 7 - 2y \)

Substitute into the second equation:
\( 3(7 - 2y) - y = 5 \)

Distribute:
\( 21 - 6y - y = 5 \)

Combine like terms:
\( 21 - 7y = 5 \)

Subtract 21 from both sides:
\( -7y = -16 \)

Divide by -7:
\( y = \frac{16}{7} \)

Substitute back into \( x = 7 - 2y \):
\( x = 7 - 2\left(\frac{16}{7}\right) = 7 - \frac{32}{7} = \frac{49}{7} - \frac{32}{7} = \frac{17}{7} \)

The solution is \( \left(\frac{17}{7}, \frac{16}{7}\right) \).

When to Use Substitution

Substitution works well when one equation is already solved for a variable, or when it is easy to isolate a variable. If one equation has a coefficient of 1 or -1 on a variable, substitution is usually efficient.

Solving Systems by Elimination

The elimination method (also called the addition method) involves adding or subtracting the two equations to eliminate one variable. This produces a single equation with one variable that can be solved easily.

Steps for Solving by Elimination

1. Write both equations in standard form \( Ax + By = C \).
2. If necessary, multiply one or both equations by a constant so that the coefficients of one variable are opposites.
3. Add the two equations to eliminate one variable.
4. Solve the resulting equation for the remaining variable.
5. Substitute this value into either original equation to find the other variable.
6. Check the solution in both original equations.

Example:  Solve the system using elimination:
\( x + y = 8 \)
\( x - y = 2 \)

What is the solution?

Solution:

Notice that the coefficients of \( y \) are already opposites (1 and -1).

Add the two equations:
\( (x + y) + (x - y) = 8 + 2 \)
\( 2x = 10 \)

Divide by 2:
\( x = 5 \)

Substitute \( x = 5 \) into the first equation:
\( 5 + y = 8 \)
\( y = 3 \)

Check: First equation: \( 5 + 3 = 8 \) ✓
Second equation: \( 5 - 3 = 2 \) ✓

The solution is (5, 3).

Example:  Solve the system using elimination:
\( 3x + 2y = 12 \)
\( 5x - 2y = 4 \)

What is the solution?

Solution:

The coefficients of \( y \) are already opposites (2 and -2).

Add the equations:
\( (3x + 2y) + (5x - 2y) = 12 + 4 \)
\( 8x = 16 \)

Divide by 8:
\( x = 2 \)

Substitute \( x = 2 \) into the first equation:
\( 3(2) + 2y = 12 \)
\( 6 + 2y = 12 \)
\( 2y = 6 \)
\( y = 3 \)

The solution is (2, 3).

Using Multiplication Before Elimination

Sometimes the coefficients of neither variable are opposites. In such cases, you must multiply one or both equations by appropriate constants before adding.

Example:  Solve the system using elimination:
\( 2x + 3y = 8 \)
\( 3x - y = 1 \)

What is the solution?

Solution:

To eliminate \( y \), multiply the second equation by 3:
\( 3(3x - y) = 3(1) \)
\( 9x - 3y = 3 \)

Now the system is:
\( 2x + 3y = 8 \)
\( 9x - 3y = 3 \)

Add the equations:
\( (2x + 3y) + (9x - 3y) = 8 + 3 \)
\( 11x = 11 \)
\( x = 1 \)

Substitute \( x = 1 \) into the second original equation:
\( 3(1) - y = 1 \)
\( 3 - y = 1 \)
\( y = 2 \)

The solution is (1, 2).

When to Use Elimination

Elimination is particularly efficient when the coefficients of one variable are the same or opposites, or can easily be made so. It is also useful when both equations are in standard form and neither variable is easily isolated.

Special Cases

Not all systems have exactly one solution. Some systems have no solution, and others have infinitely many solutions. Recognizing these cases is important.

No Solution (Inconsistent System)

When you solve a system and reach a contradiction such as \( 0 = 5 \) or \( 3 = -3 \), the system has no solution. Graphically, this means the lines are parallel.

Example:  Solve the system:
\( y = 2x + 3 \)
\( y = 2x - 1 \)

What is the solution?

Solution:

Substitute the first equation into the second:
\( 2x + 3 = 2x - 1 \)

Subtract \( 2x \) from both sides:
\( 3 = -1 \)

This is a contradiction. The system has no solution.

Infinitely Many Solutions (Dependent System)

When you solve a system and reach an identity such as \( 0 = 0 \) or \( 5 = 5 \), the system has infinitely many solutions. This happens when the two equations represent the same line.

Example:  Solve the system:
\( 2x + y = 6 \)
\( 4x + 2y = 12 \)

What is the solution?

Solution:

Notice that the second equation is exactly twice the first equation.

Multiply the first equation by -2:
\( -4x - 2y = -12 \)

Add to the second equation:
\( (-4x - 2y) + (4x + 2y) = -12 + 12 \)
\( 0 = 0 \)

This is an identity. The system has infinitely many solutions. Every point on the line \( 2x + y = 6 \) is a solution.

Choosing the Best Method

Each method has advantages depending on the form of the equations. Here are some guidelines:

Applications of Systems of Equations

Systems of equations are powerful tools for solving real-world problems involving two unknowns and two conditions.

Setting Up a System from a Word Problem

To translate a word problem into a system of equations:

1. Identify the two unknowns and assign variables to them.
2. Write two equations based on the information given in the problem.
3. Solve the system using an appropriate method.
4. Interpret the solution in the context of the problem.

Example:  At a school fundraiser, adult tickets cost $5 and student tickets cost $3.
A total of 200 tickets were sold, and the total revenue was $820.

How many adult tickets and how many student tickets were sold?

Solution:

Let \( a \) = number of adult tickets and \( s \) = number of student tickets.

First equation (total tickets):
\( a + s = 200 \)

Second equation (total revenue):
\( 5a + 3s = 820 \)

Solve the first equation for \( a \):
\( a = 200 - s \)

Substitute into the second equation:
\( 5(200 - s) + 3s = 820 \)
\( 1000 - 5s + 3s = 820 \)
\( 1000 - 2s = 820 \)
\( -2s = -180 \)
\( s = 90 \)

Substitute back:
\( a = 200 - 90 = 110 \)

There were 110 adult tickets and 90 student tickets sold.

Example:  A chemist needs to mix a 20% acid solution with a 50% acid solution to create 30 liters of a 35% acid solution.

How many liters of each solution should be used?

Solution:

Let \( x \) = liters of 20% solution and \( y \) = liters of 50% solution.

First equation (total volume):
\( x + y = 30 \)

Second equation (amount of acid):
\( 0.20x + 0.50y = 0.35(30) \)
\( 0.20x + 0.50y = 10.5 \)

Multiply the second equation by 10 to clear decimals:
\( 2x + 5y = 105 \)

Multiply the first equation by -2:
\( -2x - 2y = -60 \)

Add to the modified second equation:
\( (-2x - 2y) + (2x + 5y) = -60 + 105 \)
\( 3y = 45 \)
\( y = 15 \)

Substitute back:
\( x + 15 = 30 \)
\( x = 15 \)

The chemist should use 15 liters of the 20% solution and 15 liters of the 50% solution.

Summary

Systems of equations allow us to solve problems with multiple unknowns and multiple conditions. The three main solution methods-graphing, substitution, and elimination-each have strengths depending on the form and context of the equations. Graphing provides visual insight, substitution is ideal when a variable is isolated, and elimination works well with standard form equations. Recognizing special cases such as no solution or infinitely many solutions is also essential. By mastering these techniques, you can solve a wide variety of mathematical and real-world problems efficiently and accurately.