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Systems of linear equations | Ninth Grade

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Systems of Linear Equations - Ninth Grade Math

Introduction to Systems of Linear Equations

System of Linear Equations: Two or more linear equations with the same variables
Solution: An ordered pair $(x, y)$ that satisfies ALL equations in the system simultaneously
Standard Form:
$$\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}$$
Three Possible Outcomes:
• One Solution: Lines intersect at exactly one point
• Infinitely Many Solutions: Lines coincide (same line)
• No Solution: Lines are parallel (never intersect)

1. Is (x, y) a Solution to the System?

Steps to Check if $(x, y)$ is a Solution:
Step 1: Substitute the x-value into BOTH equations
Step 2: Substitute the y-value into BOTH equations
Step 3: Simplify both equations
Step 4: Check if BOTH equations are true
    • If BOTH are true → $(x, y)$ IS a solution ✓
    • If ANY is false → $(x, y)$ is NOT a solution ✗
Important: A point must satisfy ALL equations to be a solution to the system!
Example 1: Is $(2, 3)$ a solution to the system?
$$\begin{cases} x + y = 5 \\ 2x - y = 1 \end{cases}$$

Check Equation 1: $x + y = 5$
$2 + 3 = 5$ → $5 = 5$ ✓ TRUE

Check Equation 2: $2x - y = 1$
$2(2) - 3 = 1$ → $4 - 3 = 1$ → $1 = 1$ ✓ TRUE

Answer: YES, $(2, 3)$ is a solution (both equations are satisfied)
Example 2: Is $(1, 4)$ a solution?
$$\begin{cases} 3x + y = 7 \\ x - 2y = 5 \end{cases}$$

Check Equation 1: $3(1) + 4 = 7$ → $7 = 7$ ✓
Check Equation 2: $1 - 2(4) = 5$ → $1 - 8 = 5$ → $-7 = 5$ ✗ FALSE

Answer: NO (second equation is not satisfied)

2-3. Solve a System by Graphing

Graphing Method: Plotting both lines and finding their intersection point
Solution: The point where the two lines intersect $(x, y)$
Steps to Solve by Graphing:
Step 1: Write both equations in slope-intercept form ($y = mx + b$)
Step 2: Graph the first equation
    • Plot y-intercept
    • Use slope to find second point
    • Draw the line
Step 3: Graph the second equation on same axes
Step 4: Find the intersection point
Step 5: Write solution as ordered pair $(x, y)$
Step 6: Check solution in both original equations
Example 1: Solve by graphing
$$\begin{cases} y = 2x - 1 \\ y = -x + 5 \end{cases}$$

Graph Equation 1: $y = 2x - 1$
• Y-intercept: $(0, -1)$
• Slope: 2 → up 2, right 1

Graph Equation 2: $y = -x + 5$
• Y-intercept: $(0, 5)$
• Slope: -1 → down 1, right 1

Lines intersect at $(2, 3)$

Verify:
Equation 1: $3 = 2(2) - 1 = 3$ ✓
Equation 2: $3 = -(2) + 5 = 3$ ✓

Solution: $(2, 3)$
Example 2 (Word Problem): Adult tickets cost $8 and child tickets cost $5. If 10 tickets were sold for $65, how many of each type were sold?

Define variables:
• Let $x$ = number of adult tickets
• Let $y$ = number of child tickets

Write equations:
Total tickets: $x + y = 10$
Total money: $8x + 5y = 65$

Solve first for y:
From equation 1: $y = -x + 10$
From equation 2: $8x + 5y = 65$ → $y = -\frac{8}{5}x + 13$

Graph and find intersection: $(5, 5)$

Answer: 5 adult tickets and 5 child tickets

4-7. Number of Solutions & Classification

Three Types of Systems:

1. Consistent Independent (One Solution):
• Lines intersect at exactly ONE point
• Different slopes: $m_1 \neq m_2$
• One ordered pair solution

2. Consistent Dependent (Infinite Solutions):
• Lines coincide (same line)
• Same slope AND same y-intercept: $m_1 = m_2$ and $b_1 = b_2$
• Infinitely many solutions

3. Inconsistent (No Solution):
• Lines are parallel
• Same slope, different y-intercepts: $m_1 = m_2$ but $b_1 \neq b_2$
• No solution
System TypeGraphSlopesNumber of Solutions
Consistent IndependentLines intersect (×)$m_1 \neq m_2$One solution
Consistent DependentLines coincide (same line)$m_1 = m_2$, $b_1 = b_2$Infinitely many solutions
InconsistentLines parallel (∥)$m_1 = m_2$, $b_1 \neq b_2$No solution
Example 1: Classify the system
$$\begin{cases} y = 3x + 2 \\ y = -2x + 7 \end{cases}$$

Compare slopes: $m_1 = 3$, $m_2 = -2$
Different slopes → lines intersect

Classification: Consistent Independent (one solution)
Example 2: Classify the system
$$\begin{cases} y = 2x + 3 \\ y = 2x - 1 \end{cases}$$

Compare: Same slope ($m = 2$), different y-intercepts (3 and -1)
Lines are parallel

Classification: Inconsistent (no solution)
Example 3: Classify the system
$$\begin{cases} 2x + y = 5 \\ 4x + 2y = 10 \end{cases}$$

Convert to slope-intercept:
Equation 1: $y = -2x + 5$
Equation 2: $2y = -4x + 10$ → $y = -2x + 5$

Same equation! Lines coincide

Classification: Consistent Dependent (infinite solutions)

8-9. Solve Using Substitution Method

Substitution Method: Solving one equation for a variable, then substituting into the other equation
Best When: One variable has coefficient of 1 or is already isolated
Steps for Substitution Method:
Step 1: Solve one equation for one variable (choose the easiest)
Step 2: Substitute that expression into the other equation
Step 3: Solve for the remaining variable
Step 4: Substitute back to find the other variable
Step 5: Write solution as ordered pair
Step 6: Check solution in both original equations
Example 1: Solve using substitution
$$\begin{cases} y = 2x - 3 \\ 3x + y = 12 \end{cases}$$

Step 1: First equation already solved for $y$: $y = 2x - 3$

Step 2: Substitute into second equation
$3x + (2x - 3) = 12$

Step 3: Solve for $x$
$3x + 2x - 3 = 12$
$5x - 3 = 12$
$5x = 15$
$x = 3$

Step 4: Find $y$
$y = 2(3) - 3 = 6 - 3 = 3$

Solution: $(3, 3)$
Example 2: Solve
$$\begin{cases} x + 2y = 7 \\ 3x - y = 5 \end{cases}$$

Step 1: Solve first equation for $x$
$x = 7 - 2y$

Step 2: Substitute into second equation
$3(7 - 2y) - y = 5$
$21 - 6y - y = 5$
$21 - 7y = 5$
$-7y = -16$
$y = \frac{16}{7}$

Step 3: Find $x$
$x = 7 - 2(\frac{16}{7}) = 7 - \frac{32}{7} = \frac{49-32}{7} = \frac{17}{7}$

Solution: $\left(\frac{17}{7}, \frac{16}{7}\right)$
Example 3 (Word Problem): A store sells pencils for $0.25 and pens for $0.75. If 20 items were sold for $11, how many pencils and pens were sold?

Variables: Let $p$ = pencils, $n$ = pens

Equations:
$p + n = 20$ (total items)
$0.25p + 0.75n = 11$ (total cost)

Solve first for p: $p = 20 - n$

Substitute:
$0.25(20 - n) + 0.75n = 11$
$5 - 0.25n + 0.75n = 11$
$5 + 0.5n = 11$
$0.5n = 6$
$n = 12$ pens

$p = 20 - 12 = 8$ pencils

Answer: 8 pencils and 12 pens
Special Cases in Substitution:
• If you get a TRUE statement (like $5 = 5$ or $0 = 0$) → Infinite solutions
• If you get a FALSE statement (like $5 = 3$ or $0 = 7$) → No solution

10-11. Solve Using Elimination Method

Elimination Method: Adding or subtracting equations to eliminate one variable
Also Called: Addition method or linear combination
Best When: Coefficients are easy to match or equations are in standard form
Steps for Elimination Method:
Step 1: Write both equations in standard form ($Ax + By = C$)
Step 2: Multiply one or both equations to make coefficients of one variable opposites
Step 3: Add (or subtract) the equations to eliminate that variable
Step 4: Solve for the remaining variable
Step 5: Substitute back into either original equation to find other variable
Step 6: Write solution and check
Example 1: Solve using elimination
$$\begin{cases} 2x + 3y = 13 \\ 2x - y = 5 \end{cases}$$

Notice: Both have $2x$, so we can eliminate $x$ by subtraction

Subtract second from first:
$(2x + 3y) - (2x - y) = 13 - 5$
$2x + 3y - 2x + y = 8$
$4y = 8$
$y = 2$

Substitute $y = 2$ into first equation:
$2x + 3(2) = 13$
$2x + 6 = 13$
$2x = 7$
$x = \frac{7}{2}$

Solution: $\left(\frac{7}{2}, 2\right)$
Example 2: Solve (need to multiply first)
$$\begin{cases} 3x + 2y = 12 \\ 5x - 3y = 1 \end{cases}$$

Strategy: Eliminate $y$ by making coefficients opposites

Multiply first equation by 3: $9x + 6y = 36$
Multiply second equation by 2: $10x - 6y = 2$

Add equations:
$(9x + 6y) + (10x - 6y) = 36 + 2$
$19x = 38$
$x = 2$

Substitute $x = 2$:
$3(2) + 2y = 12$
$6 + 2y = 12$
$2y = 6$
$y = 3$

Solution: $(2, 3)$
Example 3 (Word Problem): The sum of two numbers is 50, and their difference is 10. Find the numbers.

Let $x$ = first number, $y$ = second number

Equations:
$x + y = 50$
$x - y = 10$

Add equations to eliminate $y$:
$(x + y) + (x - y) = 50 + 10$
$2x = 60$
$x = 30$

Substitute:
$30 + y = 50$
$y = 20$

Answer: The numbers are 30 and 20
Tips for Elimination:
• Look for coefficients that are already opposites or equal
• Choose to eliminate the variable that's easier to match
• You can multiply one or both equations
• If coefficients are fractions, multiply to clear them first

12-13. Solve Using Augmented Matrices

Augmented Matrix: A matrix representation of a system of equations
Row Operations: Operations that transform the matrix while preserving solutions
Row-Echelon Form: Simplified matrix form for easy solution reading
System to Augmented Matrix:
System: $\begin{cases} ax + by = e \\ cx + dy = f \end{cases}$

Augmented Matrix: $\left[\begin{array}{cc|c} a & b & e \\ c & d & f \end{array}\right]$

Goal: Transform to: $\left[\begin{array}{cc|c} 1 & 0 & x \\ 0 & 1 & y \end{array}\right]$
where $x$ and $y$ are the solutions
Three Row Operations (Elementary Operations):
1. Swap rows: $R_i \leftrightarrow R_j$
2. Multiply row by nonzero constant: $kR_i → R_i$
3. Add multiple of one row to another: $R_i + kR_j → R_i$
Steps to Solve Using Matrices:
Step 1: Write the augmented matrix
Step 2: Use row operations to get 1 in top-left position
Step 3: Use row operations to get 0 below the 1
Step 4: Use row operations to get 1 in second row, second column
Step 5: Use row operations to get 0 above that 1
Step 6: Read solution from final column
Example 1: Solve using matrices
$$\begin{cases} x + 2y = 5 \\ 3x - y = 4 \end{cases}$$

Step 1: Write augmented matrix
$\left[\begin{array}{cc|c} 1 & 2 & 5 \\ 3 & -1 & 4 \end{array}\right]$

Step 2: Eliminate 3 in position (2,1)
$R_2 - 3R_1 → R_2$
$\left[\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & -7 & -11 \end{array}\right]$

Step 3: Make second row's second element = 1
$-\frac{1}{7}R_2 → R_2$
$\left[\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & 1 & \frac{11}{7} \end{array}\right]$

Step 4: Eliminate 2 in position (1,2)
$R_1 - 2R_2 → R_1$
$\left[\begin{array}{cc|c} 1 & 0 & \frac{13}{7} \\ 0 & 1 & \frac{11}{7} \end{array}\right]$

Solution: $x = \frac{13}{7}$, $y = \frac{11}{7}$ or $\left(\frac{13}{7}, \frac{11}{7}\right)$
Example 2 (Simpler):
$$\begin{cases} 2x + y = 7 \\ x - y = 2 \end{cases}$$

Augmented matrix:
$\left[\begin{array}{cc|c} 2 & 1 & 7 \\ 1 & -1 & 2 \end{array}\right]$

Swap rows (easier to have 1 in top-left):
$\left[\begin{array}{cc|c} 1 & -1 & 2 \\ 2 & 1 & 7 \end{array}\right]$

$R_2 - 2R_1 → R_2$:
$\left[\begin{array}{cc|c} 1 & -1 & 2 \\ 0 & 3 & 3 \end{array}\right]$

$\frac{1}{3}R_2 → R_2$:
$\left[\begin{array}{cc|c} 1 & -1 & 2 \\ 0 & 1 & 1 \end{array}\right]$

$R_1 + R_2 → R_1$:
$\left[\begin{array}{cc|c} 1 & 0 & 3 \\ 0 & 1 & 1 \end{array}\right]$

Solution: $(3, 1)$

14-15. Solve Using Any Method

Choosing the Best Method:

Use GRAPHING when:
• You need a visual representation
• Equations are already in slope-intercept form
• Approximate solution is acceptable
• Teaching/demonstrating concepts

Use SUBSTITUTION when:
• One variable is already isolated (or easy to isolate)
• One variable has coefficient of 1
• Equations are: $y = ...$ or $x = ...$

Use ELIMINATION when:
• Equations are in standard form
• Coefficients can be easily matched
• Neither variable is isolated
• Both equations have similar structure

Use MATRICES when:
• Multiple equations or higher-level work
• Systematic approach needed
• Using calculator or computer
Example 1: Choose best method and solve
$$\begin{cases} y = 3x - 2 \\ 2x + y = 8 \end{cases}$$

Best Method: SUBSTITUTION (first equation already solved for $y$)

$2x + (3x - 2) = 8$
$5x - 2 = 8$
$5x = 10$
$x = 2$

$y = 3(2) - 2 = 4$

Solution: $(2, 4)$
Example 2: Choose method and solve
$$\begin{cases} 3x + 2y = 16 \\ 5x - 2y = 8 \end{cases}$$

Best Method: ELIMINATION (y coefficients are already opposites!)

Add equations:
$8x = 24$
$x = 3$

$3(3) + 2y = 16$
$9 + 2y = 16$
$y = 3.5$

Solution: $(3, 3.5)$
Example 3 (Word Problem): A movie theater sells adult tickets for $12 and student tickets for $8. One day, 100 tickets were sold for a total of $1040. How many of each type were sold?

Variables: Let $a$ = adult tickets, $s$ = student tickets

System:
$a + s = 100$ (total tickets)
$12a + 8s = 1040$ (total money)

Method Choice: SUBSTITUTION or ELIMINATION
Let's use substitution: $a = 100 - s$

$12(100 - s) + 8s = 1040$
$1200 - 12s + 8s = 1040$
$1200 - 4s = 1040$
$-4s = -160$
$s = 40$ student tickets

$a = 100 - 40 = 60$ adult tickets

Answer: 60 adult tickets and 40 student tickets

Method Comparison Summary

MethodBest ForProsCons
GraphingVisual learners, slope-intercept form• Visual
• Shows all possibilities
• Good for understanding		• Not exact
• Time-consuming
• Hard with fractions
SubstitutionVariable already isolated• Works for any system
• Direct
• Good when y = or x =		• Can get messy
• Lots of algebra
• Errors in substitution
EliminationStandard form, matching coefficients• Efficient
• Less error-prone
• Works well for integers		• Need to multiply
• Requires planning
• Not intuitive at first
MatricesAdvanced, calculator use, 3+ variables• Systematic
• Works for large systems
• Calculator-friendly		• Complex
• Abstract
• More steps

Quick Reference Guide

System Classifications:
• Consistent Independent: One solution (lines intersect)
• Consistent Dependent: Infinite solutions (same line)
• Inconsistent: No solution (parallel lines)
Checking Solutions:
A point $(x, y)$ is a solution ONLY if it satisfies ALL equations in the system
Substitution Steps:
1. Solve one equation for one variable
2. Substitute into other equation
3. Solve for remaining variable
4. Back-substitute to find other variable
Elimination Steps:
1. Align equations in standard form
2. Multiply to match coefficients
3. Add or subtract to eliminate
4. Solve for one variable
5. Substitute back
Special Results:
• If you get $0 = 0$ or true statement → Infinite solutions
• If you get $0 = 5$ or false statement → No solution
• These apply to ALL methods

Common Word Problem Types

Problem TypeVariablesTypical Equations
Ticket SalesNumber of each ticket type• Total tickets
• Total revenue
Mixture ProblemsAmounts of each substance• Total amount
• Total value or concentration
Number ProblemsTwo unknown numbers• Sum equation
• Difference or product equation
Rate/Speed ProblemsSpeeds or rates• Distance = rate × time
• Relationship between rates
Age ProblemsCurrent ages• Current relationship
• Future/past relationship
Success Tips for Systems of Equations:
✓ Always check your solution in BOTH original equations
✓ For word problems, define variables clearly
✓ Choose the method that makes the problem easiest
✓ Substitution is best when variable is already isolated
✓ Elimination is best when coefficients match easily
✓ Watch for special cases: no solution or infinite solutions
✓ Keep work organized to avoid errors
✓ Label your final answer clearly
✓ Include units in word problem answers
✓ Practice all three algebraic methods
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