Solving Linear Systems in Two Variables

Fundamental Concepts of Linear Systems in Two Variables

• System of Equations: A collection of two or more equations involving the same set of variables, where a solution must satisfy every equation in the system simultaneously.

• Linear Equation in Two Variables: An equation that can be written in the form $Ax + By = C$, where $A$, $B$, and $C$ are constants, and $A$ and $B$ are not both zero. The graph of a linear equation in two variables is a straight line in a two-dimensional coordinate system.

• Solution to a Linear System: An ordered pair $(x, y)$ that makes every equation in the system a true statement when substituted into the respective variables.

Checking System Solutions

To determine whether an ordered pair $(x, y)$ is a solution to a system of two linear equations, substitute the given $x$ and $y$ values into both equations. If both equations yield true statements, the ordered pair is a solution. If either equation fails, the pair is not a solution.

Consider the system of equations:

$x - y = 1$

$2x + 3y = 17$

• Evaluating $(a) \quad x = 5$ and $y = 6$:

  • First equation check:     $5 - 6 = -1 \neq 1$

  • Second equation check:     $2(5) + 3(6) = 10 + 18 = 28 \neq 17$

  • Conclusion: The pair $(5, 6)$ fails both equations and is not a solution.

• Evaluating $(b) \quad x = -7$ and $y = -8$:

  • First equation check:     $-7 - (-8) = -7 + 8 = 1$

  • Second equation check:     $2(-7) + 3(-8) = -14 - 24 = -38 \neq 17$

  • Conclusion: Although $(-7, -8)$ satisfies the first equation, it fails the second equation. Thus, it is not a solution.

• Evaluating $(c) \quad x = 4$ and $y = 3$:

  • First equation check:     $4 - 3 = 1$

  • Second equation check:     $2(4) + 3(3) = 8 + 9 = 17$

  • Conclusion: The pair $(4, 3)$ satisfies both equations simultaneously and is a valid solution.

Solving Linear Systems Using Substitution

The substitution method is an algebraic technique used to solve systems of equations by isolating one variable and substituting its equivalent expression into the other equation.

Procedure for Substitution

1. Isolate a variable: Solve one of the equations for one variable in terms of the other (choose whichever variable is easiest to isolate).

2. Substitute: Replace that variable in the second equation with the algebraic expression obtained in Step 1, producing a single linear equation in one variable. Solve for that variable.

3. Back-substitute: Substitute the numerical value found in Step 2 back into the expression from Step 1 (or any original equation) to solve for the remaining variable.

4. Verify: Check the resulting ordered pair $(x, y)$ in both original equations.

Standard Example of Substitution

Given the system:

$x - y = 1$

$2x + 3y = 17$

• Step 1: Solve the first equation for $x$:   $x = y + 1$

• Step 2: Substitute $x = y + 1$ into the second equation:   $2(y + 1) + 3y = 17$   $2y + 2 + 3y = 17$   $5y + 2 = 17$   $5y = 15$   $y = 3$

• Step 3: Back-substitute $y = 3$ into $x = y + 1$:   $x = 3 + 1 = 4$

• Step 4: Check $(4, 3)$ in both original equations:   $4 - 3 = 1$   $2(4) + 3(3) = 8 + 9 = 17$

  • Solution: $(x, y) = (4, 3)$

Exact Fractional Solutions via Substitution

Given the system:

$5x + 4y = 8$

$6x + 2y = 14$

• Step 1: Solve the second equation for $y$:   $2y = 14 - 6x$   $y = 7 - 3x$

• Step 2: Substitute $y = 7 - 3x$ into the first equation:   $5x + 4(7 - 3x) = 8$   $5x + 28 - 12x = 8$   $-7x + 28 = 8$   $-7x = -20$   $x = \frac{20}{7}$

• Step 3: Back-substitute $x = \frac{20}{7}$ into $y = 7 - 3x$:   $y = 7 - 3\left(\frac{20}{7}\right)$   $y = 7 - \frac{60}{7}$   $y = \frac{49}{7} - \frac{60}{7}$   $y = -\frac{11}{7}$

• Exact Solution:   $(x, y) = \left(\frac{20}{7}, -\frac{11}{7}\right)$

Solving Linear Systems Using Elimination

The elimination (or addition) method involves multiplying one or both equations by suitable constants so that the coefficients of one variable are additive inverses (opposite signs). Adding the equations together eliminates that variable.

Procedure for Elimination

1. Align and Adjust Coefficients: Select a variable to eliminate. Use multiplication (distribution) on one or both equations so that the chosen variable has matching coefficients with opposite signs.

2. Add Equations: Add the modified equations together to cancel the target variable and solve the resulting single-variable equation.

3. Back-substitute: Substitute the solved value back into either of the original equations to compute the value of the eliminated variable.

4. Verify: Check the resulting pair in both original equations.

Standard Example of Elimination

Given the system:

$x - y = 1$

$2x + 3y = 17$

• Step 1: Choose to eliminate $y$. Multiply the first equation by $3$:   $3(x - y) = 3(1) \implies 3x - 3y = 3$

• Step 2: Add this to the second equation ($2x + 3y = 17$):   $(3x - 3y) + (2x + 3y) = 3 + 17$   $5x = 20$   $x = 4$

• Step 3: Back-substitute $x = 4$ into $x - y = 1$:   $4 - y = 1$   $y = 3$

• Step 4: Check $(4, 3)$ in both original equations:   $4 - 3 = 1$   $2(4) + 3(3) = 17$

  • Solution: $(x, y) = (4, 3)$

Exact Fractional Solutions via Elimination

Given the system:

$5x + 4y = 8$

$6x + 2y = 14$

• Step 1: Choose to eliminate $y$. Multiply the second equation by $-2$:   $-2(6x + 2y) = -2(14) \implies -12x - 4y = -28$

• Step 2: Add this modified equation to the first equation ($5x + 4y = 8$):   $(5x + 4y) + (-12x - 4y) = 8 + (-28)$   $-7x = -20$   $x = \frac{20}{7}$

• Step 3: Back-substitute $x = \frac{20}{7}$ into $6x + 2y = 14$:   $6\left(\frac{20}{7}\right) + 2y = 14$   $\frac{120}{7} + 2y = 14$   $2y = 14 - \frac{120}{7}$   $2y = \frac{98}{7} - \frac{120}{7}$   $2y = -\frac{22}{7}$   $y = -\frac{11}{7}$

• Exact Solution:   $(x, y) = \left(\frac{20}{7}, -\frac{11}{7}\right)$

Categorization and Cases of Linear Systems

Every two-variable linear system falls into exactly one of three geometric and algebraic cases:

1. Exactly One Solution: The two lines intersect at a single unique point $(x, y)$. The system is consistent and independent.

2. No Solutions: The two lines are distinct and parallel; they never intersect. Algebraically, solving the system results in a false statement (such as $0 = 1$). The system is inconsistent.

3. Infinitely Many Solutions: The two equations represent the exact same line (coincident lines). Algebraically, solving the system results in an identity (such as $0 = 0$). The system is consistent and dependent.

Understanding Infinitely Many Solutions

"Infinitely many solutions" does not mean that any arbitrary pair of real numbers $(x, y)$ is a valid solution. Instead, it means there are infinitely many points that satisfy the specific functional relationship defined by the line.

Consider the system:

$x + 2y = 3$

$3x + 6y = 9$

• Relationship between equations:

  • Multiplying the first equation by $3$ yields the second equation:     $3(x + 2y) = 3(3) \implies 3x + 6y = 9$

  • Thus, both equations describe the exact same line in the plane.

• Identifying Specific Solution Points:

  • Point 1: Set $x = 1$:     $1 + 2y = 3 \implies 2y = 2 \implies y = 1 \quad \implies (1, 1)$

  • Point 2: Set $x = 3$:     $3 + 2y = 3 \implies 2y = 0 \implies y = 0 \quad \implies (3, 0)$

  • Point 3: Set $x = -1$:     $-1 + 2y = 3 \implies 2y = 4 \implies y = 2 \quad \implies (-1, 2)$

• Key Distinction:

  • An arbitrary point such as $(0, 0)$ gives $0 + 2(0) = 0 \neq 3$, showing that arbitrary real numbers for $x$ and $y$ do not solve the system. Only points of the form $\left(x, \frac{3 - x}{2}\right)$ belong to the set of infinitely many solutions.

Analysis of Special Cases: Parallel and Dependent Lines

System with No Solutions (Inconsistent System)

Find all possible solutions for:

$3x + 6y = 12$

$7x + 14y = 21$

• Algebraic Derivation:

  • Divide the first equation by $3$:     $x + 2y = 4$

  • Divide the second equation by $7$:     $x + 2y = 3$

  • Subtract the simplified second equation from the simplified first equation:     $(x + 2y) - (x + 2y) = 4 - 3$     $0 = 1$

• Explanation:

  • The algebraic statement $0 = 1$ is a contradiction. Converting both equations to slope-intercept form gives $y = -\frac{1}{2}x + 2$ and $y = -\frac{1}{2}x + \frac{3}{2}$. These lines have the same slope ($-\frac{1}{2}$) but different $y$-intercepts ($2$ vs $\frac{3}{2}$). Because they are distinct parallel lines, they never intersect.

  • Result: No solution.

System with Infinitely Many Solutions (Dependent System)

Find all possible solutions for:

$3x + 6y = 12$

$7x + 14y = 28$

• Algebraic Derivation:

  • Divide the first equation by $3$:     $x + 2y = 4$

  • Divide the second equation by $7$:     $x + 2y = 4$

  • Subtracting the two equations yields:     $0 = 0$

• Explanation:

  • The algebraic statement $0 = 0$ is a true identity. Both equations simplify to the exact same equation, $x + 2y = 4$ (or $y = 2 - \frac{1}{2}x$). Graphically, the two lines lie directly on top of one another.

  • Result: Infinitely many solutions, consisting of all ordered pairs $(x, y)$ satisfying $x + 2y = 4$

Crucial Caution Regarding Special Cases

• Do not be too quick to claim the special cases (No Solutions or Infinitely Many Solutions).

• Always complete the algebraic steps rigorously. Similar-looking coefficients do not automatically indicate a special case until the constants on the right-hand side are fully simplified and checked.