Solving Systems of Linear Equations by Addition

Fundamental Principles of the Addition Method

  • The addition method, also commonly referred to as the elimination method, is an algebraic technique used to solve systems of linear equations.
  • The foundational principle behind this method is the Addition Property of Equality:
    • Adding equal quantities to both sides of an equation preserves equality, producing an equivalent equation.
    • Formally, if A=BA = B and C=DC = D, then A+C=B+DA + C = B + D
  • The primary goal of the addition method is to combine two linear equations in such a way that one of the variable terms drops out (is eliminated), leaving a single linear equation in one variable.

Systematic Algorithm for Solving Linear Systems by Addition

To solve any system of two linear equations in two variables using the addition method, follow these six steps:

  • Step 1: Write in Standard Form

    • Rewrite each equation in standard linear form: Ax+By=CAx + By = C

  • Step 2: Transform Coefficients into Additive Inverses

    • If necessary, multiply one or both equations by non-zero real numbers so that the coefficients of one chosen variable are additive inverses (opposites, such as 55 and −5-5).

  • Step 3: Add the Equations

    • Add the two equations together. The variable with opposite coefficients will sum to 00, effectively eliminating that variable.

  • Step 4: Solve for the First Variable

    • Solve the resulting single-variable linear equation obtained in Step 3.

  • Step 5: Solve for the Second Variable

    • Substitute the numerical value obtained in Step 4 into either of the original equations and solve for the remaining variable.

  • Step 6: Verify the Solution

    • Check the proposed ordered pair (x,y)(x, y) in both original equations of the system to ensure it satisfies both equations.

Direct Elimination and Single-Step Transformations

Direct Variable Elimination

When a system already contains terms with opposite coefficients, the equations can be added immediately.

  • Consider the system:   {x+y=7 x−y=9\begin{cases} x + y = 7 \ x - y = 9 \end{cases}
  • Add the two equations together to eliminate yy:   (x+y)+(x−y)=7+9(x + y) + (x - y) = 7 + 92x=162x = 16
  • Divide both sides by 22:   x=8x = 8
  • Substitute x=8x = 8 into the first original equation x+y=7x + y = 7:   8+y=78 + y = 7
  • Subtract 88 from both sides:   y=−1y = -1
  • Check the ordered pair (8,−1)(8, -1) in both original equations:
    • Equation 1: 8+(−1)=78 + (-1) = 7 (True)
    • Equation 2: 8−(−1)=98 - (-1) = 9 (True)
  • The solution set of the system is (8,−1)(8, -1).

Simultaneous Equation Multiplication

When neither variable has opposite coefficients, both equations must be multiplied by appropriate non-zero values.

  • Consider the system:   {6x−3y=−3 4x+5y=−9\begin{cases} 6x - 3y = -3 \ 4x + 5y = -9 \end{cases}
  • To eliminate yy, create opposite coefficients of −15-15 and 1515 for the yy-terms:
    • Multiply the first equation by 55:     5(6x−3y)=5(−3)5(6x - 3y) = 5(-3)30x−15y=−1530x - 15y = -15
    • Multiply the second equation by 33:     3(4x+5y)=3(−9)3(4x + 5y) = 3(-9)12x+15y=−2712x + 15y = -27
  • Add the resulting equivalent equations:   (30x−15y)+(12x+15y)=−15+(−27)(30x - 15y) + (12x + 15y) = -15 + (-27)42x=−4242x = -42
  • Divide both sides by 4242:   x=−1x = -1
  • Substitute x=−1x = -1 into the first original equation 6x−3y=−36x - 3y = -3:   6(−1)−3y=−36(-1) - 3y = -3−6−3y=−3-6 - 3y = -3
  • Add 66 to both sides:   −3y=3-3y = 3
  • Divide both sides by −3-3:   y=−1y = -1
  • The solution set of the system is (−1,−1)(-1, -1).

Special Cases: Inconsistent and Dependent Systems

Inconsistent Systems (No Solution)

An inconsistent system occurs when two linear equations represent distinct parallel lines that never intersect.

  • Consider the system:   {x−3y=−2 −3x+9y=5\begin{cases} x - 3y = -2 \ -3x + 9y = 5 \end{cases}
  • Multiply the first equation by 33 to attempt eliminating xx:   3(x−3y)=3(−2)3(x - 3y) = 3(-2)3x−9y=−63x - 9y = -6
  • Add the transformed first equation to the second equation:   (3x−9y)+(−3x+9y)=−6+5(3x - 9y) + (-3x + 9y) = -6 + 50=−10 = -1
  • Analysis:
    • Both variables xx and yy are eliminated simultaneously, resulting in a false numerical statement (0=−10 = -1).
    • A false statement indicates that there are no ordered pairs (x,y)(x, y) that satisfy both equations simultaneously.
    • Conclusion: The system has no solution. Geometrically, the graphs of these equations are parallel lines.

Dependent Systems (Infinitely Many Solutions)

A dependent system occurs when two linear equations represent the exact same geometric line.

  • Consider the system:   {2x+5y=1 −4x−10y=−2\begin{cases} 2x + 5y = 1 \ -4x - 10y = -2 \end{cases}
  • Multiply the first equation by 22 to eliminate xx:   2(2x+5y)=2(1)2(2x + 5y) = 2(1)4x+10y=24x + 10y = 2
  • Add the transformed equation to the second equation:   (4x+10y)+(−4x−10y)=2+(−2)(4x + 10y) + (-4x - 10y) = 2 + (-2)0=00 = 0
  • Analysis:
    • Both variables are eliminated, leaving a true numerical statement (0=00 = 0).
    • A true identity indicates that every solution of one equation is also a solution of the other equation.
    • Conclusion: The system has an infinite number of solutions. Geometrically, both equations describe the exact same line.

Advanced Systems and Clearing Fractions

Standard Elimination Example

  • Consider the system:   {4x+5y=14 3x−2y=−1\begin{cases} 4x + 5y = 14 \ 3x - 2y = -1 \end{cases}
  • To eliminate yy, target a common coefficient magnitude of 1010:
    • Multiply equation 1 by 22:     2(4x+5y)=2(14)2(4x + 5y) = 2(14)8x+10y=288x + 10y = 28
    • Multiply equation 2 by 55:     5(3x−2y)=5(−1)5(3x - 2y) = 5(-1)15x−10y=−515x - 10y = -5
  • Add the two equations:   23x=2323x = 23
  • Divide by 2323:   x=1x = 1
  • Substitute x=1x = 1 into equation 1:   4(1)+5y=144(1) + 5y = 144+5y=144 + 5y = 145y=105y = 10y=2y = 2
  • The solution set is (1,2)(1, 2).

Systems containing Fractional Coefficients

When a system contains fractions, first clear all fractions by multiplying each term of an equation by the Least Common Denominator (LCD) of its fractional coefficients.

  • Consider the system:   {23x+14y=−3212x−14y=−2\begin{cases} \frac{2}{3}x + \frac{1}{4}y = -\frac{3}{2} \\ \frac{1}{2}x - \frac{1}{4}y = -2 \end{cases}

  • Clear fractions in the first equation:

    • The denominators are 33, 44, and 22. The LCD is 1212
    • Multiply both sides by 1212:     12(23x+14y)=12(−32)12\left(\frac{2}{3}x + \frac{1}{4}y\right) = 12\left(-\frac{3}{2}\right)8x+3y=−188x + 3y = -18

  • Clear fractions in the second equation:

    • Multiply both sides by 1212 (or 44):     12(12x−14y)=12(−2)12\left(\frac{1}{2}x - \frac{1}{4}y\right) = 12(-2)6x−3y=−246x - 3y = -24

  • Add the two resulting simplified linear equations:   {8x+3y=−18 6x−3y=−24\begin{cases} 8x + 3y = -18 \ 6x - 3y = -24 \end{cases}(8x+3y)+(6x−3y)=−18+(−24)(8x + 3y) + (6x - 3y) = -18 + (-24)14x=−4214x = -42

  • Divide both sides by 1414:   x=−3x = -3

  • Substitute x=−3x = -3 back into the original first equation to solve for yy:   23(−3)+14y=−32\frac{2}{3}(-3) + \frac{1}{4}y = -\frac{3}{2}−2+14y=−32-2 + \frac{1}{4}y = -\frac{3}{2}

    • Add 22 to both sides:     14y=−32+2\frac{1}{4}y = -\frac{3}{2} + 214y=12\frac{1}{4}y = \frac{1}{2}
    • Multiply both sides by 44:     y=2y = 2

  • Verify $x = -3$ and $y = 2$ in original second equation:   12(−3)−14(2)=−32−12=−42=−2\frac{1}{2}(-3) - \frac{1}{4}(2) = -\frac{3}{2} - \frac{1}{2} = -\frac{4}{2} = -2

  • The solution set is (−3,2)(-3, 2).