System of Linear Equations in Two Variables

Solving Systems of Linear Equations Using Substitution

  • Substitution Method Steps:
    • Step 1: Isolate one variable in one equation (preferably the one with the least complex expression).
    • Step 2: Substitute this isolated variable's expression into the other equation.
    • Step 3: Solve the resultant equation for the remaining variable.
    • Step 4: Substitute this found value back to find the value of the other variable using the expression obtained in Step 1.
    • Step 5: Check both equations using the found values.
Example of Substitution
  • Given the system:
    • 2x+3y=102x + 3y = 10
    • 5xy=265x - y = -26
  1. Solve for yy from the second equation:

    • 5x - y = -26 </li>
    • -y = -5x - 26 </li>
    • y = 5x + 26
  2. Substitute yy in the first equation:

    • 2x+3(5x+26)=102x + 3(5x + 26) = 10
    • 2x+15x+78=102x + 15x + 78 = 10
    • 17x+78=1017x + 78 = 10
    • 17x=107817x = 10 - 78
    • 17x=6817x = -68
    • x=4x = -4
  3. Substitute x=4x = -4 back into the expression for yy:

    • y=5(4)+26y = 5(-4) + 26
    • y=20+26y = -20 + 26
    • y=6y = 6
  4. The solution is: (-4, 6).

  5. Check:

    • First equation: 2(-4) + 3(6) = 10 </li>
    • -8 + 18 = 10
    • Works!
    • Second equation: 5(4)6=265(-4) - 6 = -26
    • -20 - 6 = -26
    • Works!

Solving Systems Using Elimination (Addition Method)

  • Elimination Method Steps:
    • Step 1: Convert both equations to the form Ax+By=CAx + By = C.
    • Step 2: Multiply equations to create coefficient opposites.
    • Step 3: Add the equations to eliminate one variable.
    • Step 4: Solve the new equation for the remaining variable.
    • Step 5: Substitute back into an original equation to find the other variable.
    • Step 6: Check solutions.
Example of Elimination
  • Given the system:
    • 2x+3y=102x + 3y = 10
    • 5xy=265x - y = -26
  1. Rewrite as:

    • 2x+3y=102x + 3y = 10
    • 5x+1y=265x + 1y = -26
  2. Multiply the first equation by 5:

    • 10x+15y=5010x + 15y = 50
    • Multiply the second equation by 3:
    • 15x+3y=7815x + 3y = -78
  3. Subtract to eliminate yy:

    • 10x+15y(15x+3y)=50(78)10x + 15y - (15x + 3y) = 50 - (-78)
    • 5x+12y=128-5x + 12y = 128
    • Solve for xx,
    • x=4x = -4.
  4. Substitute xx into one of the original equations to solve for yy.

  5. Check solutions in both original equations.


Identifying Solutions: Inconsistent or Dependent

  • Systems can have:
    • One solution: unique intersection point.
    • No solution: parallel lines - inconsistent.
    • Infinitely many solutions: same line - dependent.
Example Problem
  • Given the equations:
    • 10x5y=3010x - 5y = 30
    • y=2x6y = 2x - 6
  1. Substitute yy:
    • 10x5(2x6)=3010x - 5(2x - 6) = 30
    • Rearrange:
    • 10x10x+30=3010x - 10x + 30 = 30
    • True; thus, infinitely many solutions represented as ext(x,y)y=2x6ext{{(x, y)| y = 2x - 6}}.

Solving Word Problems Using Systems of Linear Equations

  • Example 1: The sum of two numbers is 38. The larger number is 16 more than the smaller.
    • Let larger number = xx and smaller = yy,
    • Equations:
    • x+y=38x + y = 38
    • x=y+16x = y + 16
  1. Substitute:
    • y+16+y=38y + 16 + y = 38
    • Solve: 2y + 16 = 38 </li>
    • 2y = 22 </li>
    • y = 11
    • x=27x = 27

Numbers are: 27 and 11.

  • Example 2: A fruit company delivers in two sizes.
    • Equations:
    • 2x+3y=512x + 3y = 51 (weight of boxes)
    • 6x+5y=1276x + 5y = 127

Solution found via substitution or elimination yields:

  • Large box = 15.75 kg; Small box = 6.5 kg.

Solving 3x3 Systems of Linear Equations

  • General form for three variables x,y,zx, y, z.
  • Example equations:
    • 5x2y+2z=35x - 2y + 2z = -3
    • 2x+3y=512x + 3y = 51
    • Additional equation needed for a unique solution.
Determine solutions following similar methods as above:
  • Substitution or elimination through [[3 Equations]].
  • Check with all original equations.