W3 Busn Math - Solving Systems of Equations and Breakeven Analysis

Solving Algebraic Equations

  • Solving equations with multiple variables requires having the same number of equations as there are unknowns.
    • For example, with x, y, and z, you need three equations.

Methods for Solving Equations

  • One approach involves manipulating the coefficients of variables to be the same in multiple equations.

Example 1

  • Given equations:
    • 5x+4y=75x + 4y = 7
    • 3x4y=173x - 4y = 17
  • Add the two equations:
    • (5x+3x)+(4y4y)=7+17(5x + 3x) + (4y - 4y) = 7 + 17
    • 8x=248x = 24
  • Solve for x:
    • x=248=3x = \frac{24}{8} = 3
  • Substitute x=3x = 3 into one of the original equations to solve for y:
    • 5(3)+4y=75(3) + 4y = 7
    • 15+4y=715 + 4y = 7
    • 4y=84y = -8
    • y=2y = -2
  • Solution:
    • x=3x = 3
    • y=2y = -2

Example 2

  • Given equations:
    • 4x3y=74x - 3y = -7
    • 2x+8y=442x + 8y = 44
  • Multiply the second equation by 2 to match the x coefficient in the first equation.
    • 2(2x+8y)=2442 * (2x + 8y) = 2 * 44
    • 4x+16y=884x + 16y = 88
  • Multiply the second equation by -2:
    • 2(2x+8y)=244-2 * (2x + 8y) = -2 * 44
    • 4x16y=88-4x - 16y = -88
  • Add the modified second equation to the first equation to eliminate x:
    • (4x4x)+(3y16y)=788(4x - 4x) + (-3y - 16y) = -7 - 88
    • 19y=95-19y = -95
  • Solve for y:
    • y=9519=5y = \frac{-95}{-19} = 5
  • Substitute y=5y = 5 into one of the original equations to solve for x.

Isolating Variables

  • Isolate x in the second equation:
    • 2x=448y2x = 44 - 8y
    • x=224yx = 22 - 4y
  • Substitute this expression for x into the first equation:
    • 4(224y)3y=74(22 - 4y) - 3y = -7
  • Solve for y:
    • 8816y3y=788 - 16y - 3y = -7
    • 19y=95-19y = -95
    • y=5y = 5

Complex Example with Decimals

  • Given equations:
    • 1.5x+0.8y=1.21.5x + 0.8y = 1.2
    • 0.7x+1.2y=4.40.7x + 1.2y = -4.4
  • Multiply both equations by 10 to remove decimals:
    • 15x+8y=1215x + 8y = 12
    • 7x+12y=447x + 12y = -44
  • Multiply the first equation by 3 and the second equation by 2 to make the y coefficients a multiple of 24:
    • 3(15x+8y)=3123 * (15x + 8y) = 3 * 12
    • 45x+24y=3645x + 24y = 36
    • 2(7x+12y)=2(44)2 * (7x + 12y) = 2 * (-44)
    • 14x+24y=8814x + 24y = -88
  • Subtract the second modified equation from the first:
    • (45x14x)+(24y24y)=36(88)(45x - 14x) + (24y - 24y) = 36 - (-88)
    • 31x=12431x = 124
      *Solve for x:
    • x=12431=4x = \frac{124}{31} = 4

Equations With Fractions

  • Given equations with fractions, eliminate the denominators.
  • Example:
    • 9x42y3=136\frac{9x}{4} - \frac{2y}{3} = - \frac{13}{6}
    • 4x5+3y4=12310\frac{4x}{5} + \frac{3y}{4} = \frac{123}{10}
  • Multiply the first equation by 12 (LCM of 4, 3, and 6):
    • 12(3x4)12(2y3)=12(136)12 * (\frac{3x}{4}) - 12 * (\frac{2y}{3}) = 12 * (-\frac{13}{6})
    • 9x8y=269x - 8y = -26
  • Multiply the second equation by 20 (LCM of 5, 4, and 10):
    • 20(4x5)+20(3y4)=20(12310)20 * (\frac{4x}{5}) + 20 * (\frac{3y}{4}) = 20 * (\frac{123}{10})
    • 16x+15y=24616x + 15y = 246
  • Solving for the unknown variables
    • Multiply the first equation by 1.875
    • This will allow to easily eliminate the y variables.

Word Problems

  • Word problems require translating words into algebraic equations.
General Steps
  1. Assign variables: Define variables for the unknowns.
  2. Create Equations: Formulate equations based on the problem.
Example 1: Car Sales
  • Barbie sold twice as many cars as Ken, and together they sold 15 cars.
  • Let:
    • c = number of cars Ken sold
    • 2c = number of cars Barbie sold
  • Equation:
    • c+2c=15c + 2c = 15
    • 3c=153c = 15
    • c=5c = 5
    • Ken sold 5 cars and Barbie sold 10 cars.
Example 2: Daycare Purchases
  • A daycare purchases milk and orange juice weekly. Prices increase, affecting the weekly bill.
  • Let:
    • mm = liters of milk
    • jj = cans of orange juice
  • Before price increase:
    • 1.10m+0.98j=84.401.10m + 0.98j = 84.40
  • After price increase:
    • 1.15m+1.14j=91.701.15m + 1.14j = 91.70
  • Multiply by 100 to remove decimals:
    • 110m+98j=8440110m + 98j = 8440
    • 115m+114j=9170115m + 114j = 9170
  • Double check calculations by plugging them into another equation.
Example 3: Ball Purchases
  • One equation is based on the total cost.
  • The other equation will be based on the number of balls.

Business Applications: Breakeven Analysis

  • Understanding costs, revenue, and profit.
  • Determining the breakeven point.
Key Questions for Starting a Business
  • Initial setup and running costs.
  • Potential revenue and profit.
  • Pricing strategy.
  • Units to sell to break even.
Cost Types
  1. Fixed Costs: Constant costs regardless of sales volume (e.g., rent, insurance).
  2. Variable Costs: Costs that vary with sales volume (e.g., materials, hourly wages).
Examples of Costs
  • Fixed costs: rent, salaries, property taxes.
  • Variable Costs: materials, hourly wages, commissions.
Total Cost Function
  • TC=FC+TVCTC = FC + TVC
    • Where:
      • TCTC = Total Cost
      • FCFC = Fixed Costs
      • TVCTVC = Total Variable Costs
  • TC=FC+VCxTC = FC + VC * x
    • Where:
      • VCVC = Variable Cost per unit
      • xx = Number of units
Breakeven Point
  • Total Revenue = Total Cost.
  • TRTC=0TR - TC = 0
Profit Calculation
  • Profit = Total Revenue - Total Cost.
  • Profit=SPX(FC+VCX)Profit = SP * X - (FC + VC * X)
    • Where:
      • SPSP = selling price
      • FCFC = fixed cost
      • VCVC = variable cost
Breakeven units
  • X=FCSPVCX = \frac{FC}{SP-VC}
Breakeven Point as percentage
  • BreakevenAmountMaximumCapacity\frac{Breakeven \, Amount}{Maximum \, Capacity}
    • the lower this amount is, the more potentially profitable the business is.