Statistical equations are essential tools in cooperative management for analyzing data, forecasting trends, and making informed decisions. Mastery of solving linear and quadratic equations allows cooperative managers to model financial scenarios, optimize resource allocation, and evaluate member contributions effectively. This chapter builds skills in solving these equations with practical cooperative management contexts, providing a foundation for data-driven decision making.

2.1 Solving linear equations with one or more variables

Linear equations represent relationships where variables appear to the first power only. In cooperative management, these equations help model budgets, loan repayments, and membership growth. A linear equation with one variable takes the form \( ax + b = 0 \), while systems with multiple variables involve simultaneous equations that require coordinated solutions.

2.1.1 Solving single-variable linear equations

A single-variable linear equation has the form:

$$ ax + b = 0 $$

where \( a \neq 0 \). The solution isolates \( x \) by rearranging the equation.

Worked Examples

Example 1: A SACCO wants to find the amount \( x \) that balances its monthly income and expenses given by \( 5x + 20 = 0 \).
Given: \( a = 5 \), \( b = 20 \)$$ 5x + 20 = 0 $$$$ 5x = -20 $$$$ x = \frac{-20}{5} $$$$ x = -4 $$ Answer: x = -4

Example 2: The membership fee \( f \) plus a fixed charge of Ksh 150 totals Ksh 750. Expressed as \( f + 150 = 750 \). Find \( f \).
Given: \( f + 150 = 750 \)$$ f + 150 = 750 $$$$ f = 750 - 150 $$$$ f = 600 $$ Answer: f = 600 Ksh

Example 3: A cooperative’s monthly loan repayment \( r \) satisfies \( 12r - 3600 = 0 \). Find \( r \).
Given: \( 12r - 3600 = 0 \)$$ 12r = 3600 $$$$ r = \frac{3600}{12} $$$$ r = 300 $$ Answer: r = 300 Ksh

2.1.2 Solving two-variable linear equations by substitution

Systems of two linear equations with two variables can be solved by substitution, which involves expressing one variable from one equation and substituting into the other.

Given the system:

$$ \begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases} $$

Solve one equation for \( x \) or \( y \), then substitute.

Worked Examples

Example 1: A cooperative sells two products. Product A sells for Ksh 50, Product B for Ksh 30. Total sales were Ksh 2300 from 60 units. Find the quantity of each product sold.
Given:
\( 50x + 30y = 2300 \)
\( x + y = 60 \)

From second equation:$$ x = 60 - y $$

Substitute into first:$$ 50(60 - y) + 30y = 2300 $$$$ 3000 - 50y + 30y = 2300 $$$$ 3000 - 20y = 2300 $$$$ -20y = 2300 - 3000 $$$$ -20y = -700 $$$$ y = \frac{-700}{-20} $$$$ y = 35 $$

Then$$ x = 60 - 35 = 25 $$ Answer: Product A sold 25 units, Product B sold 35 units

Example 2: A cooperative’s monthly income \( x \) plus twice the monthly expenses \( y \) equals Ksh 30,000. Monthly income minus monthly expenses equals Ksh 5,000. Find \( x \) and \( y \).
Given:$$ x + 2y = 30000 $$$$ x - y = 5000 $$

From second equation:$$ x = y + 5000 $$

Substitute into first:$$ (y + 5000) + 2y = 30000 $$$$ y + 5000 + 2y = 30000 $$$$ 3y + 5000 = 30000 $$$$ 3y = 30000 - 5000 $$$$ 3y = 25000 $$$$ y = \frac{25000}{3} $$$$ y = 8333.33 $$

Then$$ x = 8333.33 + 5000 = 13333.33 $$ Answer: Income = Ksh 13,333.33, Expenses = Ksh 8,333.33

Example 3: A cooperative divides profits \( P \) among two member groups. The profit for group A plus 3 times group B’s profit equals Ksh 120,000. Twice A’s profit minus B’s profit equals Ksh 20,000. Find profits for A and B.
Given:$$ P_A + 3P_B = 120000 $$$$ 2P_A - P_B = 20000 $$

From second equation:$$ 2P_A = 20000 + P_B $$$$ P_A = \frac{20000 + P_B}{2} $$

Substitute into first:$$ \frac{20000 + P_B}{2} + 3P_B = 120000 $$ Multiply both sides by 2:$$ 20000 + P_B + 6P_B = 240000 $$$$ 20000 + 7P_B = 240000 $$$$ 7P_B = 240000 - 20000 $$$$ 7P_B = 220000 $$$$ P_B = \frac{220000}{7} $$$$ P_B = 31428.57 $$

Then$$ P_A = \frac{20000 + 31428.57}{2} $$$$ P_A = \frac{51428.57}{2} $$$$ P_A = 25714.29 $$ Answer: Group A profit = Ksh 25,714.29, Group B profit = Ksh 31,428.57

2.1.3 Solving two-variable linear equations by elimination

Elimination involves adding or subtracting equations to eliminate one variable, making it easier to solve.

Given system:

$$ \begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases} $$

Multiply equations if necessary, then add or subtract.

Worked Examples

Example 1: A cooperative sells 40 units of two products. Product A sells for Ksh 60, Product B for Ksh 40. Total revenue is Ksh 2000. Find units sold of each product.
Given:$$ x + y = 40 $$$$ 60x + 40y = 2000 $$

Multiply first by 40:$$ 40x + 40y = 1600 $$

Subtract from second:$$ (60x + 40y) - (40x + 40y) = 2000 - 1600 $$$$ 20x = 400 $$$$ x = \frac{400}{20} $$$$ x = 20 $$

Then$$ y = 40 - 20 = 20 $$ Answer: 20 units each of product A and B sold

Example 2: Monthly savings \( x \) and loan repayments \( y \) for a cooperative satisfy:$$ 3x + 2y = 18000 $$$$ 5x + 3y = 27000 $$

Multiply first by 3:$$ 9x + 6y = 54000 $$

Multiply second by 2:$$ 10x + 6y = 54000 $$

Subtract first from second:$$ (10x + 6y) - (9x + 6y) = 54000 - 54000 $$$$ x = 0 $$

Substitute \( x=0 \) into first:$$ 3(0) + 2y = 18000 $$$$ 2y = 18000 $$$$ y = 9000 $$ Answer: Savings = Ksh 0, Loan repayments = Ksh 9000

Example 3: A cooperative has two types of shares. The number of shares of type A and B satisfy:$$ 4x + 5y = 9000 $$$$ 3x + 2y = 4800 $$

Multiply second by 2.5:$$ 7.5x + 5y = 12000 $$

Subtract first from this:$$ (7.5x + 5y) - (4x + 5y) = 12000 - 9000 $$$$ 3.5x = 3000 $$$$ x = \frac{3000}{3.5} $$$$ x = 857.14 $$

Substitute into second:$$ 3(857.14) + 2y = 4800 $$$$ 2571.43 + 2y = 4800 $$$$ 2y = 4800 - 2571.43 $$$$ 2y = 2228.57 $$$$ y = 1114.29 $$ Answer: Type A shares = 857.14 units, Type B shares = 1114.29 units

2.1.4 Solving three-variable linear equations

Systems with three variables model complex cooperative scenarios such as resource allocation or cost distribution. These require elimination or substitution extended to three equations.

Given system:

$$ \begin{cases} a_1x + b_1y + c_1z = d_1 \\ a_2x + b_2y + c_2z = d_2 \\ a_3x + b_3y + c_3z = d_3 \end{cases} $$

Eliminate one variable to reduce to two equations, then solve.

Worked Examples

Example 1: A cooperative produces three products. The total units and revenues satisfy:$$ x + y + z = 100 $$$$ 10x + 15y + 20z = 1500 $$$$ 2x + 3y + 4z = 320 $$

From first:$$ x = 100 - y - z $$

Substitute into second:$$ 10(100 - y - z) + 15y + 20z = 1500 $$$$ 1000 - 10y - 10z + 15y + 20z = 1500 $$$$ 1000 + 5y + 10z = 1500 $$$$ 5y + 10z = 500 $$$$ y + 2z = 100 $$

Substitute into third:$$ 2(100 - y - z) + 3y + 4z = 320 $$$$ 200 - 2y - 2z + 3y + 4z = 320 $$$$ 200 + y + 2z = 320 $$$$ y + 2z = 120 $$

Contradiction: \( y + 2z = 100 \) and \( y + 2z = 120 \) simultaneously is impossible. Re-examine data or problem setup.

Example 2: A cooperative’s three member groups contribute shares \( x, y, z \) with:$$ x + y + z = 1500 $$$$ 2x + 3y + 4z = 4700 $$$$ 3x + y + 2z = 4000 $$

Eliminate \( x \) using first and third:

From first:$$ x = 1500 - y - z $$

Substitute into third:$$ 3(1500 - y - z) + y + 2z = 4000 $$$$ 4500 - 3y - 3z + y + 2z = 4000 $$$$ 4500 - 2y, z = 4000 $$$$ -2y - z = -500 $$$$ 2y + z = 500 $$

Substitute \( x \) into second:$$ 2(1500 - y - z) + 3y + 4z = 4700 $$$$ 3000 - 2y - 2z + 3y + 4z = 4700 $$$$ 3000 + y + 2z = 4700 $$$$ y + 2z = 1700 $$

Solve system:$$ 2y + z = 500 $$$$ y + 2z = 1700 $$

Multiply second by 2:$$ 2y + 4z = 3400 $$

Subtract first:$$ (2y + 4z) - (2y + z) = 3400 - 500 $$$$ 3z = 2900 $$$$ z = \frac{2900}{3} $$$$ z = 966.67 $$

Substitute into \( 2y + z = 500 \):$$ 2y + 966.67 = 500 $$$$ 2y = 500 - 966.67 $$$$ 2y = -466.67 $$$$ y = -233.33 $$

Substitute into \( x = 1500 - y - z \):$$ x = 1500 - (-233.33) - 966.67 $$$$ x = 1500 + 233.33 - 966.67 $$$$ x = 766.66 $$

Answer: x = 766.66, y = -233.33, z = 966.67 (negative y indicates inconsistency or debt)

Example 3: A cooperative divides Ksh 10,000 among three projects \( x, y, z \) with constraints:$$ x + y + z = 10000 $$$$ 2x + y + 3z = 22000 $$$$ 4x + 3y + 5z = 43000 $$

Eliminate \( x \) using first and second:$$ x = 10000 - y - z $$

Substitute into second:$$ 2(10000 - y - z) + y + 3z = 22000 $$$$ 20000 - 2y - 2z + y + 3z = 22000 $$$$ 20000 - y + z = 22000 $$$$ -y + z = 2000 $$$$ y = z - 2000 $$

Substitute \( x \) and \( y \) into third:$$ 4(10000 - y - z) + 3y + 5z = 43000 $$$$ 40000 - 4y - 4z + 3y + 5z = 43000 $$$$ 40000 - y + z = 43000 $$ Substitute \( y = z - 2000 \):$$ 40000 - (z - 2000) + z = 43000 $$$$ 40000 - z + 2000 + z = 43000 $$$$ 42000 = 43000 $$

Contradiction indicates incorrect data or inconsistent system.

Worked Example 4:

Example 4: Three cooperatives, A, B, and C, contribute to a joint project. The following equations represent their contributions (in thousands of Ksh):

\[ \begin{cases} x + y + z = 60 \\ 2x - y + 3z = 100 \\ 3x + 2y - z = 80 \end{cases} \]

Step 1: Use the first equation to express \(z\) in terms of \(x\) and \(y\):
\[ z = 60 - x - y \]

Step 2: Substitute \(z\) into the second and third equations:

Second equation:
\[ 2x - y + 3z = 100 \\ 2x - y + 3(60 - x - y) = 100 \\ 2x - y + 180 - 3x - 3y = 100 \\ (2x - 3x) + (-y - 3y) + 180 = 100 \\ -x - 4y = 100 - 180 \\ -x - 4y = -80 \\ x + 4y = 80 \]

Third equation:
\[ 3x + 2y - z = 80 \\ 3x + 2y - (60 - x - y) = 80 \\ 3x + 2y - 60 + x + y = 80 \\ (3x + x) + (2y + y) - 60 = 80 \\ 4x + 3y = 80 + 60 \\ 4x + 3y = 140 \]

Step 3: Now solve the two-variable system:
\[ \begin{cases} x + 4y = 80 \\ 4x + 3y = 140 \end{cases} \]

Multiply the first equation by 4:
\[ 4(x + 4y) = 4 \times 80 \\ 4x + 16y = 320 \]

Subtract the second equation:
\[ (4x + 16y) - (4x + 3y) = 320 - 140 \\ (4x - 4x) + (16y - 3y) = 180 \\ 13y = 180 \\ y = \frac{180}{13} \approx 13.85 \]

Substitute \(y\) back into the first equation:
\[ x + 4(13.85) = 80 \\ x + 55.38 = 80 \\ x = 80 - 55.38 = 24.62 \]

Now substitute \(x\) and \(y\) into the expression for \(z\):
\[ z = 60 - 24.62 - 13.85 = 21.53 \]

Answer:

• Cooperative A: \(x \approx 24.62\) thousand Ksh
• Cooperative B: \(y \approx 13.85\) thousand Ksh
• Cooperative C: \(z \approx 21.53\) thousand Ksh

Chapter Summary

This chapter covered the application of statistical equations in business mathematics, starting with solving linear equations involving one or more variables to find unknown values. It then explored quadratic equations using the formula method and factorization to determine the roots. Simultaneous equations were addressed through substitution, elimination, and matrix methods to solve systems of equations efficiently. The chapter also introduced breakeven analysis to identify the point where total costs equal total revenue. Calculus concepts were applied with differentiation to find rates of change and integration to calculate areas under curves. Finally, the formulation of total revenue, total cost, and profit equations was examined to support business decision-making and financial analysis. Together, these topics equip students with essential mathematical tools for solving practical business problems.