Math Problem Statement

The solution of the equation ax + by + c = 0 and bx - ay - 12 = 0 is (2, -3). Find the value of a and b.

Solution

We are tasked with solving for aa and bb based on the system of equations where the solution point is (2,3)(2, -3).

The equations are:

  1. ax+by+c=0ax + by + c = 0
  2. bxay12=0bx - ay - 12 = 0

Substitute x=2x = 2 and y=3y = -3 into both equations:

Step 1: Substituting into the first equation

a(2)+b(3)+c=0a(2) + b(-3) + c = 0 2a - 3b + c = 0 \tag{1}

Step 2: Substituting into the second equation

b(2)a(3)12=0b(2) - a(-3) - 12 = 0 2b + 3a - 12 = 0 \tag{2}

Step 3: Solving the system of equations

We now solve for aa, bb, and cc from the two equations:

From Equation (1): c = -2a + 3b \tag{3}

Substitute c=2a+3bc = -2a + 3b into Equation (2): 2b+3a12=02b + 3a - 12 = 0

Simplify: 3a + 2b = 12 \tag{4}

We now have the system of equations:

  1. 2a3b+c=02a - 3b + c = 0
  2. 3a+2b=123a + 2b = 12

Solving these equations will yield the values of aa and bb. Let me compute this for you. sympy import symbols, Eq, solve

Define variables

a, b, c = symbols('a b c')

Define equations

eq1 = Eq(2a - 3b + c, 0) # 2a - 3b + c = 0 eq2 = Eq(3a + 2b, 12) # 3a + 2b = 12

Solve for a, b, c

solution = solve((eq1, eq2), (a, b, c)) solutionThe solutions for aa and bb in terms of cc are:

a=36132c13a = \frac{36}{13} - \frac{2c}{13} b=2413+3c13b = \frac{24}{13} + \frac{3c}{13}

To find the specific values of aa and bb, we would need a particular value for cc. If more details about cc are provided, I can calculate exact values.

Would you like to specify cc, or should we proceed with additional steps or a general explanation? Here are five follow-up questions:

  1. Should we assume cc is zero for simplicity?
  2. Would you like a deeper explanation of how these equations were solved?
  3. Are there additional constraints for this problem, such as cc?
  4. Do you need this solution verified through substitution?
  5. Should we plot the solutions or explore alternate approaches?

Tip: Providing constraints on all unknowns can make solving linear systems easier.

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Equations
Simultaneous Equations

Formulas

Substitution of points into linear equations
Simplifying and solving systems of linear equations

Theorems

System of linear equations has a unique solution when determinant ≠ 0

Suitable Grade Level

Grades 8-10