Math Problem Statement

b + ac + e = 2, 2b + ae = -2, 3b -c +a^2d + e = 3, ab + 2c + 7d + e = a^3. find the value of a by substitution method

Solution

We are given the following system of equations:

  1. b+ac+e=2b + ac + e = 2
  2. 2b+ae=22b + ae = -2
  3. 3bc+a2d+e=33b - c + a^2d + e = 3
  4. ab+2c+7d+e=a3ab + 2c + 7d + e = a^3

To find the value of aa using the substitution method, we will proceed step-by-step.

Step 1: Solve for ee from Equation (1)

From Equation (1): b+ac+e=2b + ac + e = 2 Solve for ee: e = 2 - b - ac \tag{5}

Step 2: Substitute ee into Equation (2)

Substitute the expression for ee from Equation (5) into Equation (2): 2b+ae=22b + ae = -2 2b+a(2bac)=22b + a(2 - b - ac) = -2 Expand: 2b+a(2)aba2c=22b + a(2) - ab - a^2c = -2 Simplify: 2b+2aaba2c=22b + 2a - ab - a^2c = -2 2b - ab + 2a - a^2c = -2 \tag{6}

Step 3: Substitute ee into Equation (3)

Substitute the expression for ee from Equation (5) into Equation (3): 3bc+a2d+e=33b - c + a^2d + e = 3 3bc+a2d+(2bac)=33b - c + a^2d + (2 - b - ac) = 3 Simplify: 3bc+a2d+2bac=33b - c + a^2d + 2 - b - ac = 3 (3bb)c+a2dac+2=3(3b - b) - c + a^2d - ac + 2 = 3 2bc+a2dac+2=32b - c + a^2d - ac + 2 = 3 Simplify further: 2b - c + a^2d - ac = 1 \tag{7}

Step 4: Substitute ee into Equation (4)

Substitute the expression for ee from Equation (5) into Equation (4): ab+2c+7d+e=a3ab + 2c + 7d + e = a^3 ab+2c+7d+(2bac)=a3ab + 2c + 7d + (2 - b - ac) = a^3 Simplify: ab+2c+7d+2bac=a3ab + 2c + 7d + 2 - b - ac = a^3 (abb)+(2cac)+7d+2=a3(ab - b) + (2c - ac) + 7d + 2 = a^3 b(a - 1) + c(2 - a) + 7d + 2 = a^3 \tag{8}

Now we have the following simplified equations:

  • Equation (6): 2bab+2aa2c=22b - ab + 2a - a^2c = -2
  • Equation (7): 2bc+a2dac=12b - c + a^2d - ac = 1
  • Equation (8): b(a1)+c(2a)+7d+2=a3b(a - 1) + c(2 - a) + 7d + 2 = a^3

Step 5: Solve the system of equations

At this point, we can attempt to solve for bb, cc, and dd in terms of aa using these equations, then substitute into one of the equations to solve for aa.

Let me continue solving this system of equations.The solutions for bb, cc, and dd in terms of aa are as follows:

b=a72a325a228a14a54a317a2+7a+14b = \frac{a^7 - 2a^3 - 25a^2 - 28a - 14}{a^5 - 4a^3 - 17a^2 + 7a + 14}

c=a6+2a5+2a4+2a36a221a42a54a317a2+7a+14c = \frac{-a^6 + 2a^5 + 2a^4 + 2a^3 - 6a^2 - 21a - 42}{a^5 - 4a^3 - 17a^2 + 7a + 14}

d=3a5+a4+5a3+4a24a+6a54a317a2+7a+14d = \frac{-3a^5 + a^4 + 5a^3 + 4a^2 - 4a + 6}{a^5 - 4a^3 - 17a^2 + 7a + 14}

These expressions give bb, cc, and dd as functions of aa. To determine the specific value of aa, we need to further analyze or simplify these expressions. Would you like me to continue simplifying or solving for aa numerically, or provide more details about the process?


Here are 5 questions to consider:

  1. What role does substitution play in solving non-linear systems like this?
  2. How can the denominator a54a317a2+7a+14a^5 - 4a^3 - 17a^2 + 7a + 14 impact the solution for aa?
  3. Why might factoring be useful when working with these kinds of polynomial expressions?
  4. How can we verify if these solutions are valid for specific values of aa?
  5. What happens if we try specific integer values for aa in these expressions?

Tip: When solving complex systems, symmetry or special values (like zero or small integers) can sometimes simplify the problem.

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Math Problem Analysis

Mathematical Concepts

Algebra
Systems of Equations
Polynomial Equations
Substitution Method

Formulas

Substitution method: Solving one equation for a variable and substituting into other equations
Polynomial equations of degree n: ax^n + bx^(n-1) + ... = 0

Theorems

Substitution Method: Solving systems of equations
Factoring Theorem: Simplifying polynomial equations

Suitable Grade Level

College Level (Advanced Algebra or Pre-Calculus)