Math Problem Statement

a+f+e = 0 -5a+e = -7 -4f-e = 4 Find the values of a, f, e

Solution

We have the system of equations:

  1. a+f+e=0a + f + e = 0
  2. 5a+e=7-5a + e = -7
  3. 4fe=4-4f - e = 4

Step 1: Express ee in Terms of aa

From Equation (2):
e=5a7e = 5a - 7

Step 2: Express ee in Terms of ff

From Equation (3):
e=4f4e = -4f - 4

Step 3: Equating the Expressions for ee

Since both expressions represent ee, we set them equal to each other:
5a7=4f45a - 7 = -4f - 4 Rearrange:
5a+4f=35a + 4f = 3

Step 4: Solve for aa and ff

Using Equation (1):
a+f+e=0a + f + e = 0 Substituting e=5a7e = 5a - 7:
a+f+(5a7)=0a + f + (5a - 7) = 0 6a+f=76a + f = 7

Now, we solve the system:

  1. 6a+f=76a + f = 7
  2. 5a+4f=35a + 4f = 3

Using the first equation, solve for ff:
f=76af = 7 - 6a

Substituting into the second equation:
5a+4(76a)=35a + 4(7 - 6a) = 3 5a+2824a=35a + 28 - 24a = 3 19a+28=3-19a + 28 = 3 19a=25-19a = -25 a=2519a = \frac{25}{19}

Step 5: Find ff and ee

f=76(2519)f = 7 - 6\left(\frac{25}{19}\right) f=715019f = 7 - \frac{150}{19} f=1331915019f = \frac{133}{19} - \frac{150}{19} f=1719f = \frac{-17}{19}

e=5(2519)7e = 5\left(\frac{25}{19}\right) - 7 e=1251913319e = \frac{125}{19} - \frac{133}{19} e=819e = \frac{-8}{19}

Final Answer:

a=2519,f=1719,e=819a = \frac{25}{19}, \quad f = \frac{-17}{19}, \quad e = \frac{-8}{19}

Would you like a detailed verification or explanation?

Related Questions:

  1. How can we verify the solution using substitution?
  2. What happens if we change one of the equations slightly?
  3. Can this system be solved using matrices?
  4. How does elimination compare to substitution in solving this?
  5. What are real-world applications of solving such systems?

Tip: Always double-check your solutions by substituting them back into the original equations!

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Math Problem Analysis

Mathematical Concepts

Systems of Equations
Algebra
Substitution Method
Linear Equations

Formulas

a + f + e = 0
-5a + e = -7
-4f - e = 4

Theorems

Substitution method for solving systems of linear equations

Suitable Grade Level

Grades 9-11