Math Problem Statement

solve 2a + c + d = 18 a + 2c + d = 16 2a + b + c + d = 24

Solution

We have the system of equations:

1.2a+c+d=181. \quad 2a + c + d = 18 2.a+2c+d=162. \quad a + 2c + d = 16 3.2a+b+c+d=243. \quad 2a + b + c + d = 24

We need to solve for aa, bb, cc, and dd.

Step 1: Subtract the first equation from the second

(a+2c+d)(2a+c+d)=1618(a + 2c + d) - (2a + c + d) = 16 - 18 Simplifying: a+2c+d2acd=2a + 2c + d - 2a - c - d = -2 a+c=2a=c+2-a + c = -2 \quad \Rightarrow \quad a = c + 2

Step 2: Substitute a=c+2a = c + 2 into the first and third equations

Substitute into Equation 1: 2(c+2)+c+d=182(c + 2) + c + d = 18 2c+4+c+d=183c+d=14(Equation 4)2c + 4 + c + d = 18 \quad \Rightarrow \quad 3c + d = 14 \quad \text{(Equation 4)}

Substitute into Equation 3: 2(c+2)+b+c+d=242(c + 2) + b + c + d = 24 2c+4+b+c+d=243c+b+d=20(Equation 5)2c + 4 + b + c + d = 24 \quad \Rightarrow \quad 3c + b + d = 20 \quad \text{(Equation 5)}

Step 3: Solve Equations 4 and 5

We now have the system: 4.3c+d=144. \quad 3c + d = 14 5.3c+b+d=205. \quad 3c + b + d = 20

Subtract Equation 4 from Equation 5: (3c+b+d)(3c+d)=2014(3c + b + d) - (3c + d) = 20 - 14 Simplifying: b=6b = 6

Step 4: Find dd and cc

Now, substitute b=6b = 6 into Equation 5: 3c+6+d=203c+d=143c + 6 + d = 20 \quad \Rightarrow \quad 3c + d = 14 This is exactly the same as Equation 4, so it confirms that d=143cd = 14 - 3c.

Step 5: Substitute d=143cd = 14 - 3c into Equation 4

Substitute into 3c+d=143c + d = 14: 3c+(143c)=143c + (14 - 3c) = 14 Simplifying: 3c+143c=1414=143c + 14 - 3c = 14 \quad \Rightarrow \quad 14 = 14 This is always true, so it confirms that the value of cc can be any number.

Step 6: Final Answer

Thus, a=c+2a = c + 2, b=6b = 6, and d=143cd = 14 - 3c.

Since cc can be any number, the system has infinitely many solutions where aa, bb, and dd depend on cc.

Solution:

a=c+2,b=6,d=143c,c is freea = c + 2, \quad b = 6, \quad d = 14 - 3c, \quad c \text{ is free}


Would you like to see how this works with a specific value for cc, or do you have any further questions?

Here are some related questions for you:

  1. How would you solve a system of equations with three variables and one equation missing?
  2. What is the concept of free variables in a system of equations?
  3. How can you represent solutions graphically in a 3D coordinate system?
  4. What methods can be used for solving non-linear systems of equations?
  5. How can you check the correctness of a solution to a system of equations?

Tip: When solving systems with multiple variables, keep an eye out for any dependent or free variables, as they often indicate infinite solutions!

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Math Problem Analysis

Mathematical Concepts

Algebra
Systems of Linear Equations
Free Variables

Formulas

Substitution Method
Elimination Method
Linear Equation Representation: Ax + By + Cz = D

Theorems

Consistency and Dependency in Linear Systems

Suitable Grade Level

Grades 9-12