Exercise 1 ( 5 points): Let \( a, b \) be real numbers. Consider the following system of equations: \[ \begin{array}{l}X+Y+2 Z=a \\ \qquad 2 X+2 Y+3 Z=b \\ \qquad 3 X+3 Y+4 Z=a+b\end{array} \] (1) Determine all possible values of \( a, b \) for which the above system has a solution. When the system has a solution, describe all solutions in terms of \( a \) and \( b \). (2) Are there any real numbers \( a, b \) for which the system of equations above has exactly one solution?

Let's analyze the given system of equations. We have: 1. \( X + Y + 2Z = a \) — (Equation 1) 2. \( 2X + 2Y + 3Z = b \) — (Equation 2) 3. \( 3X + 3Y + 4Z = a + b \) — (Equation 3) ### Step (1): Finding Conditions for Solutions To find the solutions, we first express everything in terms of a suitable variable. Let's manipulate the equations: 1. From Equation 1, we get \( Z = \frac{a - X - Y}{2} \). 2. Substitute \( Z \) into Equation 2: \[ 2X + 2Y + 3\left(\frac{a - X - Y}{2}\right) = b \] Multiplying through by 2 to eliminate the fraction gives: \[ 4X + 4Y + 3(a - X - Y) = 2b \] Which simplifies to: \[ 4X + 4Y + 3a - 3X - 3Y = 2b \quad \Rightarrow \quad X + Y + 3a = 2b \] Thus, we derive: \[ X + Y = 2b - 3a \quad \text{— (Equation 4)} \] 3. Substitute \( X + Y = 2b - 3a \) into Equation 3: Now we have: \[ 3(2b - 3a) + 4Z = a + b \] Substituting for \( Z \): \[ 3(2b - 3a) + 4\left(\frac{a - (2b - 3a)}{2}\right) = a + b \] This simplifies to: \[ 6b - 9a + 2(a + 3b - 2b) = a + b \] Which leads to: \[ 6b - 9a + a + 3b - 2b = a + b \] And finally: \[ 7b - 8a = a + b \quad \Rightarrow \quad 6b - 9a = 0 \quad \Rightarrow \quad 2b = 3a \] Thus, for the system to have solutions, the condition on \( a \) and \( b \) is: \[ b = \frac{3}{2}a \] ### Describing Solutions Using the relation \( b = \frac{3}{2}a \), we substitute back into previously derived equations: 1. From \( X + Y = 2\left(\frac{3}{2}a\right) - 3a = 3a - 3a = 0 \), we find: \[ X + Y = 0 \implies Y = -X \] 2. We can then substitute \( Y = -X \) into the first equation: \[ X - X + 2Z = a \implies 2Z = a \implies Z = \frac{a}{2} \] The solutions can be stated as: \[ (X, Y, Z) = (t, -t, \frac{a}{2}), \, \text{where} \, t \text{ is any real number.} \] ### Step (2): Unique Solution Condition For the system to have a unique solution, we would need the equations to not be dependent. This means we would want an inconsistency among the equations that defines a unique set of values for \( X, Y, Z \). Looking back at our relations, since we derived \( b = \frac{3}{2}a \) as a necessary condition for any solution, there's no point at which one particular value of \( a \) yields exactly one solution. Instead, the system forms a plane in three-dimensional space, suggesting infinitely many solutions given a proper pair \( (a, b) \). In conclusion, no values of \( a, b \) will lead to exactly one solution. The result establishes parameters wherein there can either be no solutions or infinitely many solutions (defined along the lines of our parameters).