\begin{tabular}{l} EXFRCISF \( 1.12: \) Substitution \\ \( 1 . \quad \) If \( a=2 \) and \( b=10 \) ealeulate the value of \( a \), \\ \( 1.1 \quad a+b=c \) \\ \( 1.2 \quad c=a b \) \\ \( 1.3 \quad \frac{c}{a}=b+4 \) \\ \hline \( 1.4 \quad a+b+c=b-2 a \) \\ \hline\end{tabular}

Certainly! Let's solve the given system of equations using the substitution method. Here's the step-by-step process. ## Problem Statement Given: 1. \( a = 2 \) and \( b = 10 \) (However, we'll later verify if these values are consistent with the system) 2. \( a + b = c \) \quad (Equation 1.1) 3. \( c = a \cdot b \) \quad (Equation 1.2) 4. \( \frac{c}{a} = b + 4 \) \quad (Equation 1.3) 5. \( a + b + c = b - 2a \) \quad (Equation 1.4) **Note:** There seems to be a potential inconsistency with the initial values \( a = 2 \) and \( b = 10 \). We'll proceed to solve the system to check for consistency. ## Step-by-Step Solution ### **1. From Equation 1.1: \( c = a + b \)** We already have: \[ c = a + b \] ### **2. From Equation 1.2: \( c = a \cdot b \)** Setting the two expressions for \( c \) equal to each other: \[ a + b = a \cdot b \] \[ \Rightarrow a \cdot b - a - b = 0 \] \[ \Rightarrow (a - 1)(b - 1) = 1 \] *(This factors the equation for easier substitution later.)* ### **3. From Equation 1.3: \( \frac{c}{a} = b + 4 \)** Substitute \( c = a + b \) from Equation 1.1: \[ \frac{a + b}{a} = b + 4 \] \[ \Rightarrow 1 + \frac{b}{a} = b + 4 \] \[ \Rightarrow \frac{b}{a} = b + 3 \] \[ \Rightarrow b = a(b + 3) \] \[ \Rightarrow b = a \cdot b + 3a \] \[ \Rightarrow b - a \cdot b = 3a \] \[ \Rightarrow b(1 - a) = 3a \] \[ \Rightarrow b = \frac{3a}{1 - a} \quad (Equation 3) \] ### **4. From Equation 1.4: \( a + b + c = b - 2a \)** Substitute \( c = a + b \) from Equation 1.1: \[ a + b + (a + b) = b - 2a \] \[ \Rightarrow 2a + 2b = b - 2a \] \[ \Rightarrow 4a + b = 0 \] \[ \Rightarrow b = -4a \quad (Equation 4) \] ### **5. Substitute Equation 4 into Equation 3** From Equation 3: \[ b = \frac{3a}{1 - a} \] From Equation 4: \[ b = -4a \] Set them equal: \[ -4a = \frac{3a}{1 - a} \] \[ \Rightarrow -4a(1 - a) = 3a \] \[ \Rightarrow -4a + 4a^2 = 3a \] \[ \Rightarrow 4a^2 - 7a = 0 \] \[ \Rightarrow a(4a - 7) = 0 \] **Solutions:** 1. \( a = 0 \) 2. \( 4a - 7 = 0 \Rightarrow a = \frac{7}{4} \) **Checking \( a = 0 \):** - From Equation 4: \( b = -4 \times 0 = 0 \) - From Equation 1.1: \( c = 0 + 0 = 0 \) - Check Equation 1.3: \( \frac{0}{0} \) is undefined. **Thus, \( a = 0 \) is not a valid solution.** **Valid Solution:** \[ a = \frac{7}{4} \] ### **6. Find \( b \) and \( c \) Using \( a = \frac{7}{4} \)** From Equation 4: \[ b = -4a = -4 \times \frac{7}{4} = -7 \] From Equation 1.1: \[ c = a + b = \frac{7}{4} + (-7) = \frac{7}{4} - \frac{28}{4} = -\frac{21}{4} \] ### **7. Verification** - **Equation 1.2:** \( c = a \cdot b \) \[ -\frac{21}{4} = \frac{7}{4} \times (-7) \] \[ -\frac{21}{4} = -\frac{49}{4} \] **Mismatch!** **Oops! There's an inconsistency here. Let's re-examine our steps.** ### **Re-evaluating Equation 1.2** Given \( a = \frac{7}{4} \) and \( b = -7 \): \[ c = a \cdot b = \frac{7}{4} \times (-7) = -\frac{49}{4} \] However, from Equation 1.1: \[ c = a + b = \frac{7}{4} - 7 = -\frac{21}{4} \] **There's a contradiction:** \[ -\frac{49}{4} \neq -\frac{21}{4} \] ### **Conclusion** The system of equations provided is **inconsistent** because it leads to contradictory values for \( c \). This means there is **no solution** that satisfies all the equations simultaneously with the given conditions. **Possible Reasons for Inconsistency:** - **Typographical Errors:** There might be mistakes in the equations provided. - **Over-Determined System:** Too many equations for the number of variables, leading to no possible solution. **Recommendation:** - **Review the Problem Statement:** Ensure that all equations are correctly stated. - **Check for Additional Constraints:** There might be missing information or constraints that can resolve the inconsistency.