Question

Solve the system: \[ x + 2y + 2z = 1 \] \[ 2x + 3y + 2z = 2 \] \[ ax + 5y + bz = b \] Find $a + b$ for infinite solutions.

Hint:

For infinite solutions, the third equation must be a linear combination of the first two. Match coefficients carefully including constant term.

Answer 1

Correct Answer: 6.66

Step 1: Condition for infinite solutions.
For infinite solutions, the third equation must be a linear combination of the first two equations.
Let \[ \lambda (x + 2y + 2z = 1) + \mu (2x + 3y + 2z = 2) = ax + 5y + bz = b. \]
Step 2: Equate coefficients.
Coefficient of $x$:
\[ \lambda + 2\mu = a. \]
Coefficient of $y$:
\[ 2\lambda + 3\mu = 5. \]
Coefficient of $z$:
\[ 2\lambda + 2\mu = b. \]
Constant term:
\[ \lambda(1) + \mu(2) = b. \]
Step 3: Solve for $\lambda, \mu$.
From $y$-coefficient: \[ 2\lambda + 3\mu = 5. \]
From constant equation: \[ \lambda + 2\mu = b. \]
Also from $z$-coefficient: \[ 2\lambda + 2\mu = b. \]
Now compare last two equations:
Multiply $\lambda + 2\mu = b$ by 2: \[ 2\lambda + 4\mu = 2b. \]
But we also have: \[ 2\lambda + 2\mu = b. \]
Subtracting: \[ (2\lambda + 4\mu) - (2\lambda + 2\mu) = 2b - b. \]
\[ 2\mu = b. \]
\[ b = 2\mu. \]
Substitute in $2\lambda + 2\mu = b$:
\[ 2\lambda + 2\mu = 2\mu. \]
\[ 2\lambda = 0. \]
\[ \lambda = 0. \]
Step 4: Find $\mu$.
From \[ 2\lambda + 3\mu = 5. \]
Since $\lambda=0$:
\[ 3\mu = 5. \]
\[ \mu = \frac{5}{3}. \]
Thus, \[ b = 2\mu = \frac{10}{3}. \]
Step 5: Find $a$.
\[ a = \lambda + 2\mu = 0 + 2\left(\frac{5}{3}\right) = \frac{10}{3}. \]
Step 6: Final Answer.
\[ a = \frac{10}{3}, \quad b = \frac{10}{3}. \]
\[ \boxed{ a + b = \frac{20}{3} }. \]