Question

Given that a, b, and c are real numbers satisfying the equations:

\(2a - 5b + 11c = 0\)
\(11a + 10b - 2c = 5\)
Find the value of the expression \( (a^2 - b^2 + c^2) \).

• A. 5/11
• B. 2/13
• C. 1
• D. CBD (Cannot be determined)

Hint:

With 3 unknowns and only 2 equations, first check whether the target expression could still be pinned down: try substituting one convenient value (like c=0) to get one candidate value, then try another (like b=0); if the results differ, the expression cannot be determined without needing the full general algebra.

Answer 1

The Correct Option is D

Approach: With two equations and three unknowns, isolate the symmetric combination by squaring-and-adding so the cross term \(ac\) cancels. Then read off honestly whether the target is pinned down.

Step 1: Rearrange. From \(2a - 5b + 11c = 0\): \(2a + 11c = 5b\). From \(11a + 10b - 2c = 5\): \(11a - 2c = 5 - 10b\).

Step 2: Square both and add. The cross terms are \(+44ac\) and \(-44ac\), which cancel: \[ (2a+11c)^2 + (11a-2c)^2 = 125a^2 + 125c^2. \] On the right, \((5b)^2 + (5-10b)^2 = 25b^2 + 25 - 100b + 100b^2 = 125b^2 - 100b + 25.\)

Step 3: Equate and divide by \(25\): \(5a^2 + 5c^2 = 5b^2 - 4b + 1\), so \[ 5(a^2 - b^2 + c^2) = 1 - 4b. \]
Step 4 (the honest read): The right side still contains \(b\). Since the system has \(3\) unknowns and only \(2\) equations, \(b\) is free, so \(a^2 - b^2 + c^2\) is not fixed. Check: with \(b=0\), a valid solution \(a = \tfrac{11}{25},\, c = -\tfrac{2}{25}\) gives \(a^2 - b^2 + c^2 = \tfrac{1}{5}\); with \(b=-1\) it gives \(1\). Two different values are possible.

Answer: The expression cannot be determined \(\Rightarrow \boxed{\text{CBD (option 4)}}\). (Note: the stored key marks option 3 = 1; this appears to be an error and is flagged.)

Answer 2

Approach: Solve the linear system directly for \(a\) and \(c\) in terms of the free parameter \(b\) (via elimination), then see how the target expression depends on \(b\) explicitly \(-\) a closed-form route, rather than the squaring trick or plugging in numbers.

Write the system as \(2a+11c=5b\) and \(11a-2c=5-10b\). Solving this \(2\times2\) system for \(a,c\) gives \[ a=\frac{11-20b}{25}, \qquad c=\frac{15b-2}{25}. \]
Now compute \(a^2+c^2\) directly: \[ a^2+c^2=\frac{(11-20b)^2+(15b-2)^2}{625}=\frac{625b^2-500b+125}{625}=b^2-\frac{4}{5}b+\frac{1}{5}. \]
So \[ a^2-b^2+c^2=(a^2+c^2)-b^2=-\frac{4}{5}b+\frac{1}{5}=\frac{1-4b}{5}, \] a genuine linear function of the free parameter \(b\) \(-\) not a constant.

Since \(b\) is not pinned down by only two equations in three unknowns, the expression takes different values for different valid \((a,b,c)\), so the answer is \[ \boxed{\text{CBD} - \text{cannot be determined (option 4)}} \]

Answer 3

Concept:

• Instead of manipulating the two equations symbolically, plug in one convenient value for one of the variables at a time to generate two different valid triples $(a,b,c)$ satisfying both equations.
• If the target expression genuinely has a single fixed value, every valid triple must give the same number; finding even two triples that give different values is enough to prove the expression cannot be determined.

Step 1: Find one valid triple by setting $c=0$.
With $c=0$: $2a-5b=0 \Rightarrow a=2.5b$. Substitute into the second equation: $11(2.5b)+10b-0=5 \Rightarrow 37.5b=5 \Rightarrow b=\dfrac{2}{15}$, so $a=\dfrac{1}{3}$.
Compute the target: $a^2-b^2+c^2=\left(\dfrac13\right)^2-\left(\dfrac{2}{15}\right)^2+0=\dfrac{1}{9}-\dfrac{4}{225}=\dfrac{25}{225}-\dfrac{4}{225}=\dfrac{21}{225}=\dfrac{7}{75}$.

Step 2: Find a second valid triple by setting $b=0$.
With $b=0$: $2a+11c=0 \Rightarrow a=-5.5c$. Substitute into the second equation: $11(-5.5c)-2c=5 \Rightarrow -62.5c=5 \Rightarrow c=-\dfrac{2}{25}$, so $a=\dfrac{11}{25}$.
Compute the target: $a^2-b^2+c^2=\left(\dfrac{11}{25}\right)^2-0+\left(-\dfrac{2}{25}\right)^2=\dfrac{121+4}{625}=\dfrac{125}{625}=\dfrac15$.

Step 3: Compare the two results.
The first triple gives $\dfrac{7}{75}$, the second gives $\dfrac15=\dfrac{15}{75}$. These are different numbers, even though both triples fully satisfy the two given equations.

Step 4: Conclude.
Since two equally valid triples give two different values of $a^2-b^2+c^2$, the expression genuinely depends on which solution you pick, so its value cannot be pinned down from the given information.

Final Answer: The expression cannot be determined (CBD).