Question

65. Consider the system of linear equations:
\[ 2x + 3y + 4z = 16 \]
\[ 4x + 4y + 5z = 26 \]
\[ ax + by + cz = r \]
For \(r = 5\) and \(a = 1\), the system of linear equations will have an infinite number of solutions if \(c = ?\)

• A. \(\dfrac{3}{2}\)
• B. 1
• C. \(\dfrac{1}{2}\)
• D. 0

Hint:

Solve the first two equations for x and y in terms of z, then require the third equation to hold for every value of z.

Answer 1

The Correct Option is C

Step 1: Understand what "infinite solutions" means here.
Three planes in x, y, z space meet in infinitely many points only when the third plane is not really a new, independent plane. It must be built out of the first two, so the third equation should equal \(\lambda\) times the first equation plus \(\mu\) times the second equation, for some numbers \(\lambda\) and \(\mu\).

Step 2: Write out that combination.
\[ \lambda(2x+3y+4z) + \mu(4x+4y+5z) = \lambda(16) + \mu(26) \]
Matching coefficients with \(ax+by+cz=r\):
\[ 2\lambda + 4\mu = a, \qquad 3\lambda+4\mu=b, \qquad 4\lambda+5\mu=c, \qquad 16\lambda+26\mu=r \]

Step 3: Substitute the given values a = 1, r = 5.
\[ 2\lambda+4\mu=1 \quad (i) \]
\[ 16\lambda+26\mu=5 \quad (ii) \]
From (i): \(\lambda = \dfrac{1-4\mu}{2}\). Substitute this into (ii):
\[ 16\left(\frac{1-4\mu}{2}\right)+26\mu=5 \]
\[ 8(1-4\mu)+26\mu=5 \]
\[ 8-32\mu+26\mu=5 \]
\[ 8-6\mu=5 \ \Rightarrow\ \mu=\frac{1}{2} \]
Then \(\lambda = \dfrac{1-4(1/2)}{2}=\dfrac{1-2}{2}=-\dfrac{1}{2}\).

Step 4: Find c.
\[ c = 4\lambda+5\mu = 4\left(-\frac{1}{2}\right)+5\left(\frac{1}{2}\right)=-2+\frac{5}{2}=\frac{1}{2} \]

Step 5: Why the other options are wrong.
Options 1 (3/2) and 2 (1) come from mixing up the roles of \(\lambda\) and \(\mu\), or from using only one of the a and r conditions instead of both together. Option 4 (0) would only be right if the third equation needed none of the second equation at all, but that does not fit \(a=1\) here (using only the first equation would need \(a=2\), not 1). Only \(c=\tfrac12\) is consistent with every given number in the problem.

Final Answer:
\(c=\dfrac{1}{2}\) makes the third equation exactly \(-\tfrac12\) times the first equation plus \(\tfrac12\) times the second one, so all three planes coincide along the same line, and the system has infinitely many solutions.
\[ \boxed{c=\frac{1}{2}} \]