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: Find the line where the first two planes meet.
Solve the first two equations for x and y in terms of z. From $2x+3y+4z=16$: $2x+3y=16-4z$. From $4x+4y+5z=26$: $4x+4y=26-5z$.


Step 2: Eliminate x.
Multiply the first reduced equation by 2: $4x+6y=32-8z$. Subtract the second reduced equation from this:
\[ (4x+6y)-(4x+4y) = (32-8z)-(26-5z) \]
\[ 2y = 6-3z \ \Rightarrow\ y = 3-1.5z \]


Step 3: Solve for x.
Put $y=3-1.5z$ back into $2x+3y=16-4z$:
\[ 2x + 3(3-1.5z) = 16-4z \]
\[ 2x + 9 - 4.5z = 16-4z \]
\[ 2x = 7+0.5z \ \Rightarrow\ x = 3.5+0.25z \]
So every point where the first two planes meet has this form, whatever value z takes.


Step 4: Force the third plane to contain this whole line.
For infinitely many solutions, the third equation $x+by+cz=5$ (using $a=1$, $r=5$) must hold for every z on that line, not just for one point. Substitute:
\[ (3.5+0.25z) + b(3-1.5z) + cz = 5 \]
\[ (3.5+3b) + z(0.25-1.5b+c) = 5 \]
This can only be true for every z if the constant part equals 5 and the part multiplying z equals 0.


Step 5: Solve the two conditions.
\[ 3.5+3b=5 \ \Rightarrow\ b=\frac{1}{2} \]
\[ 0.25-1.5b+c=0 \ \Rightarrow\ 0.25-0.75+c=0 \ \Rightarrow\ c=\frac{1}{2} \]
This matches the linear-combination method exactly: c = 1/2.


Step 6: Check the answer against the original equations.
With $b=\frac{1}{2}$ and $c=\frac{1}{2}$, the third equation is $x+\frac{1}{2}y+\frac{1}{2}z=5$, which is the same as $2x+y+z=10$. Multiplying this by 2 gives $4x+2y+2z=20$, and it is worth checking that this line of reasoning is consistent with both original equations along the shared solution line found in Step 3: taking, say, $z=0$ gives $x=3.5$, $y=3$ from Steps 2 and 3, and indeed $2(3.5)+3(3)+4(0)=7+9=16$ (matches equation 1) and $4(3.5)+4(3)+5(0)=14+12=26$ (matches equation 2) and $1(3.5)+\frac{1}{2}(3)+\frac{1}{2}(0)=3.5+1.5=5$ (matches the third equation with $r=5$). All three equations agree at this point, and since the same holds for every value of z, the three planes share the entire line, confirming infinitely many solutions.


Step 7: Why the other choices fail.
If $c=\frac{3}{2}$, $c=1$ or $c=0$ instead, the coefficient of z in Step 4 would not cancel out, so the third equation would only match the shared line at one particular value of z rather than for every z; that gives a single unique solution (or no solution at all), not infinitely many. Only $c=\frac{1}{2}$ makes the z-coefficient vanish completely. \[ \boxed{c=\frac{1}{2}} \]