Question

The system of equations \(2x+y=5,\; x-3y=-1,\; 3x+4y=k\) is consistent when \(k\) is

• A. \(1\)
• B. \(2\)
• C. \(5\)
• D. \(10\)

Hint:

For consistency questions: \[ \boxed{ \text{Solve the independent equations first, then substitute into the remaining equation.} } \] If the obtained values satisfy the remaining equation, the system is consistent.

Answer 1

The Correct Option is D

Concept: A system of linear equations is said to be consistent if there exists at least one common solution satisfying all the equations. For three equations, \[ \begin{aligned} 2x+y&=5 \\ x-3y&=-1 \\ 3x+4y&=k \end{aligned} \] the first two equations determine the values of \(x\) and \(y\). The third equation will be consistent only if these values satisfy it.

Step 1: Solve the first two equations.
Given, \[ 2x+y=5 \] From this, \[ y=5-2x \] Substitute into \[ x-3y=-1 \] \[ x-3(5-2x)=-1 \] \[ x-15+6x=-1 \] \[ 7x=14 \] \[ \boxed{x=2} \] Now, \[ y=5-2(2)=1 \] Hence, \[ \boxed{(x,y)=(2,1)} \]

Step 2: Substitute into the third equation.
The third equation is \[ 3x+4y=k \] Substituting \(x=2,\;y=1\), \[ k=3(2)+4(1) \] \[ =6+4 \] \[ \boxed{k=10} \]

Step 3: Identify the correct option.
The value of \(k\) for which all three equations are satisfied is \[ \boxed{k=10} \] Therefore, \[ \boxed{Option (D) is correct \] \[ \boxed{k=10} \] Hence, the question contains an error in its answer key.