Question 464·Hard·Algebra

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(k+2)x + (k-1)y = k\\[1.4em] 3x + \frac{k-1}{2}\,y = 2

In the system of equations above, $k$ is a real number. For which value of $k$ does the system have infinitely many solutions?

In a linear system, infinitely many solutions means the two equations draw the same line. Every term, including the constant, must match after one equation is scaled. Match the $y$ coefficients, then subtract to see whether the equations still restrict $x$ . If the $y$ terms disappear entirely, check that case separately before deciding the lines overlap.

Hints

Hint 1
For two equations to draw the same line, multiplying one entire equation by a number must make it match the other. Double the second equation to match its $y$ coefficient to the first. What happens to its constant?
Hint 2
Subtract equations with matching $y$ terms to eliminate $y$ . If the result forces one value of $x$ , ask whether either original equation also forces a value of $y$ .
Hint 3
A zero coefficient removes its variable. Check the value that makes both $y$ terms disappear: do the remaining equations agree about $x$ ?

Step-by-step

Match and subtract the equations
Step 1
Match the $y$ coefficients
The second equation's $y$ coefficient, the number multiplying $y$ , is half the first equation's. Multiply both sides of the second equation by $2$ :
$6x+(k-1)y=4.$
Step 2
Subtract to cancel $y$
Subtract the doubled equation from the first to eliminate the matching $y$ terms: $\begin{aligned}[(k+2)x+(k-1)y]-[6x+(k-1)y]&=k-4\\(k+2-6)x&=k-4\\(k-4)x&=k-4.\end{aligned}$ The remaining equation tells you whether the pair restricts $x$ .
Step 3
Rule out a fixed value of $x$
If $k\ne4$ , divide by $k-4$ : $x=1$ . If $k\ne1$ as well, the $y$ coefficient $k-1$ is nonzero. Substituting this fixed $x$ into either original equation then fixes $y$ , so there cannot be infinitely many solutions. The case $k=1$ needs a separate check.
Step 4
Check when $y$ disappears
At $k=1$ , both $y$ terms vanish. The original equations become $\begin{aligned}3x&=1\\3x&=2.\end{aligned}$ These cannot both hold, so this case has no solutions.
Step 5
Find and verify the overlapping lines
The remaining possibility makes the leftover restriction vanish. Set its coefficient to zero: $k-4=0$ . Add $4$ to both sides: $k=4$ . Type $k=4$ and both original equations into Desmos. They draw the same line; turning either graph off and on confirms that neither line moves. Every point on that line solves both equations, so the system has infinitely many solutions. Choice C.

Question 464·Hard·Algebra

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