Animated Solution for Mathematics - Matrices and Determinants: The number of values of $k$ for which the system of equations $(k+1)x+8y=4k$; $kx+(k+3)y=3k-1$ has infinitely many solutions is

Imagine you are standing before two lines on a graph. Usually, two lines either intersect at a single point or they run parallel, never meeting.

But there is a third, more elusive possibility: they could be the exact same line, resting perfectly on top of one another. This is the state of coincidence, and it is the key to unlocking the mystery of infinitely many solutions.

When we say a system of equations has infinitely many solutions, we are saying that every single point on the first line is also a point on the second. They are, for all intents and purposes, the same equation.

The Algebraic Soul

To translate this geometric reality into algebra, we look at the coefficients. For two equations $a_{1}x+b_{1}y=c_1$ and $a_{2}x+b_{2}y=c_2$, the lines are coincident if and only if their coefficients are proportional:

Our system is $(k+1)x+8y=4k$ and $kx+(k+3)y=3k-1$. By identifying our coefficients, we get $a_1=k+1$, $b_1=8$, $c_1=4k$ for the first line, and $a_2=k$, $b_2=k+3$, $c_2=3k-1$ for the second.

Plugging these into our ratio condition, we arrive at the triple equality:

Solving the System

We cannot solve all three ratios at once, so we break the problem down into pairs. First, let us tackle the pair:

Cross-multiplying gives us $(k+1)(k+3)=8k$. Expanding this, we get $k^2+4k+3=8k$, which simplifies beautifully to the quadratic $k^2-4k+3=0$.

Factoring this, we find $(k-1)(k-3)=0$, leading us to $k=1$ or $k=3$.

Next, we must ensure these values also satisfy the second pair:

Cross-multiplying here yields $8(3k-1)=4k(k+3)$, which simplifies to $24k-8=4k^2+12k$. Rearranging, we get $4k^2-12k+8=0$.

Dividing by 4, we find $k^2-3k+2=0$, which factors into $(k-1)(k-2)=0$. This gives us $k=1$ or $k=2$.

We have two sets of potential values for $k$: from the first pair, $k\in \{1,3\}$, and from the second pair, $k\in \{1,2\}$. For the lines to be truly coincident, $k$ must satisfy both conditions simultaneously.

The intersection of these two sets is simply $k=1$. The question asks for the number of values of $k$.

Since $k=1$ is the only value that works, there is exactly one such value. The final answer is 1.