Animated Solution for Mathematics - Matrices and Determinants: If the system of linear equations $2x-3y=\gamma +5$, $\alpha x+5y=\beta +1$, where $\alpha ,\beta ,\gamma \in \mathbb{R}$ has infinitely many solutions, then the value of $|9\alpha +3\beta +5\gamma |$ is equal to

Welcome, fellow traveler on the path to JEE excellence! Today, we are not just solving a system of linear equations; we are exploring the geometric harmony of lines.

Imagine you are standing in a 2D plane, looking at two lines, $L_1$ and $L_2$. Usually, lines intersect at a single point, or they run parallel, never meeting. But what if they were destined to be the same?

What if they occupied the exact same space, sharing every single point? That is the beautiful, rare condition of infinitely many solutions.

When we say a system of linear equations has infinitely many solutions, we are describing a scenario where the two equations are essentially describing the same line. Geometrically, if you were to plot $2x-3y=\gamma +5$ and $\alpha x+5y=\beta +1$, they would overlap perfectly.

They are coincident. This isn't just a coincidence of terminology; it is a fundamental geometric reality.

If they are the same line, their slopes must be identical, and their intercepts must be identical. This leads us to the powerful algebraic tool we call the ratio condition.

For two lines $a_{1}x+b_{1}y=c_1$ and $a_{2}x+b_{2}y=c_2$ to be coincident, their coefficients must be proportional. We express this as the elegant ratio:

Let's apply this to our specific system. We have $a_1=2$, $b_1=-3$, and $c_1=\gamma +5$. For the second line, we have $a_2=\alpha$, $b_2=5$, and $c_2=\beta +1$.

Substituting these into our ratio condition, we get:

This is the bridge that connects our geometry to our algebra. Now, we just need to walk across it.

First, let's isolate $\alpha$. By equating the first two ratios, $\frac{2}{\alpha }=\frac{-3}{5}$, we can cross-multiply to find $-3\alpha =10$, which gives us $\alpha =-\frac{10}{3}$.

Now, look at our target expression: $|9\alpha +3\beta +5\gamma |$. We already have $\alpha$, so let's calculate $9\alpha$ immediately:

That is one piece of the puzzle solved!

Next, we need to find the value of $3\beta +5\gamma$. We use the remaining part of our ratio condition:

Cross-multiplying gives us $-3(\beta +1)=5(\gamma +5)$. Expanding this, we get $-3\beta -3=5\gamma +25$.

Rearranging the terms to group the variables, we move $-3\beta$ to the right and $25$ to the left: $-3-25=3\beta +5\gamma$. This simplifies beautifully to $3\beta +5\gamma =-28$.

We have all the components! Our target expression is $|9\alpha +3\beta +5\gamma |$. Substituting our findings, we get $|-30+(-28)|$.

This simplifies to $|-58|$. The modulus function, that wonderful mathematical filter, strips away the negative sign, leaving us with the final, elegant answer of 58.

It is a testament to the beauty of mathematics that such a complex-looking system collapses into such a clean, positive integer. Keep practicing, keep visualizing, and remember: every equation tells a story!