# How do you determine the solution in terms of a System of Linear Equations for -2x + 3y =5 and ax - y = 1?

Oct 7, 2015

This system has:

• No solution if $a = \frac{2}{3}$
• One solution : $\left\{\begin{matrix}x = \frac{- 8}{2 - 3 a} \\ y = \frac{- 2 - 5 a}{2 - 3 a}\end{matrix}\right.$ if $a \ne \frac{2}{3}$

#### Explanation:

To find the connection between value of a parameter $a$ and the number of solutions of a linear system you can use the Cramer's Rule.

It can be written as follows:

Let there be a system of 2 linear equations:

$\left\{\begin{matrix}{a}_{1} x + {b}_{1} y = {c}_{1} \\ {a}_{2} x + {b}_{2} y = {c}_{2}\end{matrix}\right.$

Let

$W = {a}_{1} \cdot {b}_{2} - {b}_{1} \cdot {a}_{2}$

${W}_{x} = {c}_{1} {b}_{2} - {b}_{1} {c}_{2}$

${W}_{y} = {a}_{1} {c}_{2} - {c}_{1} {a}_{2}$

Then the system has:

• One solution $\left\{\begin{matrix}x = {W}_{x} / W \\ y = {W}_{y} / W\end{matrix}\right. \iff W \ne 0$
• No solutions $\iff W = 0$ and ${W}_{x} \ne 0$ or ${W}_{y} \ne 0$
• Infinitely many solutions $\iff W = 0$ and ${W}_{x} = 0$ and ${W}_{y} = 0$

This rule can be expanded for any system of $n$ equations with $n$ variables, then you have $W$ and for every variable ${x}_{i}$ you have a corresponding determinant ${W}_{{x}_{i}}$, and the rule says that:

1. If $W \ne 0$ system has exactly one solution: ${x}_{i} = \frac{{W}_{{x}_{i}}}{W}$ for $1 \le i \le n$

2. If $W = 0$ and any of ${W}_{{x}_{i}}$ is not zero, then system has no solutions

3. If $W = 0$ and all ${W}_{{x}_{i}}$ are zeros, then the system has infinitely many solutions.