Determine how many solutions the following system have?
$\begin{cases}~\dfrac{5}{4} x -\dfrac{2}{3} y – 3 = 0\\
~\dfrac{1}{4} x + \dfrac{5}{3} y – 6 = 0\end{cases}$
The correct answer is: "one solution".
How did they determine this?
When I apply the method of elimination to determine it, I come to the answer "infinitely many solutions".
Can someone explain me why the correct answer is "one solution"?
Thank you.
Best Answer
When we use the elimination method to find a solution, the equation we have after we eliminate one variable tell us the nature of the solution.
If there is infinitly many solutions, we will have an equation such as $$0x +0y+0=0$$ It mean that the two equations are the same and no mather what value we gave to $x$, there is a value $y$ such that $(x,y)$ satisfy both equation.
If there are no solution, we Will have an equation such as$$0x+0y+c=0 \qquad c\neq0$$ No mather what values we gave to $x$ and $y$, this equation will never be true.
After we eliminate a variable, if the coefficient of the other is different of $0$, then the solution is unique.
In your example $$\begin{cases}~\dfrac{5}{4} x -\dfrac{2}{3} y - 3 = 0\\ ~\dfrac{1}{4} x + \dfrac{5}{3} y - 6 = 0\end{cases}$$ Multiplying the second équation by $5$ $$\begin{cases}~\dfrac{5}{4} x -\dfrac{2}{3} y - 3 = 0\\ ~\dfrac{5}{4} x + \dfrac{25}{3} y - 30= 0\end{cases}$$ Substract both equations give $$0x-\frac{27}{3}y+27=0$$ Thus the solution is unique and $y=3$. By replacing this value in one of the equation, we find that $x=4$.