Solution of a linear system

Question

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.

Answer 1

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 $x=4$.

Answer 2

Note that the system is given in matrix form by $$ \begin{bmatrix} 5/4 & -2/3 \\ 1/4 & 5/3 \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}3\\6\end{bmatrix}, $$ and $$ \det\begin{bmatrix} 5/4 & -2/3 \\ 1/4 & 5/3 \end{bmatrix} = \frac{5}{4}\cdot\frac{5}{3} + \frac{2}{3}\cdot \frac{1}{4} = \frac{9}{4} \not= 0, $$ so that the matrix is invertible.

Answer 3

The algorithmic way: Use row operations on the augmented matrix, until is left-hand side is the $2{\times}2$ unit matrix. The solution is the last column: \begin{align} &\begin{bmatrix} \begin{array}{rr|r} \frac 54&-\frac23 &3 \\ \frac14&\frac53& 6 \end{array} \end{bmatrix} \rightsquigarrow \begin{bmatrix} \begin{array}{rr|r} 1&-\frac73 &-3 \\ \frac14&\frac53& 6 \end{array} \end{bmatrix}\rightsquigarrow \begin{bmatrix} \begin{array}{rr|r} 1&-\frac73 &-3 \\ 0&\frac94& \frac{27}4 \end{array} \end{bmatrix}\\[1ex]\rightsquigarrow &\begin{bmatrix} \begin{array}{rr|r} 1&-\frac73 &-3 \\ 0& 1& 3 \end{array} \end{bmatrix}\rightsquigarrow \begin{bmatrix} \begin{array}{rr|r} 1 &0 & \color{red}4 \\ 0& 1& \color{red}3 \end{array} \end{bmatrix} \end{align}