Given,
The equation is
[tex]\begin{gathered} 5x+3y=1.............(1) \\ 3x+4y=-6............(2) \end{gathered}[/tex]
To find: Identify the systems with no solution and systems with infinitely many solutions.
Solutions: We will use the substitution method. After moving 3y to the right, we get:
[tex]\begin{gathered} 3y=1-5x \\ y=\frac{1-5x}{3} \\ y=\frac{1}{3}-\frac{5x}{3} \end{gathered}[/tex]
Substitute this in the second equation.
[tex]\begin{gathered} 3x+4y=-6 \\ 3x+4\times(\frac{1}{3}-\frac{5x}{3})=-6 \\ 3x+\frac{4}{3}-\frac{20x}{3}=-6 \\ \frac{9x-20x}{3}=-6-\frac{4}{3} \end{gathered}[/tex]
Further solved as,
[tex]\begin{gathered} \frac{-11x}{3}=\frac{-18-4}{3} \\ \frac{-11x}{3}=\frac{-22}{3} \\ x=2 \end{gathered}[/tex]
Put the value of x in equation (1)
[tex]\begin{gathered} 5\times2+3y=1 \\ 3y=1-10 \\ 3y=-9 \\ y=-3 \end{gathered}[/tex]
Thus, the value of x and y is (2,-3)
