Question

solve by the method of your choice. Identify systems with no solution and systems with infinitely many solutions, using set notation to express their solution sets

Answer 1

Given,

The equation is

`\begin{gathered} 5x+3y=1.............(1) \\ 3x+4y=-6............(2) \end{gathered}`

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:

`\begin{gathered} 3y=1-5x \\ y=(1-5x)/(3) \\ y=(1)/(3)-(5x)/(3) \end{gathered}`

Substitute this in the second equation.

`\begin{gathered} 3x+4y=-6 \\ 3x+4*((1)/(3)-(5x)/(3))=-6 \\ 3x+(4)/(3)-(20x)/(3)=-6 \\ (9x-20x)/(3)=-6-(4)/(3) \end{gathered}`

Further solved as,

`\begin{gathered} (-11x)/(3)=(-18-4)/(3) \\ (-11x)/(3)=(-22)/(3) \\ x=2 \end{gathered}`

Put the value of x in equation (1)

`\begin{gathered} 5*2+3y=1 \\ 3y=1-10 \\ 3y=-9 \\ y=-3 \end{gathered}`

Thus, the value of x and y is (2,-3)