Answer:
D) Infinitely many
Step-by-step explanation:
We have the following linear equations system :
[tex]\left \{ {{y=(\frac{5}{2}})x+2 \atop {2y=5x+4}} \right.[/tex]
Given that this is a linear equations system :
It can have no solution
It can have only one solution
If it have more than one solution, it have infinite solutions.
One way to solve it is to put the equations in a matrix and work with it :
[tex]y=(\frac{5}{2})x+2[/tex]
[tex]y-(\frac{5}{2})x=2[/tex]
[tex]-(\frac{5}{2})x+y=2[/tex] (I)
[tex]2y=5x+4[/tex]
[tex]2y-5x=4[/tex]
[tex]-5x+2y=4[/tex] (II)
Putting (I) and (II) into a matrix :
[tex]\left[\begin{array}{ccc}(-\frac{5}{2})&1&2\\-5&2&4\\\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}1&(-\frac{2}{5})&(-\frac{4}{5})\\-5&2&4\\\end{array}\right][/tex]
[tex]\left[\begin{array}{ccc}1&(-\frac{2}{5})&(-\frac{4}{5})\\0&0&0\\\end{array}\right][/tex]
The new equivalent system is
[tex]x-(\frac{2}{5})y=-\frac{4}{5}[/tex]
We can write ''x'' in terms of ''y'' :
[tex]x=(\frac{2}{5})y-\frac{4}{5}[/tex]
The solution will be the points with the form [tex]\left[\begin{array}{c}x&y\end{array}\right][/tex] in IR2 ⇒
[tex]\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{c}(\frac{2}{5})y-\frac{4}{5} &y\end{array}\right]=y\left[\begin{array}{c}\frac{2}{5}&1\end{array}\right]+\left[\begin{array}{c}-\frac{4}{5}&0\end{array}\right][/tex]
⇒
The solutions are the points on the line :
[tex]IL=c\left[\begin{array}{c}\frac{2}{5}&1\end{array}\right]+\left[\begin{array}{c}-\frac{4}{5}&0\end{array}\right][/tex]
c ∈ IR
Given that the quantity of points on a line are infinite ⇒
The answer is D) infinitely many
