We are given the following system of equations:
[tex]\begin{gathered} y=\frac{3}{4}x+2,(1) \\ -6x+8y=16,(2) \end{gathered}[/tex]
We will solve for "y" in equation (2), first by adding "6x" to both sides:
[tex]8y=6x+16[/tex]
Now we will divide both sides by 8:
[tex]y=\frac{6x}{8}+\frac{16}{8}[/tex]
Simplifying we get:
[tex]y=\frac{3x}{4}+2[/tex]
The new equivalent system is:
[tex]\begin{gathered} y=\frac{3}{4}x+2,(1) \\ y=\frac{3}{4}x+2,(2) \end{gathered}[/tex]
This system of the form:
[tex]\begin{gathered} y=m_1x+b_1 \\ y=m_2x+b_2 \end{gathered}[/tex]
Where "m" represents the slopes of each equation and "b" represents the y-intercepts. From the given system we have:
[tex]\begin{gathered} m_1=\frac{3}{4} \\ m_2=\frac{3}{4} \\ b_1=2 \\ b_2=2 \end{gathered}[/tex]
Since the slopes are the same this means that the system has infinite solutions.
