We have to solve the following system of equations:
[tex]\begin{gathered} \frac{1}{2}x+\frac{3}{4}y=1 \\ 2x-3y=4 \end{gathered}[/tex]We have to graph the equations and, as they are written in standard form, we are going to calculate the intercepts for both.
We will write the equations in slope-intercept form.
For the first equation we have:
[tex]\begin{gathered} \frac{1}{2}x+\frac{3}{4}y=1 \\ \frac{3}{4}y=-\frac{1}{2}x+1 \\ y=\frac{4}{3}(-\frac{1}{2}x+1) \\ y=-\frac{2}{3}x+\frac{4}{3} \end{gathered}[/tex]For the second equation we have:
[tex]\begin{gathered} 2x-3y=4 \\ 2x-4=3y \\ y=\frac{1}{3}(2x-4) \\ y=\frac{2}{3}x-\frac{4}{3} \end{gathered}[/tex]Both slopes are different, what means that the lines are not parallel and will intersect, so we already know that the system is independent.
Using the slopes and the y-intercepts, we can graph the equations as:
The solution to the system is the intersection point which is (2,0).
Answer:
The system is independent and its solution is (x,y) = (2,0)
