Given the system of equations:
8x - 6y = 3
-3y + 4x = 4
To find how many times the lines will intersect. let's solve the system of equation using substitution method.
8x - 6y = 3 ........................................1
-3y + 4x = 4 ......................................2
From equation 1, make x the subject:
8x - 6y = 3
8x = 3 + 6y
[tex]\begin{gathered} x=\frac{3}{8}+\frac{6}{8}y \\ \\ x=\frac{3}{8}+\frac{3}{4}y \end{gathered}[/tex][tex]\text{Substitute (}\frac{3}{8}+\frac{3}{4}y)\text{ for x in equation 2}[/tex]We have:
[tex]\begin{gathered} -3y+4(\frac{3}{8}+\frac{3}{4}y)=4 \\ \\ -3y+\frac{3}{2}+3y=4 \\ \\ \end{gathered}[/tex]Multiply through by 2 to eliminate the fraction:
[tex]\begin{gathered} -3y(2)+\frac{3}{2}(2)+3y(2)=4(2) \\ \\ -6y+3+6y=8 \end{gathered}[/tex][tex]\begin{gathered} -6y+6y=8-3 \\ \\ 0=5 \end{gathered}[/tex]Since we have 0 = 5, it means the system of equations has no solution.
Therefore, the lines will not intersect, because this system has no solution.
ANSWER:
A. The lines will not intersect, because this system has no solution.
