The graph shows one of the linear equations for a system of equations. Which equation represents the second linear equation for the system of equations that has the solution which corresponds to a point at (16/5,14/5)?A. 2x - 3y = 2B. 2x + 3y = -2C. x/2 +2y = 4D. x/2 + 2y = -4

Answer with explanation:Equation of line shown in the graph is [tex]\frac{x}{1} +\frac{y}{0.7}=1\\\\ \frac{x}{1} +\frac{10 y}{7}=1\\\\ 7 x + 10 y=7\text{using intercept form of line}, \frac{x}{a}+\frac{y}{b}=1[/tex]Where, a and b are intercept cut by line on x axis and y axis.  Now, we have to find equation of line which passes through, [tex](\frac{16}{5},\frac{14}{5})[/tex]Substituting the value of , x and y in equation of all the lines , from 1 to 4.  [tex]A. LHS=2 \times \frac{16}{5} +3 \times \frac{14}{5}\\\\=\frac{32}{5}-\frac{42}{5}\\\\=\frac{-10}{5}\\\\=-2,\neq RHS\\\\ B. LHS=2 \times \frac{16}{5} +3 \times \frac{14}{5}\\\\=\frac{32}{5}+\frac{42}{5}\\\\=\frac{74}{5}\neq RHS\\\\ C.LHS=\frac{16}{5}\times \frac{1}{2}+2 \times \frac{14}{5}\\\\=\frac{8}{5}+\frac{28}{5}\\\\=\frac{36}{5}\neq RHS\\\\D.LHS= =\frac{16}{5}\times \frac{1}{2}+2 \times \frac{14}{5}\\\\=\frac{8}{5}+\frac{28}{5}\\\\=\frac{36}{5}\neq RHS[/tex]→→ Point of intersection of two lines,that is one in graph and other in option has not point of intersection equal to  [tex](\frac{16}{5},\frac{14}{5})[/tex]. None of the option