Write a system of linear equations in which (3,-5) is a solution of equations 1 but not a solution of equation 2 and (-1,7) is a solution of the system

Y+5=m(x-3), or y=mx-3m-5, satisfies Equation 1. But Equation 1 and Equation 2 are satisfied when y-7=n(x+1).
(-1,7) is graphically represented by the intersection of 2 lines. So Equation 1 must pass through (-1,7) and (3,-5). The slope is (-5-7)/(3-(-1))=-12/4=-3=m, so y=-3x+4.

Equation 2 is y=x+8 when n=1 (arbitrary choice). Note that m is not equal to n.

A typical example of the required system is Eqn 1: y=-3x+4 and Eqn 2: y=x+8.

6xy and -16xy000000000