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hichkok12 [17]
4 years ago
7

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
Mathematics
1 answer:
AnnZ [28]4 years ago
7 0
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.
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