Which ordered pair is a solution to the given system of inequalities? y >2x+4 and y≤x+3

Answer 1

try some simple integers

x=0, but that won't satisfy both inequalities

x=1 won't work

x=-1, y>2 and < or = 2, another contradiction

x=-2 works

y>0 and y < or = 1

(x,y) = (-2,1) satisfies both inequalities

another approach is graph both inequalities, and see where in the graph both are satisfied. That's an area partly in the 2nd quadrant, but mostly in the 3rd quadrant, with an infinite number of solutions

Or algebraically see where the boundaries of the inequalities intersect. That's at (-1,2) which doesn't satisfy the inequalities, but is on the boundary of the area that does satisfy them.

2x+4 < x+3

subtract x+4

x < -1

and y < or = -1+3 or -2

All the ordered pairs that satisfy have x values less than -1

if the x value is less than -3, all the corresponding y values are also negative, the ordered pairs are all in quadrant III

such as (-4,-3)