Question 334522
I would if I had too :)

Just plug in the x values given and solve for y
{{{y = 2x +5}}}
x = 8
{{{y = 2*8 + 5}}}
{{{y = 16+5}}}
{{{y = 21}}}
(8,21)

x=-1
{{{y = 2*(-1) + 5}}}
{{{y = -2 + 5}}}
{{{y = 3}}}
(-1,3)

The last one gives you y and requires you solve for x
y = 1
{{{y = 2x +5}}}
{{{1 = 2x +5}}}
{{{-4 = 2x}}}
{{{-2 = x}}}
(-2,1)