Answer:
The solution is (4, 1).
Step-by-step explanation:
Our system is
[tex]\left \{ {{2d-e=7} \atop {d+e=5}} \right.[/tex]
Since the coefficients of e are the same, we will add the two equations together to cancel it:
[tex]\left \{ {{2d-e=7} \atop {+(d+e=5)}} \right. \\\\3d=12[/tex]
Divide both sides by 3:
3d/3 = 12/3
d = 4
Substitute this into the second equation:
4+e=5
Subtract 4 from each side:
4+e-4=5-4
e = 1
This makes the point (4, 1).