Question 180582
y=4/5x..........eq 1
y=-2x+3.........eq 2
:
take the y value in eq 1 and plug it into eq 2
:
(4/5)x=-2x+3
:
4x=-10x+15.......muliplied all terms by 5 to eliminate the fraction
:
14x=15....added 10x to both sides
:
{{{highlight(x=15/14)}}}
:
now take x's found value and plug it back into either equation. I choose eq 1
:
y=4/5(15/14)=60/70={{{highlight(6/7)}}}
:
so the hive is at (15/14,6/7)