Centroid of triangle with vertices (0,0) ,(3,0) & (0,-3) is……. [ ]
2 answers:
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First simplify
x-4y-8=3/4-10y
add 10y both sides and add 8
x+6y=8+3/4
times 4 both sides
4x+24y=32+3
4x+24y=35
5y-3(6-7/8)=5x-4y+3
add 4y both sides
9y-3(6-7/8)=5x+3
distribute
9y-18+21/8=5x+3
minus 5x both sides
9y-5x-18+21/8=3
add 18 to both sides and minus 21/8 both sides
9y-5x=21-21/8
times 8 both sides
72y-40x=147
72y-40x=147
so we got
24y+4x=35
72y-40x=147
multiply first equation by -3 and add to 2nd
-72y-12x=-105
72y<span>-40x=147 +</span>
0y-52x=42
-52x=42
divide both sides by -52
x=-21/26
sub back
24y+4x=35
24y+4(-21/26)=35
using math
y=497/312
the solution is
x-4y-8=3/4-10y
add 10y both sides and add 8
x+6y=8+3/4
times 4 both sides
4x+24y=32+3
4x+24y=35
5y-3(6-7/8)=5x-4y+3
add 4y both sides
9y-3(6-7/8)=5x+3
distribute
9y-18+21/8=5x+3
minus 5x both sides
9y-5x-18+21/8=3
add 18 to both sides and minus 21/8 both sides
9y-5x=21-21/8
times 8 both sides
72y-40x=147
72y-40x=147
so we got
24y+4x=35
72y-40x=147
multiply first equation by -3 and add to 2nd
-72y-12x=-105
72y<span>-40x=147 +</span>
0y-52x=42
-52x=42
divide both sides by -52
x=-21/26
sub back
24y+4x=35
24y+4(-21/26)=35
using math
y=497/312
the solution is