Asked by Jasmine20
Find four solutions for the equation
3x+5y=15
so the equation will turn to be
y=-(3)/(5)x+ (3)
and when i do the table i get these points for the solution
(-2,4.2),(-1,3.60),(0,3),(1,1.8)
ok
Let me give you a help.
Take the slope, denominator. 5. Now make each x an multiple of 5, say 0,5, 10, 15, -5
It makes it so easy to multiply in your head.
Find four solutions for the equation
3x+5y=15
Assuming you are seeking integer answers:
Dividing through by 3 yields x + y + 2y/3 = 5
2y/2 must be an integer as does y/3 so y/3 = k making y = 3k.
Substituting, 3x + 15k = 15 making x 5 - 5k
k.....0.....1.....2.....3
x.....5.....0....-5....-10
y.....0.....3.....6.....9
The only positive solutions exist between (5,0) and (0,3)on the line defined by y = (15 - 3x)/5
There are no integer solutions between these two points but there are an infinite number of rational solutions.
x.....0.....1.....2.....3.....4.....5
y.....5....2.4...1.8...1.2....6.....0
x....5....3.33..1.66...0
y....o.....1.....2.....3
3x+5y=15
so the equation will turn to be
y=-(3)/(5)x+ (3)
and when i do the table i get these points for the solution
(-2,4.2),(-1,3.60),(0,3),(1,1.8)
ok
Let me give you a help.
Take the slope, denominator. 5. Now make each x an multiple of 5, say 0,5, 10, 15, -5
It makes it so easy to multiply in your head.
Find four solutions for the equation
3x+5y=15
Assuming you are seeking integer answers:
Dividing through by 3 yields x + y + 2y/3 = 5
2y/2 must be an integer as does y/3 so y/3 = k making y = 3k.
Substituting, 3x + 15k = 15 making x 5 - 5k
k.....0.....1.....2.....3
x.....5.....0....-5....-10
y.....0.....3.....6.....9
The only positive solutions exist between (5,0) and (0,3)on the line defined by y = (15 - 3x)/5
There are no integer solutions between these two points but there are an infinite number of rational solutions.
x.....0.....1.....2.....3.....4.....5
y.....5....2.4...1.8...1.2....6.....0
x....5....3.33..1.66...0
y....o.....1.....2.....3
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