SOLUTION: I have a system of two linear equations in the form ax+by=c. I am trying to convert them to the form y=ax+b for graphing. System:
2/x - 3/y = 4
1/x + 4/y = -10
I have tried
Algebra ->
Coordinate Systems and Linear Equations
-> SOLUTION: I have a system of two linear equations in the form ax+by=c. I am trying to convert them to the form y=ax+b for graphing. System:
2/x - 3/y = 4
1/x + 4/y = -10
I have tried
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Question 777397: I have a system of two linear equations in the form ax+by=c. I am trying to convert them to the form y=ax+b for graphing. System:
2/x - 3/y = 4
1/x + 4/y = -10
I have tried:
2/x - 3/y = 4 ......................-3/y = -2/x +4....................1/y=2/3x - 1 1/3
- 2/x.......- 2/x....................x -1/3....x -1/3
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Also tried;
2/x - 3/y = 4 __________________ -3/y = -2/x + 4 ________________-3 = -2/x x y +4
- 2/x ___- 2/x _________________ x y ________x y
I did these processes on the 1/x + 4/y = -10, too. I am stuck on the fractions with the variable in the denominator, i do not know how to handle them.
-Alexandrea
You can put this solution on YOUR website! I have a system of two linear equations in the form ax+by=c. I am trying to convert them to the form y=ax+b for graphing. System:
2/x - 3/y = 4
1/x + 4/y = -10
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Those are not linear equations.
The graph of each is a hyperbola.
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You can solve for y in order to make a table of points to graph them.
2/x - 3/y = 4
2y/x - 3 = 4y
2y/x - 4y = 3
y*(2/x - 4) = 3
y = 3/(2/x - 4)
y = 3/(2/x - 4x/x) = 3/((2-4x)/x)
y = 3x/(2-4x)
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dl the FREE graph software at
http://www.padowan.dk
Use F6 to enter the equations as they are.
