# How can we transform other forms of linear equations into general linear equations?

The straight line equation has oblique section: y==kx+m

Point-skew type: y-y0==k (x-x0)

Intercept type: x/a+y/b==1

Two-point formula: (x-x1)/(x2-x1)==(y-y1)/(y2-y1)

How to unify the above form into general equation form: Ax+By+c==0

The actual example is:

y==3x+6

x/3+y/6==1

y-8==2(x-3)

(x-2)/(7-2)==(y-1)/(9-1)

6x+8y+10==0

-x+7y+10==0



The above is unified into the form of Ax+By+C==0

The coefficient in front of x is required to be positive, and the constant terms of x and y are the simplest integers.

6x+8y+10==0 Its final result is 3x+4y+5==0

-x+7y+10==0 Its final result is x-7y-10==0

• Again, your choice of tags is confusing. Where are the differential equations? Also, this y=3x+6 should have been y==3x+6 and likewise for the rest
– bmf
Feb 25 at 10:46
• You could also use CoefficientArrays, see e.g. 185668. For instance {1, x, y} . Flatten@CoefficientArrays[#, {x, y}] & /@ eqs Feb 25 at 14:34

To get an expression f[x,y]==0, you may first simplify the equations and then transform the equation into an expression by subtracting the left side from the right side.

The equations:

eqs = {y == 3 x + 6, x/3 + y/6 == 1,
y - 8 == 2 (x - 3), (x - 2)/(7 - 2) == (y - 1)/(9 - 1),
6 x + 8 y + 10 == 0, -x + 7 y + 10 == 0}


may be transformed by:

eqs= Simplify[eq] /. Equal -> Subtract

{6 + 3 x - y, -6 + 2 x + y, 2 + 2 x - y, -11 + 8 x - 5 y,
5 + 3 x + 4 y, -10 + x - 7 y}


If you want to write them again in equation form:

(# == 0) & /@ eqs

{6 + 3 x - y == 0, -6 + 2 x + y == 0,
2 + 2 x - y == 0, -11 + 8 x - 5 y == 0,
5 + 3 x + 4 y == 0, -10 + x - 7 y == 0}


For any equation eq=A==B, then eq[] is A and eq[] is B. This is the same as saying First@eq is A and Last@eq is B.

So First@eq is the left hand side of equation, and Last@eq is the right hand side of equation.

So to move all terms to one side of an equation, just do

 myNewEquation = First@eq - Last@eq == 0


Using the above

eqs = {y == 3 x + 6, x/3 + y/6 == 1,
y - 8 == 2 (x - 3), (x - 2)/(7 - 2) == (y - 1)/(9 - 1),
6 x + 8 y + 10 == 0, -x + 7 y + 10 == 0} Now do

((Expand[First[#] - Last[#]] == 0) & /@ eqs) // Column SubtractSides can also be used if you prefer, but need to add Simplify for a reason I do not know yet. Like this

 (SubtractSides[Simplify[#]] & /@ eqs) // Column If we do the following, then one equation fail to be moved to one side. I do not know why. May be this needs separate question.

 SubtractSides /@ eqs // Column 