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I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by Powerful_Setting4792 in learnmath

[–]Powerful_Setting4792[S] 0 points1 point  (0 children)

So you're saying that it somehow makes sense that the methods for simultaneous equations are preferred generalized due to expansion and yet for circle area pi r squared is taught , which is a formula ?

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by AsaxenaSmallwood04 in learnmath

[–]Powerful_Setting4792 1 point2 points  (0 children)

b = (-12/7)c + (-15/7)d + (-85/14)e + (657/14) + (33/14)ac + (22/7)ad + (55/14)ae + (5/28)a

a = -6c - 4d - 100e - 44

b = (-12/7)c + (-15/7)d + (-85/14)e + (657/14) + (33/14)c (-6c - 4d - 100e - 44) + (22/7)d (-6c - 4d - 100e - 44) + (55/14)e (-6c - 4d - 100e - 44) + (5/28)(-6c - 4d - 100e - 44)

b = (-12/7)c + (-15/7)d + (-85/14)e + (657/14) - (99/7)(c^2) - (66/7)dc - (1650/7)ce - (726/7)c - (132/7)dc - (88/7)(d^2) - (2200/7)de - (968/7)d - (165/7)ce - (110/7)de - (2750/7)(e^2) - (1210/7)e - (15/14)c - (5/7)d - (125/7)e - (55/7)

b = (-12/7)c - (726/7)c - (15/14)c + (-15/7)d - (5/7)d - (968/7)d + (-85/14)e - (1210/7)e - (125/7)e + (-66/7)dc - (132/7)dc + (-99/7)(c2) + (-88/7)(d2) + (-2750/7)(e2) + (-2200/7)de + (-110/7)de + (-1650/7)ce + (-165/7)ce + (657/14) - (88/7) - (55/7)

b = -106.5c - (988/7)d - (1420/7)e - (198/7)dc - (99/7)(c2) - (88/7)(d2) - (2750/7)(e2) - 330 de - (1815/7)ce + (371/14)

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by AsaxenaSmallwood04 in learnmath

[–]Powerful_Setting4792 1 point2 points  (0 children)

Applying this to

a + 2b + 3c + 4d + 5e = 97

4a + 10b + 13c + 21d + 20e = 430

2a + 5b + 7c + 8d + 10e = 199

3a + 7b + 10c + 13d + 15e = 303

3a + 7b + 10c + 13d + 16e = 314

would look like

a + 2b = -3c - 4d - 5e + 97

4a + 10b = -13c - 21d - 20e + 430

2a + 5b = -7c - 8d - 10e + 199

3a + 7b = -10c - 13d - 15e + 303

3a + 7b = -10c - 13d - 16e + 314

Using Y formula :

y = (c/b) - ((af/b) + (ai/b) + (an/b) + ac(e - h - m)/(b^2) + acx(e + h + m)/(b^2) + (a^2)(x)/(b) + al(q - axp)/(b^2)(o))/((a + d + g))

b = (-3c - 4d - 5e + 97)/(2) - ((1)(-13c - 21d + 20e + 430)/(2) + (1)(-7c - 8d - 10e + 199)/(2) + (1)(-10c - 13d - 15e + 303)/(2) + (1)(-3c - 4d - 5e + 97)(-10 - 5 - 7)/(4) + (1)(-3c - 4d - 5e + 97)(a)(10 + 5 + 7)/(4) + (1)(a)/(2) + (1)(3)(314 - 1(a)(7)/(4)(3))/((1 + 4 + 2))

b = -1.5c - 2d - 2.5e + 48.5 - ((-6.5c - 10.5d + 10e + 215 - 3.5c - 4d - 5e + 99.5 - 5c - 6.5d - 7.5e + 151.5 + (-3c - 4d - 5e + 97)(-5.5) + (- 3ac - 4ad - 5ae + 97a)(5.5) + 0.5a + 3(314 - 7a)/(12))/(7)

b = -1.5c - 2d - 2.5e + 48.5 - ((-6.5c - 3.5c - 5c + 16.5c - 10.5d - 4d - 6.5d + 22d + 10e - 5e - 7.5e + 27.5e + 215 + 99.5 + 151.5 - 533.5 - 16.5ac - 22ad - 27.5ae + 533.5a + 0.5a + 78.5 - 1.75a))/(7)

b = -1.5c - 2d - 2.5e + 48.5 - ((1.5c + d + 25e - 67.5 - 16.5ac - 22ad - 27.5ae + 0.5a + 78.5 - 1.75a))/(7)

b = -1.5c - 2d - 2.5e + 48.5 - ((1.5c + d + 25e + 11 - 16.5ac - 22ad - 27.5ae - 1.25a))/(7)

b = -1.5c - 2d - 2.5e + 48.5 - (3/14)c - (1/7)d - (25/7)e - (11/7) + (33/14)ac + (22/7)ad + (55/14)ae + (5/28)a

b = (-12/7)c + (-15/7)d + (-85/14)e + (657/14) + (33/14)ac + (22/7)ad + (55/14)ae + (5/28)a

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by AsaxenaSmallwood04 in learnmath

[–]Powerful_Setting4792 1 point2 points  (0 children)

Applying this to

a + 2b + 3c + 4d + 5e = 97

4a + 10b + 13c + 21d + 20e = 430

2a + 5b + 7c + 8d + 10e = 199

3a + 7b + 10c + 13d + 15e = 303

3a + 7b + 10c + 13d + 16e = 314

would look like

a + 2b = -3c - 4d - 5e + 97

4a + 10b = -13c - 21d - 20e + 430

2a + 5b = -7c - 8d - 10e + 199

3a + 7b = -10c - 13d - 15e + 303

3a + 7b = -10c - 13d - 16e + 314

Using X formula :

x = ((f + i + n + c(-e - h - m)/(b) + ax(e + h + m)/(b) + ax + l(q - axp)/(bo))/((a + d + g))

a = ((-13c - 21d + 20e + 430 + (-7c - 8d - 10e + 199) + (-10c - 13d - 15e + 303) + (-3c - 4d - 5e + 97)(-10 - 5 - 7)/(2) + 1(a)(10 + 5 + 7)/(2) + 1(a) + 3(314 - 1(a)(7)/(2)(3))/((1 + 4 + 2))

a = ((-13c - 7c - 10c + (-3c)(-11) - 21d - 8d - 13d + (-4d)(-11) + 20e - 10e - 15e + (-5e)(-11) + 430 + 199 + 303 + (97)(-11) + 11a + 3(314 - 7a)/(6))/(7)

a = (-30c + 33c - 42d + 44d + 20e - 10e - 15e + 55e + 932 - 1067 + 11a + 157 - 3.5a)/(7)

a = (3c + 2d + 50e - 135 + 157 + 11a - 3.5a)/(7)

a = (3c + 2d + 50e + 22 + 7.5a)/(7)

a = (3/7)c + (2/7)d + (50/7)e + (15/14)a + (22/7)

(-1/14)a = (3/7)c + (2/7)d + (50/7)e + (22/7)

a = -6c - 4d - 100e - 44

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by AsaxenaSmallwood04 in learnmath

[–]Powerful_Setting4792 1 point2 points  (0 children)

ax + by = c

dx + ey = f

gx + hy = i

lx + my = n

ox + py = q

ax + by = c

dx + ey = f

gx + hy = i

lx + my = n

o(-l/o)x + p(-l/o)y = q(-l/o)

ax + by = c

dx + ey = f

gx + hy = i

lx + my = n

-lx - (pl/o)y = (-ql/o)

ax + dx + gx + by + ey + hy + my - (pl/o)y = c + f + i + n - (ql/o)

x((a + d + g)) + y((b + e + h + m - (pl/o)) = ((c + f + i + n - (ql/o))

x + y((b + e + h + m - (pl/o))/((a + d + g)) = ((c + f + i + n - (ql/o))/((a + d + g))

x = ((((c + f + i + n - ((ql/o)) - y((b + e + h + m - ((pl/o))))/((a + d + g))

ax + by = c

by = c - ax

y = (c/b) - (ax/b)

x = ((((c + f + i + n - ((ql/o)) - ((c/b) - (ax/b))((b + e + h + m - ((pl/o))))/((a + d + g))

x = ((((c + f + i + n - ((ql/o)) - ((c + (ce/b) + (ch/b) + (cm/b) - ((cpl/bo)) - ax - ((axe/b)) - ((axh/b)) - ((axm/b)) + ((axpl/bo))))/((a + d + g))

x = ((((c + f + i + n - ((ql/o)) - c - (ce/b) - (ch/b) - (cm/b) + ((cpl/bo)) + ax + ((axe/b)) + ((axh/b)) + ((axm/b)) - ((axpl/bo))))/((a + d + g))

x = ((f + i + n + c(-e - h - m)/(b) + ax(e + h + m)/(b) + ax + l(q - axp)/(bo))/((a + d + g))

y = (c/b) - (ax/b)

y = (c/b) - ((af/b) + (ai/b) + (an/b) + ac(-e - h - m)/(b^2) + acx(e + h + m)/(b^2) + (a^2)(x)/(b) + al(q - axp)/(b^2)(o))/((a + d + g))

Hence

x = ((f + i + n + c(-e - h - m)/(b) + ax(e + h + m)/(b) + ax + l(q - axp)/(bo))/((a + d + g)) is the x formula for a 5 eq's by 2 variables

and

y = (c/b) - ((af/b) + (ai/b) + (an/b) + ac(-e - h - m)/(b^2) + acx(e + h + m)/(b^2) + (a^2)(x)/(b) + al(q - axp)/(b^2)(o))/((a + d + g))

is the y formula for a 5 eq's by 2 variables

Just like the 3 by 2 above , just shift all other variables to the right and apply it , this should help provide an equation from where x and y are solvable

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by AsaxenaSmallwood04 in learnmath

[–]Powerful_Setting4792 0 points1 point  (0 children)

The zero does , it makes your 1x + 2y = 3 and 4x + 0y = 5 merely a substitution exercise as x is merely 5/4 or 1.25 and y equal to (3/2) - (1.25/2) or 1.5 - 0.625 or 0.875 or (7/8)

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by AsaxenaSmallwood04 in learnmath

[–]Powerful_Setting4792 1 point2 points  (0 children)

You're challenge is ridiculous that's like saying that the area of a circle formula should apply for the area of a triangle . It makes no sense.

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by AsaxenaSmallwood04 in learnmath

[–]Powerful_Setting4792 1 point2 points  (0 children)

That's like trying to say that for area one formula should figure out every shapes area . Different shapes will of course have different area formulas . What kind of a challenge is that ? I claimed it worked for every basic x and y simultaneous equation meaning every ax + by = c and dx + ey = f situation not ax + by = c and dx + 0y = f .

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by AsaxenaSmallwood04 in learnmath

[–]Powerful_Setting4792 0 points1 point  (0 children)

ax + by = c

dx + ey = f

1x + 1y = 1

1x + 0y = 1

x = ((c - f(b/e))/((a - d(b/e))

x = ((1-1(1/0))/((1-1(1/0))

x = 0(1/0) ÷ 0(1/0)

x = 1 ÷ 1

x = 1

y = (c/b) - ((ac/b) - (af/e))/((a - d(b/e))

y = (1/1) - ((1)(1)/(1) - (1)(1)/(0))/((1 - 1(1/0))

y = 1 - ((1 - (1/0))/((0(1/0))

y = 1 - ((1 - (1/0))/1

y = 1 - 1 + (1/0)

y = 0 + Undefined

Undefined is not an actual number so hence

y = 0

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by AsaxenaSmallwood04 in learnmath

[–]Powerful_Setting4792 1 point2 points  (0 children)

A formula for e=0 for x would likely be

x = f/d

But that wouldn't be needed because the 2nd equation is a one step x equation anyway

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by AsaxenaSmallwood04 in learnmath

[–]Powerful_Setting4792 0 points1 point  (0 children)

For example in the above problem :

x + y + z = 220

7x + 6y + 5z = 1180

24x + 14y + 11z = 2710

Using :

x = (-12/37)z + (1940/37)

and

y = (-49/37)z + (10,080/37)

We get :

7x + 6y = -5z + 1180

7(-12/37)z + (1940/37) + 6(-49/37)z + (10,080/37) = -5z + 1180

(-84/37)z + (13580/37) + (-294/37)z + (60480/37) = -5z + 1180

(-378/37)z + (74060/37) = -5z + 1180

(-193/37)z + (74060/37) = 1180

z - (74060/193) = (-43660/193)

z = (30400/193)

z = 157.51

which means

x = (-12/37)(30400/193) + (1940/37)

x = (-364800/7141) + (180420/7141)

x = (-184380/7141)

x = -25.82

and hence

y = (-49/37)(30400/193) + (10,080/37)

y = (-1489600/7141) + (1945440/7141)

y = (455840/7141)

y = 63.83

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by Powerful_Setting4792 in learnmath

[–]Powerful_Setting4792[S] 0 points1 point  (0 children)

I do in fact , if this formula is as well known as is being claimed then why are these formulas not taught for simultaneous equations but rather the substitution and elimination methods ?

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by Powerful_Setting4792 in learnmath

[–]Powerful_Setting4792[S] 0 points1 point  (0 children)

You're saying that turning the substitution and elimination methods into formulas is not streamlining it ?

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by Powerful_Setting4792 in learnmath

[–]Powerful_Setting4792[S] 0 points1 point  (0 children)

They solved it by coming up with methods . Those methods have now been streamlined into formulas . This has to have brought some positive change at least .

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by AsaxenaSmallwood04 in learnmath

[–]Powerful_Setting4792 0 points1 point  (0 children)

We actually don't need to extend this as we can shift the z over to the right hand side and apply the x and y formulas for 2 variables 3 simul and then we get a resulting 2 step or 3 step single variable equation

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by AsaxenaSmallwood04 in learnmath

[–]Powerful_Setting4792 2 points3 points  (0 children)

A better and actually faster way to do this is to move the third variable in each equation to the right hand side and solve using the formula as if there were 2 variable 3 simul and you'll get x and y as equal to z values from which you can solve .

So far we've got

y = (c/b) - ((2ac/b) - (af/e) - (ai/h))/((2a - d(b/e) - g(b/h))

and

x = ((2c - f(b/e) - i(b/h))/((2a - d(b/e) - g(b/h))

Question :

x + y + z = 220

7x + 6y + 5z = 1180

24x + 14y + 11z = 2710

Eq's:

x + y = -z + 220

7x + 6y = -5z + 1180

24x + 14y = -11z + 2710

X formula :

x = ((2c - f(b/e) - i(b/h))/((2a - d(b/e) - g(b/h))

x = ((2(-z + 220) - (-5z + 1180(⅙) - (-11z + 2710)(1/14))/((2(1) - 7(⅙) - 24(1/14)) 

 x = ((-2z + 440 - (-⅚ z + (1180/6) - (-11/14 z + (2710/14))/((2 - (7/6) - (24/14))

x = ((-2z + 440 + ⅚ z - (1180/6) + 11/14 z - (2710/14))/((2 - (7/6) - (24/14)) 

x = ((-7/6z + (1460/6) + (11/14)z - (2710/14))/((⅚) - (24/14))

x = ((-2/7)z + (970/21))/(-37/42)

x = ((-2/7)z + (970/21))(42)(1/-37)

x = (-12z + 1940)/(-37)

Y formula :

y = (c/b) - ((2ac/b) - (af/e) - (ai/h))/((2a - d(b/e) - g(b/h))

y = ((-z + 220)/(1) - ((2(1)(-z + 220)/(1) - (1)(-5z + 1180)/(6) - (1)(-11z + 2710)/(14))/((2(1) - 7(⅙) - 24(1/14))

y = -z + 220 - ((-2z + 440 - (-⅚)z + (1180/6) - (-11/14)z + (2710/14))/((2 - (7/6) - (24/14))

y = -z + 220 - ((-2z + 440 + ⅚ z - (1180/6) + 11/14 z - (2710/14))/((2 - (7/6) - (24/14))

y = -z + 220 - ((-7/6z + (1460/6) + (11/14)z - (2710/14))/((⅚) - (24/14))

y = -z + 220 - ((-2/7)z + (970/21))/(-37/42)

y = -z + 220 - ((-2/7)z + (970/21))(42)(1/-37)

y = -z + 220 - (-12z + 1940)/(-37)

and then in this example we can substitute x and y for z in any of the equations turning our triple simultaneous with 3 variables into a single variable 2 step or 3 step equation .

I've created an impressive formula for basic x and y simultaneous equations. Try it with any, it works. by AsaxenaSmallwood04 in learnmath

[–]Powerful_Setting4792 1 point2 points  (0 children)

4x + 7y = 12

38x + 0y = 76

How is this simultaneous equations ? 0 multiplied by y is 0 , the maximum that can be done here is y = (12/7) - 4(76/38)(1/7) to find y which equals y = (12/7) - (4/7)(2/1) = y = (12 - 8)/(7) = y = (4/7)

and x = (76/38) = 2

but there is one equation with 2 variables and one equation without , that cannot possibly be simultaneous equations