The standard form, which is the form that can be generalised to 3 or more dimensional spaces, is 4x + 5y - 13 = 0
Well, the way you wrote it, you would have:(4x - 5y - 3) x 3x + 5ywhich would give12x^2 - 15xy - 9x + 5yBut, if you meant the quantity (3x + 5y), then you would have:(4x - 5y - 3) x (3x + 5y)which would give12x^2 - 15xy - 9x + 20xy - 25y^2 - 15yand combining terms12x^2 - 25y^2 + 5xy - 9x - 15y
Multiply the first by 3 and the second by -2
5y+8
The lines intersect at (3, 5)
If: 20x+5y = 15 Then: y = -4x+3
The standard form, which is the form that can be generalised to 3 or more dimensional spaces, is 4x + 5y - 13 = 0
Well, the way you wrote it, you would have:(4x - 5y - 3) x 3x + 5ywhich would give12x^2 - 15xy - 9x + 5yBut, if you meant the quantity (3x + 5y), then you would have:(4x - 5y - 3) x (3x + 5y)which would give12x^2 - 15xy - 9x + 20xy - 25y^2 - 15yand combining terms12x^2 - 25y^2 + 5xy - 9x - 15y
Multiply the first by 3 and the second by -2
x = 3, y = 3
5y+8
The lines intersect at (3, 5)
You have two simultaneous equations; solve them:2x - 3y + 5 = 0→ 2x - 3y = -54x + 5y - 1 = 0 → 4x + 5y = 1As in the second equation the x coefficient is twice that in the first equation, subtract twice the first equation from the second and solve for y; {2}-2×{1}:(4x - 2×2x) + (5y - 2×-3y) = 1 - 2×-5→ (4x - 4x) + (5y - -6y) = 1 - -10→ 5y + 6y = 1 + 10→ 11y = 11→ y = 1Now substitute back into one of the equations (say {2}):4x + 5y = 1→ 4x + 5×1 = 1→ 4x = 1 - 5 = -4→ x = -1Now substitute the values for x and y back into the other equation ({1}) as a check:2x - 3y = 2×-1 - 3×1 = -2 - 3 = -5 as required.Thus the lines intersect at the point (-1, 1).
2x + 3 = -92x = -12x = -6
9y+3 = 5y+13 9y-5y = 13-3 4y = 10 y =2.5
(x + 3)2 = 11 + (x + 2)2x2 + 6x + 9 = 11 + x2 + 4x + 46x - 4x = 11 + 4 - 92x = 6x = 3Check: (3 + 3)2 = 11 + (3 + 2)2 ? ie 36 = 11 + 25 ? Yes. There you have it.
x=(-5y-1)/3