6x- 8y- 5x + 3y?
6x- 8y- 5x + 3y
x-5y
switch x and y, then solve again for y:

x = 3y+12
x-12 = 3y
(x-12)/3 = y

or,

y = x/3 - 4 equating the expression with 0,3x2 + 5x + 2 = 0on factorising it further,3x2 + 3x + 2x + 2 = 03x(x + 1) + 2(x + 1) = 0(3x + 2) (x + 1) = 0 x = - or x = -1 the zeroes of 3x2 + 5x + 2 are – and -1. answer: `4+1+(1)/(4)+(1)/(6)+...`
solution: n/a slope of the line ⇒ slope of the lineequation of line :⇒ 3(y - 5) = 2(x - 2)⇒ 2x - 3y + 11 = 0 …equation (i)putting x = x + 3 and y = y + 2 in the above equation we get2(x + 3) - 3(y + 2) + 11 = 0⇒ 2x - 3y + 11 = 0since it again gives the same equation as (i) so it is a point on this line.hence proved y = 3x

It can be observed that x = −1, y = −3 and x = 1, y = 3 are solutions of the above equation. Therefore, the solution table is as follows.



| x | -1 | 1 |
| y | -3 | 3 |

The graph of the above equation is constructed as follows.

![](https://www.sarthaks.com/?qa=blob&qa_blobid=15681428413536225923) without calculus, the  horizontal coordinate of a quadratic of  f(x) =ax^2 + bx + c  is equal to - b/ 2a
   which in this case of
       f( x) =  - 5 x^2 + 3x + 2  
 
       equals :    - 3/[2(-5)] = -3/ -10 = 3/10 finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). f(x) = u/v....i.e. u = 5x² + 4x + 3 and v = 2x² - 2x - 14

now, f '(x) = (u'v - uv')/v²

so, f '(x) = [(10x + 4)(2x² - 2x - 14) - (5x² + 4x + 3)(4x - 2)]/(2x² - 2x - 14)²   

a bit of tidying up with the denominator gives:

f '(x) = -2[9x² + 76x + 25]/(2x² - 2x - 14)²

:)> let a (x, y) be the point on the curve y = f(x). then slope of the tangent to the curve at point a is `"dy"/"dx"`. according to the given condition, `"dy"/"dx" = "x" + 3"y" - 1` ∴ `"dy"/"dx" - 3"y" = "x - 1"` ....(1) this is the linear differential equation of the form `"dy"/"dx" + "py" = "q"`, where p = - 3 and q = x - 1 ∴ i.f. = `"e"^(int "p dx") = "e"^(int - 3"dx") = "e"^(- 3"x")` ∴ the solution of (1) is given by `"y" * ("i.f.") = int "q" * ("i.f.") "dx" + "c"` ∴ `"y" * "e"^(- 3"x") = int ("x - 1") * "e"^(-3"x") "dx" + "c"` ∴ `"e"^(- 3"x") * "y" = ("x - 1") int "e"^(- 3"x") - int ["d"/"dx" ("x - 1") * int "e"^(- 3"x") "dx"] "dx" + "c"\_1` ∴ `"e"^(- 3"x") * "y" = ("x - 1") * "e"^(- 3"x")/-3 - int 1 * "e"^(- 3"x") * "y"/-3 "dx" + "c"\_1` ∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1") * "e"^(- 3"x") + 1/3 int "e"^(- 3"x") "dx" + "c"\_1` ∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1") "e"^(- 3"x") + 1/3 * "e"^(- 3"x")/-3 + "c"\_1` ∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1")"e"^(- 3"x") - 1/9 "e"^(- 3"x") + "c"\_1` ∴ 9y = - 3(x - 1) - 1 + 9`"c"\_1 * e^("3x")` ∴ 9y + 3(x - 1) + 1 = `9"c"\_1 * e^("3x")` ∴ 9y + 3x - 3 + 1 = `9"c"\_1 * e^("3x")` ∴ 3(x + 3y) = 2 + `9"c"\_1 * e^("3x")` ∴ 3(x + 3y) = 2 + `"c" * e^("3x")` where c = 9c1 ....(2) this is the general equation of the curve. but the required curve is passing through the origin (0, 0). ∴ by putting x = 0 and y= 0 in (2), we get 0 = 2 + c ∴ c = - 2 ∴ from (2), the equation of the required curve is 3(x + 3y) = 2 - 2`e^("3x")` i.e. 3(x + 3y) = 2(1 - `e^("3x")`). 65 ÷ ( – 13) = **-5**