Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation x^3=125
- Equation x^4-15x^2-16=0
- Equation (2y)/(y-3)=(3y+3)/y
- Equation (7)/(1-cos^2(x))+(9)/(sin(x))=10
- Express {x} in terms of y where:
- (x^2+y^2-1)^3+x^2*y^3=0
- -1*x-19*y=-10
- 4*x-16*y=6
- 5*x+5*y=20
-
Identical expressions
- (2y)/(y- three)=(three y+3)/y
- (2y) divide by (y minus 3) equally (3y plus 3) divide by y
- (2y) divide by (y minus three) equally (three y plus 3) divide by y
- 2y/y-3=3y+3/y
- (2y) divide by (y-3)=(3y+3) divide by y
-
Similar expressions
- (y^3+y^2-12y)/((y-3)×(y+2))=0
- (2y)/(y-3)=(3y-3)/y
- 2-3y/y^2-9+5-2y/y-3
- (2y)/(y+3)=(3y+3)/y
(2y)/(y-3)=(3y+3)/y equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2*y 3*y + 3 ----- = ------- y - 3 y
$$\frac{2 y}{y - 3} = \frac{3 y + 3}{y}$$
Detail solution
Given the equation:
$$\frac{2 y}{y - 3} = \frac{3 y + 3}{y}$$
Multiply the equation sides by the denominators:
-3 + y and y
we get:
$$\frac{2 y \left(y - 3\right)}{y - 3} = \frac{\left(y - 3\right) \left(3 y + 3\right)}{y}$$
$$2 y = 3 y - 6 - \frac{9}{y}$$
$$y 2 y = y \left(3 y - 6 - \frac{9}{y}\right)$$
$$2 y^{2} = 3 y^{2} - 6 y - 9$$
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$2 y^{2} = 3 y^{2} - 6 y - 9$$
to
$$- y^{2} + 6 y + 9 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$y_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$y_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -1$$
$$b = 6$$
$$c = 9$$
, then
Because D > 0, then the equation has two roots.
or
$$y_{1} = 3 - 3 \sqrt{2}$$
$$y_{2} = 3 + 3 \sqrt{2}$$
$$\frac{2 y}{y - 3} = \frac{3 y + 3}{y}$$
Multiply the equation sides by the denominators:
-3 + y and y
we get:
$$\frac{2 y \left(y - 3\right)}{y - 3} = \frac{\left(y - 3\right) \left(3 y + 3\right)}{y}$$
$$2 y = 3 y - 6 - \frac{9}{y}$$
$$y 2 y = y \left(3 y - 6 - \frac{9}{y}\right)$$
$$2 y^{2} = 3 y^{2} - 6 y - 9$$
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$2 y^{2} = 3 y^{2} - 6 y - 9$$
to
$$- y^{2} + 6 y + 9 = 0$$
This equation is of the form
a*y^2 + b*y + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$y_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$y_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -1$$
$$b = 6$$
$$c = 9$$
, then
D = b^2 - 4 * a * c =
(6)^2 - 4 * (-1) * (9) = 72
Because D > 0, then the equation has two roots.
y1 = (-b + sqrt(D)) / (2*a)
y2 = (-b - sqrt(D)) / (2*a)
or
$$y_{1} = 3 - 3 \sqrt{2}$$
$$y_{2} = 3 + 3 \sqrt{2}$$
Sum and product of roots
[src]
sum
___ ___ 3 - 3*\/ 2 + 3 + 3*\/ 2
$$\left(3 - 3 \sqrt{2}\right) + \left(3 + 3 \sqrt{2}\right)$$
=
6
$$6$$
product
/ ___\ / ___\ \3 - 3*\/ 2 /*\3 + 3*\/ 2 /
$$\left(3 - 3 \sqrt{2}\right) \left(3 + 3 \sqrt{2}\right)$$
=
-9
$$-9$$
-9
Rapid solution
[src]
___ y1 = 3 - 3*\/ 2
$$y_{1} = 3 - 3 \sqrt{2}$$
___ y2 = 3 + 3*\/ 2
$$y_{2} = 3 + 3 \sqrt{2}$$
y2 = 3 + 3*sqrt(2)