Mister Exam
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How to use it?
- Equation:
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Equation x^2-3x+1=0
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Equation x^2-5*x-5=0
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Equation x^2+4*x-5=0
- Equation x+y=4
- Express {x} in terms of y where:
- 1*x+6*y=-6
- -5*x-2*y=9
- 1*x+11*y=-3
- 17*x-19*y=-12
- Sum of series:
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25
- Integral of d{x}:
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25
- Canonical form:
- 25
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Identical expressions
- (nineteen +2x)-(5x- eleven)= twenty-five
- (19 plus 2x) minus (5x minus 11) equally 25
- (nineteen plus 2x) minus (5x minus eleven) equally twenty minus five
- 19+2x-5x-11=25
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Similar expressions
- (19+2x)+(5x-11)=25
- (19+2x)-(5x+11)=25
- (19-2x)-(5x-11)=25
- (1/(x*(x-1)*(x-4)*(x-3)*(x-2)))*(((5*(x-5))*(x-6))*(x-7)+(x-8)*((5*(x-5))*(x-6)+(x-7)*(-30+5*x+5*(x-5))))+5*(x-8)*(x-7)*(x-6)*(x-5)*(-(x*(x-1))*(x-2)*(x-3)-(x-4)*((x*(x-1))*(x-2)+(x-3)*(x*(x-1)+(-1+2*x)*(x-2))))/(x^2*(x-1)^2*(x-4)^2*(x-3)^2*(x-2)^2)=0
(19+2x)-(5x-11)=25 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
19 + 2*x + -5*x + 11 = 25
$$\left(11 - 5 x\right) + \left(2 x + 19\right) = 25$$
Detail solution
Given the linear equation:
Expand brackets in the left part
Looking for similar summands in the left part:
Move free summands (without x)
from left part to right part, we given:
$$- 3 x = -5$$
Divide both parts of the equation by -3
We get the answer: x = 5/3
(19+2*x)-(5*x-11) = 25
Expand brackets in the left part
19+2*x-5*x+11 = 25
Looking for similar summands in the left part:
30 - 3*x = 25
Move free summands (without x)
from left part to right part, we given:
$$- 3 x = -5$$
Divide both parts of the equation by -3
x = -5 / (-3)
We get the answer: x = 5/3
Sum and product of roots
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sum
5/3
$$\frac{5}{3}$$
=
5/3
$$\frac{5}{3}$$
product
5/3
$$\frac{5}{3}$$
=
5/3
$$\frac{5}{3}$$
5/3
The graph