Mister Exam
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How to use it?
- Equation:
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Equation 1/(x-1)^2+2/(x-1)-3=0
-
Equation ((57.6/(150-x))^0.15)*((150+x)+10)/171.712=1
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Equation x^2+5x=36
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Equation -1(x-4)=2(x+1)
- Express {x} in terms of y where:
- 2*x+3*y=14
- 7*x-1*y=-7
- 20*x+5*y=-3
- -9*x+5*y=-2
- Graphing y =:
- 2(x+1)
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Identical expressions
- - one (x- four)= two (x+ one)
- minus 1(x minus 4) equally 2(x plus 1)
- minus one (x minus four) equally two (x plus one)
- -1x-4=2x+1
-
Similar expressions
- sin2x+16cos^2x=4
- (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
- 1(x-4)=2(x+1)
- (x-3)/((x-1)(x-4))+((x-1)(x-4))/(x-3)=(5)/(2)
- -1(x+4)=2(x+1)
- 5tg^2x+11tgx−12=0.
- l[(x-1)(x-4)]=ln(-3x+7)
- (x-1)*(x-4)
- ((2x+1)^2)/25-(x-1)/3=x
- -1(x-4)=2(x-1)
- x^2-2x+10=0
-1(x-4)=2(x+1) equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the linear equation:
Expand brackets in the left part
Expand brackets in the right part
Move free summands (without x)
from left part to right part, we given:
$$- x = 2 x - 2$$
Move the summands with the unknown x
from the right part to the left part:
$$\left(-3\right) x = -2$$
Divide both parts of the equation by -3
We get the answer: x = 2/3
-1*(x-4) = 2*(x+1)
Expand brackets in the left part
-1*x+1*4 = 2*(x+1)
Expand brackets in the right part
-1*x+1*4 = 2*x+2*1
Move free summands (without x)
from left part to right part, we given:
$$- x = 2 x - 2$$
Move the summands with the unknown x
from the right part to the left part:
$$\left(-3\right) x = -2$$
Divide both parts of the equation by -3
x = -2 / (-3)
We get the answer: x = 2/3
Sum and product of roots
[src]
sum
2/3
$$\frac{2}{3}$$
=
2/3
$$\frac{2}{3}$$
product
2/3
$$\frac{2}{3}$$
=
2/3
$$\frac{2}{3}$$
2/3
The graph