Mister Exam
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How to use it?
- Equation:
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Equation (x-3)*(x+4)=0
-
Equation x^3-x^2-5=0
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Equation x-2(3x+2)=16
- Equation x^2-1=1+(1/2)*y
- Express {x} in terms of y where:
- -10*x+16*y=-6
- -13*x+9*y=1
- -10*x+9*y=-6
- -17*x+13*y=-3
- Integral of d{x}:
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(x+2)/6
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Identical expressions
- (x+ three)/ twelve - one *(x- three)/ four =(x+ two)/ six
- (x plus 3) divide by 12 minus 1 multiply by (x minus 3) divide by 4 equally (x plus 2) divide by 6
- (x plus three) divide by twelve minus one multiply by (x minus three) divide by four equally (x plus two) divide by six
- (x+3)/12-1(x-3)/4=(x+2)/6
- x+3/12-1x-3/4=x+2/6
- (x+3) divide by 12-1*(x-3) divide by 4=(x+2) divide by 6
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Similar expressions
- (x-3)/12-1*(x-3)/4=(x+2)/6
- (x+3)/12+1*(x-3)/4=(x+2)/6
- (x+3)/12-1*(x+3)/4=(x+2)/6
- (2x-3)/4(5x+2/)6=3
- (4x-1)/(9)-(x+2)/(6)=2
- (x-2)/5-(3x+2)/6=2/3-x
- (x+3)/12-1*(x-3)/4=(x-2)/6
(x+3)/12-1*(x-3)/4=(x+2)/6 equation
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The solution
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x + 3 x - 3 x + 2 ----- - ----- = ----- 12 4 6
$$- \frac{x - 3}{4} + \frac{x + 3}{12} = \frac{x + 2}{6}$$
Detail solution
Given the linear equation:
Expand brackets in the left part
Expand brackets in the right part
Looking for similar summands in the left part:
Move free summands (without x)
from left part to right part, we given:
$$- \frac{x}{6} = \frac{x}{6} - \frac{2}{3}$$
Move the summands with the unknown x
from the right part to the left part:
$$\frac{\left(-1\right) x}{3} = - \frac{2}{3}$$
Divide both parts of the equation by -1/3
We get the answer: x = 2
(x+3)/12-1*(x-3)/4 = (x+2)/6
Expand brackets in the left part
x/12+3/12-1*x/4+1*3/4 = (x+2)/6
Expand brackets in the right part
x/12+3/12-1*x/4+1*3/4 = x/6+2/6
Looking for similar summands in the left part:
1 - x/6 = x/6+2/6
Move free summands (without x)
from left part to right part, we given:
$$- \frac{x}{6} = \frac{x}{6} - \frac{2}{3}$$
Move the summands with the unknown x
from the right part to the left part:
$$\frac{\left(-1\right) x}{3} = - \frac{2}{3}$$
Divide both parts of the equation by -1/3
x = -2/3 / (-1/3)
We get the answer: x = 2