Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation (x+3)^3=81*(x+3)
-
Equation x^4=(3x-10)^2
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Equation -7*x^2+4*x=0
- Equation (x+3)(x-2)(x+9)=0
- Express {x} in terms of y where:
- -1*x-11*y=-5
- -20*x+12*y=-2
- -12*x-1*y=6
- -9*x+2*y=3
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Identical expressions
- - zero ,2m- five (zero , four + zero , six m)= one ,8m- one ,6
- minus 0,2m minus 5(0,4 plus 0,6m) equally 1,8m minus 1,6
- minus zero ,2m minus five (zero , four plus zero , six m) equally one ,8m minus one ,6
- -0,2m-50,4+0,6m=1,8m-1,6
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Similar expressions
- -0,2m-5(0,4-0,6m)=1,8m-1,6
- 0,2m-5(0,4+0,6m)=1,8m-1,6
- -0,2m-5(0,4+0,6m)=1,8m+1,6
- -0,2m+5(0,4+0,6m)=1,8m-1,6
-0,2m-5(0,4+0,6m)=1,8m-1,6 equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
m /2 3*m\ 9*m 8 - - - 5*|- + ---| = --- - - 5 \5 5 / 5 5
$$- \frac{m}{5} - 5 \left(\frac{3 m}{5} + \frac{2}{5}\right) = \frac{9 m}{5} - \frac{8}{5}$$
Detail solution
Given the linear equation:
Expand brackets in the left part
Expand brackets in the right part
Move free summands (without m)
from left part to right part, we given:
$$- \frac{16 m}{5} = \frac{9 m}{5} + \frac{2}{5}$$
Move the summands with the unknown m
from the right part to the left part:
$$\left(-5\right) m = \frac{2}{5}$$
Divide both parts of the equation by -5
We get the answer: m = -2/25
-(1/5)*m-5*((2/5)+(3/5)*m) = (9/5)*m-(8/5)
Expand brackets in the left part
-1/5m-5*2/5+3/5m) = (9/5)*m-(8/5)
Expand brackets in the right part
-1/5m-5*2/5+3/5m) = 9/5m-8/5
Move free summands (without m)
from left part to right part, we given:
$$- \frac{16 m}{5} = \frac{9 m}{5} + \frac{2}{5}$$
Move the summands with the unknown m
from the right part to the left part:
$$\left(-5\right) m = \frac{2}{5}$$
Divide both parts of the equation by -5
m = 2/5 / (-5)
We get the answer: m = -2/25
Sum and product of roots
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sum
-2/25
$$- \frac{2}{25}$$
=
-2/25
$$- \frac{2}{25}$$
product
-2/25
$$- \frac{2}{25}$$
=
-2/25
$$- \frac{2}{25}$$
-2/25