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How to use it?
- Equation:
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Equation cos(x)=(-1/2)
-
Equation sin(x-1)=0
-
Equation x^2-36=0
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Equation sin(x)=-3/2
- Express {x} in terms of y where:
- 12*10^(-30)-0,1884*10^12*x^2-y(0,3768*10^12*x+1,183152*10^12*x^3)+y^2(2,760688*10^24*x^2)=0=0
- (x^2+y^2)×(x+y-3)=2xy
- ((42^2-22)+(67*y))/67=((21^3-22)+(67*x))/67
- sqrt(1-x)^2+(4-y)^2=0
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Identical expressions
- six x+ twelve -3x= five -(7x+6)
- 6x plus 12 minus 3x equally 5 minus (7x plus 6)
- six x plus twelve minus 3x equally five minus (7x plus 6)
- 6x+12-3x=5-7x+6
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Similar expressions
- 6x+12-3x=5-(7x-6)
- 6x+12-3x=5+(7x+6)
- 9x-5-7x+6=13
- 6x+12+3x=5-(7x+6)
- 6x-12-3x=5-(7x+6)
6x+12-3x=5-(7x+6) equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
6*x + 12 - 3*x = 5 + -7*x - 6
$$- 3 x + \left(6 x + 12\right) = \left(- 7 x - 6\right) + 5$$
Detail solution
Given the linear equation:
Expand brackets in the right part
Looking for similar summands in the left part:
Looking for similar summands in the right part:
Move free summands (without x)
from left part to right part, we given:
$$3 x = - 7 x - 13$$
Move the summands with the unknown x
from the right part to the left part:
$$10 x = -13$$
Divide both parts of the equation by 10
We get the answer: x = -13/10
6*x+12-3*x = 5-(7*x+6)
Expand brackets in the right part
6*x+12-3*x = 5-7*x-6
Looking for similar summands in the left part:
12 + 3*x = 5-7*x-6
Looking for similar summands in the right part:
12 + 3*x = -1 - 7*x
Move free summands (without x)
from left part to right part, we given:
$$3 x = - 7 x - 13$$
Move the summands with the unknown x
from the right part to the left part:
$$10 x = -13$$
Divide both parts of the equation by 10
x = -13 / (10)
We get the answer: x = -13/10
Sum and product of roots
[src]
sum
-13 ---- 10
$$- \frac{13}{10}$$
=
-13 ---- 10
$$- \frac{13}{10}$$
product
-13 ---- 10
$$- \frac{13}{10}$$
=
-13 ---- 10
$$- \frac{13}{10}$$
-13/10