Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation (x+2)^2=(x-8)^2
-
Equation (x+2)^5=32
-
Equation x^2+3x=0
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Equation (x+2)^4-4*(x+2)^2-5=0
- Express {x} in terms of y where:
- -4*x+19*y=-13
- -5*x-6*y=-1
- -16*x-2*y=4
- -7*x+15*y=18
- 4x-2y
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Identical expressions
- 4x-2y
- 4x minus 2y
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Similar expressions
- 4*x-2*y+5*z+20=0
- 4x+2y
- (x+y)*(x+y)+4*(x-2*y+5)=0
- 4*x-2*y-1*z=0
4x-2y equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the linear equation:
Looking for similar summands in the left part:
Move the summands with the other variables
from left part to right part, we given:
$$- 2 y = - 4 x$$
Divide both parts of the equation by -2
We get the answer: y = 2*x
4*x-2*y = 0
Looking for similar summands in the left part:
-2*y + 4*x = 0
Move the summands with the other variables
from left part to right part, we given:
$$- 2 y = - 4 x$$
Divide both parts of the equation by -2
y = -4*x / (-2)
We get the answer: y = 2*x
The graph
Sum and product of roots
[src]
sum
2*re(x) + 2*I*im(x)
$$2 \operatorname{re}{\left(x\right)} + 2 i \operatorname{im}{\left(x\right)}$$
=
2*re(x) + 2*I*im(x)
$$2 \operatorname{re}{\left(x\right)} + 2 i \operatorname{im}{\left(x\right)}$$
product
2*re(x) + 2*I*im(x)
$$2 \operatorname{re}{\left(x\right)} + 2 i \operatorname{im}{\left(x\right)}$$
=
2*re(x) + 2*I*im(x)
$$2 \operatorname{re}{\left(x\right)} + 2 i \operatorname{im}{\left(x\right)}$$
2*re(x) + 2*i*im(x)
Rapid solution
[src]
y1 = 2*re(x) + 2*I*im(x)
$$y_{1} = 2 \operatorname{re}{\left(x\right)} + 2 i \operatorname{im}{\left(x\right)}$$
y1 = 2*re(x) + 2*i*im(x)