Mister Exam
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How to use it?
- Equation:
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Equation x^2+x=12
-
Equation 3x-2(3x-2)=19
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Equation x^4=(x-56)^2
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Equation (-2x+1)*(-2x-7)=0
- Express {x} in terms of y where:
- -7*x-13*y=-12
- 8*x+1*y=4
- 20*x-16*y=7
- -3*x+19*y=16
- Sum of series:
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26
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Identical expressions
- twelve - two *(x+ three)= twenty-six
- 12 minus 2 multiply by (x plus 3) equally 26
- twelve minus two multiply by (x plus three) equally twenty minus six
- 12-2(x+3)=26
- 12-2x+3=26
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Similar expressions
- log2(6-x)=2logx2
- 12-2*(x-3)=26
- 2*6-y+2=0
- 12+2*(x+3)=26
- (6*x-1)^2-(5*x+2)*(6*x+5)=6*((x-1)^2)+37*x
- -x-(26/5)=12-2*(x+(3/5))-10
12-2*(x+3)=26 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
Looking for similar summands in the left part:
Move free summands (without x)
from left part to right part, we given:
$$- 2 x = 20$$
Divide both parts of the equation by -2
We get the answer: x = -10
12-2*(x+3) = 26
Expand brackets in the left part
12-2*x-2*3 = 26
Looking for similar summands in the left part:
6 - 2*x = 26
Move free summands (without x)
from left part to right part, we given:
$$- 2 x = 20$$
Divide both parts of the equation by -2
x = 20 / (-2)
We get the answer: x = -10
Sum and product of roots
[src]
sum
-10
$$\left(-10\right)$$
=
-10
$$-10$$
product
-10
$$\left(-10\right)$$
=
-10
$$-10$$
The graph