Mister Exam
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How to use it?
- Equation:
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Equation x^2-8=(x-4)^2
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Equation 36-x^2=0
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Equation 4-x=1
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Equation (2-c)+3(c-3)=-13
- Express {x} in terms of y where:
- 8*x+5*y=-12
- -7*x-4*y=-4
- -2*x-10*y=20
- -1*x+11*y=16
- Sum of series:
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-13
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Identical expressions
- (two -c)+ three (c- three)=- thirteen
- (2 minus c) plus 3(c minus 3) equally minus 13
- (two minus c) plus three (c minus three) equally minus thirteen
- 2-c+3c-3=-13
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Similar expressions
- (2+c)+3(c-3)=-13
- (2-c)-3(c-3)=-13
- 2*(-4+3*(2*x-1)/(x-1))/(x-1)^3=0
- (2-c)+3(c+3)=-13
- (2-c)+3*(c-3)=-13
- (2-c)+3(c-3)=+13
- (2*x-1)^3-4*x^2*(2*x-3)=17
(2-c)+3(c-3)=-13 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 c)
from left part to right part, we given:
$$2 c = -6$$
Divide both parts of the equation by 2
We get the answer: c = -3
(2-c)+3*(c-3) = -13
Expand brackets in the left part
2-c+3*c-3*3 = -13
Looking for similar summands in the left part:
-7 + 2*c = -13
Move free summands (without c)
from left part to right part, we given:
$$2 c = -6$$
Divide both parts of the equation by 2
c = -6 / (2)
We get the answer: c = -3
The graph