Mister Exam
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How to use it?
- Equation:
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Equation sqrt(12+x)-sqrt(1-x)=1
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Equation 4x^3-3x^2
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Equation x^4=15x^2-56
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Identical expressions
- (two z+ four): five =(3z- one):2
- (2z plus 4):5 equally (3z minus 1):2
- (two z plus four): five equally (3z minus one):2
- 2z+4:5=3z-1:2
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Similar expressions
- (2z+4):5=(3z+1):2
- (2z-4):5=(3z-1):2
(2z+4):5=(3z-1):2 equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2*z + 4 3*z - 1 ------- = ------- 5 2
$$\frac{2 z + 4}{5} = \frac{3 z - 1}{2}$$
Detail solution
Given the linear equation:
Expand brackets in the left part
Expand brackets in the right part
Move free summands (without z)
from left part to right part, we given:
$$\frac{2 z}{5} = \frac{3 z}{2} - \frac{13}{10}$$
Move the summands with the unknown z
from the right part to the left part:
$$\frac{\left(-11\right) z}{10} = - \frac{13}{10}$$
Divide both parts of the equation by -11/10
We get the answer: z = 13/11
(2*z+4)/5 = (3*z-1)/2
Expand brackets in the left part
2*z/5+4/5 = (3*z-1)/2
Expand brackets in the right part
2*z/5+4/5 = 3*z/2-1/2
Move free summands (without z)
from left part to right part, we given:
$$\frac{2 z}{5} = \frac{3 z}{2} - \frac{13}{10}$$
Move the summands with the unknown z
from the right part to the left part:
$$\frac{\left(-11\right) z}{10} = - \frac{13}{10}$$
Divide both parts of the equation by -11/10
z = -13/10 / (-11/10)
We get the answer: z = 13/11
Sum and product of roots
[src]
sum
13 -- 11
$$\frac{13}{11}$$
=
13 -- 11
$$\frac{13}{11}$$
product
13 -- 11
$$\frac{13}{11}$$
=
13 -- 11
$$\frac{13}{11}$$
13/11