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-
How to use it?
- Inequation:
- -49x^2+14x-1=>0
- -9a>-9b
- 15*x+21>5*x-2
- -5t<15
- Sum of series:
-
15
- Integral of d{x}:
-
15
- Derivative of:
-
15
-
Identical expressions
- -5t< fifteen
- minus 5t less than 15
- minus 5t less than fifteen
-
Similar expressions
- 5t<15
- (t^2-5*t+6)/(t^2-3*t-18)>0
- (t-5)*(t+4)>=0
- (t-5)(t+4)<0
-5t<15 inequation
The solution
Detail solution
Given the inequality:
$$- 5 t < 15$$
To solve this inequality, we must first solve the corresponding equation:
$$- 5 t = 15$$
Solve:
Given the linear equation:
Divide both parts of the equation by -5
$$t_{1} = -3$$
$$t_{1} = -3$$
This roots
$$t_{1} = -3$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$t_{0} < t_{1}$$
For example, let's take the point
$$t_{0} = t_{1} - \frac{1}{10}$$
=
$$-3 + - \frac{1}{10}$$
=
$$- \frac{31}{10}$$
substitute to the expression
$$- 5 t < 15$$
$$- \frac{\left(-31\right) 5}{10} < 15$$
but
Then
$$t < -3$$
no execute
the solution of our inequality is:
$$t > -3$$
$$- 5 t < 15$$
To solve this inequality, we must first solve the corresponding equation:
$$- 5 t = 15$$
Solve:
Given the linear equation:
-5*t = 15
Divide both parts of the equation by -5
t = 15 / (-5)
$$t_{1} = -3$$
$$t_{1} = -3$$
This roots
$$t_{1} = -3$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$t_{0} < t_{1}$$
For example, let's take the point
$$t_{0} = t_{1} - \frac{1}{10}$$
=
$$-3 + - \frac{1}{10}$$
=
$$- \frac{31}{10}$$
substitute to the expression
$$- 5 t < 15$$
$$- \frac{\left(-31\right) 5}{10} < 15$$
31/2 < 15
but
31/2 > 15
Then
$$t < -3$$
no execute
the solution of our inequality is:
$$t > -3$$
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t1
Solving inequality on a graph