Mister Exam
1+8x<9 inequation
The solution
Detail solution
Given the inequality:
$$8 x + 1 < 9$$
To solve this inequality, we must first solve the corresponding equation:
$$8 x + 1 = 9$$
Solve:
Given the linear equation:
Move free summands (without x)
from left part to right part, we given:
$$8 x = 8$$
Divide both parts of the equation by 8
$$x_{1} = 1$$
$$x_{1} = 1$$
This roots
$$x_{1} = 1$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 1$$
=
$$\frac{9}{10}$$
substitute to the expression
$$8 x + 1 < 9$$
$$1 + \frac{8 \cdot 9}{10} < 9$$
the solution of our inequality is:
$$x < 1$$
$$8 x + 1 < 9$$
To solve this inequality, we must first solve the corresponding equation:
$$8 x + 1 = 9$$
Solve:
Given the linear equation:
1+8*x = 9
Move free summands (without x)
from left part to right part, we given:
$$8 x = 8$$
Divide both parts of the equation by 8
x = 8 / (8)
$$x_{1} = 1$$
$$x_{1} = 1$$
This roots
$$x_{1} = 1$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 1$$
=
$$\frac{9}{10}$$
substitute to the expression
$$8 x + 1 < 9$$
$$1 + \frac{8 \cdot 9}{10} < 9$$
41/5 < 9
the solution of our inequality is:
$$x < 1$$
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