Minggu, 29 Oktober 2017
Solve the problem of one variable inequality
(Yanti Susanti, 160103061)
There are three ways to solve linear
equations of one variable by substitution, equivalent equation and switching
segments. The above three ways also apply to the linear inequality of one
variable.
How To Substitute
To solve linear inequalities one
variable by substitution is almost the same as solving a linear equation of one
variable by substitution. To understand it now consider the inequality 10 -
3x> 2, with the x variable in the set of natural numbers. To resolve the
inequality you must substitute x with any native number.
If x = 1 then:
<=> 10 - 3. 1> 2
<=> 7> 2 (true statement)
If x = 2 then:
<=> 10 - 3. 2> 2
<=> 4> 2 (true statement)
If x = 2 then:
<=> 10 - 3. 3> 2
<=> 1> 2 (statement is false)
If x = 4 then:
<=> 10 - 3. 4> 2
<=> - 2> 2 (statement is wrong)
It turns out for x = 1 and x = 2, the
10 - 3x> 2 inequality becomes the correct sentence. Thus, the set of
solutions of 10 - 3x> 2 is {1, 2}.
In general it can be written that the
solution of the linear inequality of a variable is a substitute for the
variable of an inequality, so that it becomes the correct statement.
To solidify your understanding of how
to solve linear inequalities of one variable by substitution, please note the
example below.
Example Problem 1
Find the set of p + 5 ≥ 9 inequality
settlement if the variables in the count set.
Solution:
To resolve the inequality you must
substitute x by any number of counts.
If x = 0 then:
<=> p + 5 ≥ 9
<=> 0 + 5 ≥ 9
<=> 5 ≥ 9 (statement is wrong)
If x = 1 then:
<=> p + 5 ≥ 9
<=> 1 + 5 ≥ 9
<=> 6 ≥ 9 (statement is wrong)
If x = 2 then:
<=> p + 5 ≥ 9
<=> 2 + 5 ≥ 9
<=> 7 ≥ 9 (statement is wrong)
If x = 3 then:
<=> p + 5 ≥ 9
<=> 3 + 5 ≥ 9
<=> 8 ≥ 9 (statement is wrong)
If x = 4 then:
<=> p + 5 ≥ 9
<=> 4 + 5 ≥ 9
<=> 9 ≥ 9 (true statement)
If x = 5 then:
<=> p + 5 ≥ 9
<=> 5 + 5 ≥ 9
<=> 10 ≥ 9 (true statement)
If x = 6 then:
<= > p + 5 ≥ 9
<=> 6 + 5 ≥ 9
<=> 11 ≥ 9 (true statement)
It turns out for x = 4, 5, 6,. . . the
inequality p + 5 ≥ 9 becomes the correct sentence. Thus, the settling set of p
+ 5 ≥ 9 is {4, 5, 6,. . }.
Solving linear inequalities by
substitution is rather difficult because we have to play the guess on the
numbers we will input. We know that numbers exist infinitely. So we use the
second alternative to solve the linear equations of one variable by using the
equivalent equation.
Equivalent Equations
An inequality can be expressed in
equivalent inequality in the following way: a). Increase or decrease both
segments with the same number without altering the inequality marks; b).
Multiply or divide the two segments with the same positive number without
changing the sign of inequality; c). Multiply or divide both segments with the
same negative number, but the sign of unequal change, where> becomes <,
<be>, ≤ becomes ≥, and ≥ becomes ≤.
To solidify your understanding of how
to solve a linear inequality of one variable with equivalent equations, see the
example below.
Example Problem 2
Find the set of inequality settlements
3 (2t - 1) ≤ 2t + 9 if the variables in the count set.
Completion:
<=> 3 (2t - 1) ≤ 2t + 9
<=> 6t - 3 ≤ 2t + 9
<=> 6t - 3 + 3 ≤ 2t + 9 + 3 (plus
3)
<=> 6t ≤ 2t + 12
<=> 6t - 2t ≤ 2t - 2t + 12 (minus
2t)
<=> 4t ≤ 12
<=> (¼) 4t ≤ (¼) 12 (multiplied
by ¼)
<=> t ≤ 3
Example Problem 3
Define the set completion of inequality
2 (x - 30) <4 (x - 2) if the variables in the count set.
Completion:
<=> 2 (x - 30) <4 (x - 2)
<=> 2x - 60 <4x - 8
<=> 2x - 60 + 60 <4x - 8 + 60
(plus 60)
<=> 2x <4x + 52
<=> 2x - 4x <4x - 4x + 52
(minus 4x)
<=> - 2x ≤ 52
<=> (- ½). 2x ≥ (- ½). 52
(multiplied - ½ and the sign changed because multiplied by the negative number
from ≤ to ≥) <=> x ≥ 26
How? Easy is not it? The way above is
too much time consuming and too long, then there is an alternative that arguably
the easiest is to move the segment.
Moving a Segment
To work on a linear inequality one
variable is the same way as doing a linear equation of a variable by moving a
segment. This is essentially the same as resolving inequalities with equivalent
equations. Okay, we just go to the example of the question so that you more
easily understand it.
Example Problem 4
Find the set of inequalities 6 - 2 (y -
3) ≤ 3 (2y - 4) if the variables in the chunk count set.
Completion:
<=> 6 - 2 (y - 3) ≤ 3 (2y - 4)
<=> 6 - 2y + 6 ≤ 6y - 12
<=> - 2y - 6y ≤ - 12 - 6 - 6
<=> - 8y ≤ - 24
<=> y ≥ - 24 / - 8
<=> y ≥ 3
Glossary :
linear inequality one variable = a mathematical form that contains one variable of
one rank and use the inequality sign (˃, ˂, ≤, ≥)
substitution = process or result of replacement
of language elements by other elements in larger units to obtain differentiator
elements
equivalent equations = equations
are said to be equivalent if they have the same set of settlements
the segment = marked with the sign “equals
(=)”. The left is called the left side, which is on the right is called the
right side
