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Sagot :
Solving an inequality
Having that
10 ≤ 6 - 2x < 14
It meets two statements:
10 ≤ 6 - 2x and 6 - 2x < 14
We are solving each of them separately. We have to remember that we can add or substract any amount both sides of the inequalities and multiply or divide by a positive number both sides.
First statement: 10 ≤ 6 - 2x
On one hand, we want to solve:
10 ≤ 6 - 2x
then
10 ≤ 6 - 2x
↓ adding 2x both sides
10 + 2x ≤ 6
↓ substracting 10 both sides
2x ≤ 6 -10
↓ 6 - 10 = -4
2x ≤ -4
↓ dividing by 2 both sides
2x/2 ≤ -4/2
↓ -4/2 = -2
x ≤ -2
We have that x ≤ -2
Second statement: 6 - 2x < 14
For the other hand, we want to solve
6 - 2x < 14
then
6 - 2x < 14
↓ adding 2x both sides
6 < 14 + 2x
↓ substracting 14 both sides
6 - 14 < 2x
↓ 6 - 14 = -8
-8 < 2x
↓ dividing by 2 both sides
-8/2 < 2x/2
↓ -8/2 = -4
-4 < x
We have that -4 < x
Therefore, joining both conclusions, we have that -4 < x and x ≤ -2, then
Answer: -4 < x ≤ -2
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