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Sagot :
To solve the inequality [tex]\( x + 29 < 39 \)[/tex], we first isolate [tex]\( x \)[/tex] by subtracting 29 from both sides of the inequality:
[tex]\[ x + 29 - 29 < 39 - 29 \][/tex]
This simplifies to:
[tex]\[ x < 10 \][/tex]
Now we need to determine which of the provided numbers satisfy [tex]\( x < 10 \)[/tex]. Let's check each option one by one:
A. [tex]\( 0 \)[/tex]:
- [tex]\( 0 < 10 \)[/tex] is true, so 0 belongs to the solution set.
B. [tex]\( 71 \)[/tex]:
- [tex]\( 71 < 10 \)[/tex] is false, so 71 does not belong to the solution set.
C. [tex]\( 10 \)[/tex]:
- [tex]\( 10 < 10 \)[/tex] is false, so 10 does not belong to the solution set.
D. [tex]\( 8 \)[/tex]:
- [tex]\( 8 < 10 \)[/tex] is true, so 8 belongs to the solution set.
E. [tex]\( 15 \)[/tex]:
- [tex]\( 15 < 10 \)[/tex] is false, so 15 does not belong to the solution set.
F. [tex]\( 5 \)[/tex]:
- [tex]\( 5 < 10 \)[/tex] is true, so 5 belongs to the solution set.
Therefore, the numbers that belong to the solution set of the inequality [tex]\( x + 29 < 39 \)[/tex] are:
- [tex]\( \boxed{0, 8, 5} \)[/tex]
[tex]\[ x + 29 - 29 < 39 - 29 \][/tex]
This simplifies to:
[tex]\[ x < 10 \][/tex]
Now we need to determine which of the provided numbers satisfy [tex]\( x < 10 \)[/tex]. Let's check each option one by one:
A. [tex]\( 0 \)[/tex]:
- [tex]\( 0 < 10 \)[/tex] is true, so 0 belongs to the solution set.
B. [tex]\( 71 \)[/tex]:
- [tex]\( 71 < 10 \)[/tex] is false, so 71 does not belong to the solution set.
C. [tex]\( 10 \)[/tex]:
- [tex]\( 10 < 10 \)[/tex] is false, so 10 does not belong to the solution set.
D. [tex]\( 8 \)[/tex]:
- [tex]\( 8 < 10 \)[/tex] is true, so 8 belongs to the solution set.
E. [tex]\( 15 \)[/tex]:
- [tex]\( 15 < 10 \)[/tex] is false, so 15 does not belong to the solution set.
F. [tex]\( 5 \)[/tex]:
- [tex]\( 5 < 10 \)[/tex] is true, so 5 belongs to the solution set.
Therefore, the numbers that belong to the solution set of the inequality [tex]\( x + 29 < 39 \)[/tex] are:
- [tex]\( \boxed{0, 8, 5} \)[/tex]
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