208k views
1 vote

x + 7 \leqslant \: or \: 5 + 2 * \ \textgreater \ 7 x + 7 <= 7 or 5 + 2x > 7solve equation

User Proski
by
3.7k points

1 Answer

2 votes

Answer:

( - ∞, 0] U ( 1, ∞)

Step-by-step explanation:

To solve 5 + 2x > 7, we need to subtract 5 on both sides:

5 + 2x - 5 > 7 - 5

2x > 2

Then, dividing by 2 on both sides as:

2x/2 > 2/2

x > 1

Therefore, the solution is the set of all x that are greater than 1 or:

( 1, ∞)

At the same way, to solve x + 7 ≤ 7, we need to subtract 7 on both sides as:

x + 7 - 7 ≤ 7 - 7

x ≤ 0

Therefore, the solution of this part is the set of all x that are less or equal to 0 or:

( - ∞, 0]

Finally, the solution is the join of both sets, so the solution is:

( - ∞, 0] U ( 1, ∞)

User Awongh
by
3.6k points