Answer:
We have the system:
x ≤ 7
x ≥ a
Now we want to find the possible values of a such that the system has, at least, one solution.
First, we should look at the value of a where the system has only one solution:
We can write the 2 sets as:
a ≤ x
x ≥ 7
So, writing both together:
a ≤ x ≤ 7
if a is larger than 7, we do not have solutions.
then a = 7 gives:
7 ≤ x ≤ 7
Here the only solution is 7.
Now, if a is smaller than 7, for example 5, we have:
5 ≤ x ≤ 7
Now x can take different values, so we have a lot of solutions.
Then the restrictions for a, such that the system has at least one solution, is:
a ≤ 7.
Answer: multiply the number with the number in ()
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33
Y= - x + 2 - 4
y = - x - 2
-y = x + 2
<span>6x−2(x+4)=12
6x - 2x - 8 = 12
4x = 20
x = 5</span>
