The values -3 , -6 , 3 , 8 are in the solution set of the compound inequality 4(x + 3) ≤ 0 and x + 1 > 3
Given, a compound inequality
4(x + 3) ≤ 0 and x + 1 > 3
Now, on solving the first inequality, we get
4(x + 3) ≤ 0
4x + 12 ≤ 0
On subtracting both the sides by 12, we get
4x + 12 - 12 ≤ 0 - 12
4x ≤ -12
On dividing both the sides by 4, we get
4x/4 ≤ -12/4
x ≤ -3
So, the solution set of the 4(x + 3) ≤ 0 inequality be, x ≤ -3 i.e. all the real numbers less than -3.
Now, on solving the second inequality, we get
x + 1 > 3
On subtracting both the sides by 1, we get
x + 1 - 1 > 3 - 1
x > 2
So, the solution set of the x + 1 > 3 inequality be, x > 2 i.e. all the real number greater than 2.
Now, the values in the solution set of the compound inequality
4(x + 3) ≤ 0 and x + 1 > 3 be,
x ≤ -3 and x > 2.
As, we can see -3 , -6 , 3 , 8 are satisfying the condition.
Hence, -3 , -6 , 3 , 8 are the values in the solution set of the compound inequality 4(x + 3) ≤ 0 and x + 1 > 3.
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