Find the integral values which satisfy the inequality 3x+4≤4x+6<2x+18
Starting with the left inequality, 3x + 4 ≤ 4x + 6, we can subtract 3x from both sides to get:
4 ≤ x + 6.
Then, subtracting 6 from each side gives:
-2 ≤ x.
Now, considering the right inequality, 4x + 6 < 2x + 18, we can subtract 2x from both sides to get:
2x + 6 < 18.
Subtracting 6 from each side gives:
2x < 12.
Dividing both sides by 2 gives:
x < 6.
Therefore, the solution to the inequality 3x + 4 ≤ 4x + 6 < 2x + 18 is -2 ≤ x < 6. The integral values which satisfy this inequality are -2, -1, 0, 1, 2, 3, 4, 5.
To solve the inequality 3x + 4 ≤ 4x + 6 < 2x + 18, we can break it down into two separate inequalities:
First, we solve 3x + 4 ≤ 4x + 6:
Subtract 3x from both sides:
4 ≤ x + 6
Subtract 6 from both sides:
-2 ≤ x
So, the first inequality is -2 ≤ x.
Next, we solve 4x + 6 < 2x + 18:
Subtract 2x from both sides:
2x + 6 < 18
Subtract 6 from both sides:
2x < 12
Divide by 2 on both sides:
x < 6
So, the second inequality is x < 6.
Therefore, the integral values that satisfy the given inequality are all values of x such that -2 ≤ x < 6.