Answer:
the two positive consecutive integers are 4 and 6.
Step-by-step explanation:
Let the smaller integer be s; then s^2 = (s + 2) + 10.
Simplifying, s^2 - s - 2 - 10 = 0, or
s^2 - s - 12 = 0.
Solve this by factoring: (s - 4)(s + 3) = 0.
Then s = 4 and s = -3.
If the first even integer is 4, the next is 6. We omit s = -3 because it's not even.
The smaller integer is 4. Does this satisfy the equation s^2 = (s + 2) + 10?
4^2 = (4 + 2) + 10 True or False?
16 = 6 + 10 = 16.
True.
So the two positive consecutive integers are 4 and 6.
Answer: x ≤ 10
Step-by-step explanation:
20 + 13x ≤ 150
13x ≤ 150 - 20
13x ≤ 130
x ≤ 130 ÷ 13
x ≤ 10
he solution set is
{
x
∣
x
>
1
}
.
Explanation
For each of these inequalities, there will be a set of
x
-values that make them true. For example, it's pretty clear that large values of
x
(like 1,000) work for both, and negative values (like -1,000) will not work for either.
Since we're asked to solve a "this OR that" pair of inequalities, what we'd like to know are all the
x
-values that will work for at least one of them. To do this, we solve both inequalities for
x
, and then overlap the two solution set
