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yarga [219]
3 years ago
13

The largest set of x values satisfying

Mathematics
1 answer:
velikii [3]3 years ago
7 0

Answer:

If p\ge -\dfrac{1}{3}, then x>\dfrac{2}{3}+p and mn=\dfrac{2}{3}+p=\left(\dfrac{2}{3}+p\right)\cdot 1,\ \ m+n=\dfrac{5}{3}+p

Step-by-step explanation:

Solve two inequalities for x.

1. 2,018x-p

Separate terms with x and without x into two sides:

2,018x-2,020x

Multiply by -1:

2x>-2p\\ \\x>-p

2. 7x+3p

Separate terms with x and without x into two sides:

7x-10x

Multiply by -1:

3x>2+3p\\ \\x>\dfrac{2}{3}+p

Find the largest set of x values satisfying both inequalities:

x>-p\\ \\x>\dfrac{2}{3}+p

If

-p>\dfrac{2}{3}+p\\ \\-2p>\dfrac{2}{3}\\ \\2p

then x>-p and mn=-p=p\cdot (-1),\ m+n=p-1. In this case both m and n are negative.

If p>-\dfrac{1}{3}, then x>\dfrac{2}{3}+p and mn=\dfrac{2}{3}+p=\left(\dfrac{2}{3}+p\right)\cdot 1,\ \ m+n=\dfrac{5}{3}+p

If p=-\dfrac{1}{3}, then x>\dfrac{1}{3} and mn=\dfrac{1}{3}=\dfrac{1}{3}\cdot 1,\ \ m+n=\dfrac{4}{3}

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