Question

The largest set of x values satisfying
2018x−p<2020x+p and 7x+3p<10x−2
can be written in simplest form as x>mn for positive m and n. What is the value of m+n?

Answer 1

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. $2,018x-p<2,020x+p$

Separate terms with x and without x into two sides:

$2,018x-2,020x<p+p\\ \\-2x<2p$

Multiply by -1:

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

2. $7x+3p<10x-2$

Separate terms with x and without x into two sides:

$7x-10x<-2-3p\\ \\-3x<-2-3p$

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

$-p>\dfrac{2}{3}+p\\ \\-2p>\dfrac{2}{3}\\ \\2p<-\dfrac{2}{3}\\ \\p<-\dfrac{1}{3},$

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 $mn=\dfrac{1}{3}=\dfrac{1}{3}\cdot 1,\ \ m+n=\dfrac{4}{3}$