simultaneous inequalities (7th grade)
2(x+1) > 8
3x + 1 < a + 2
"x's inequality group only has 2 integers, find a's value"
i have a test tomorrow and ive just been sitting here and crying for hours. i've been testing myself with these kind of questions since a week ago and i still get stuck everytime i try to do them. i'm so frustrated with it all .
and if anyone has some time to spare, please explain this too:
(3x-1)/2 - 3 <= (6-x)/3
2a-5x<15
"x's inequality group only has 3 integers, and those are___"
Sunny
1 Answer
Remember that you can add or subtract anything from both sides of an inequality, and multiply or divide both sides by a positive number. Here we have $$2(x+1)>8\implies x+1>4\implies x>3 $$ Now since there are only two integer values for $x$, the group must only contain $4,5$. So we must have $$ 5 \le x \implies 15\le 3x \implies 16\le 3x+1<a+2$$ So we get $$16<a+2 \implies 14<a $$ The other one is similar.
Martin
- 1 Answer
- 424 views
- Pro Bono
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