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# Compound inequalities: AND

CCSS.Math:

## Video transcript

solve for y we have 3y plus 7 is less than 2 y + 4 y plus 8 is greater than negative 48 so we have to find all the Y's that meet both of these constraints so let's just solve for y in each of the constraints and just remember that this end is here so we have 3y plus 7 is less than 2 y so let's isolate the Y's on the left-hand side so let's get rid of this 2y on the right-hand side and we can do that by subtracting 2y from both sides so we're going to subtract 2y from both sides the left hand side we have 3y minus 2y which is just y plus 7 is less than 2 y minus 2y and there's nothing else there that's just going to be 0 and then we can get rid of this 7 here by subtracting 7 from both sides so let's subtract 7 from both sides left hand side y plus 7 minus 7 those cancel out we just have a y is less than 0 minus 7 which is negative 7 so that's one of the constraints that's this constraint right over here now let's work on this constraint we have 4y plus 8 is greater than negative 48 so let's get rid of the 8 from the left-hand side so we can subtract 8 from both sides subtract 8 from both sides the left hand side we're just left with a 4y because these guys cancel out 4y is greater than negative 48 minus 8 we're going to go another 8 negative so 48 plus 8 would be would be a 56 so this is going to be negative 56 negative 56 and now to isolate the Y we can divide both sides by positive 4 and we don't have to swap the inequality since we're dividing by a positive number so it's divide both sides by 4 over here so we get y we get Y is greater than what is 56 over 4 negative 56 over 4 let's see 40 is 10 times 4 and then we have another 16 to worry about so it's 14 times 4 so Y is greater than Y is great later than negative 14 is that right 4 times 10 is 40 4 times 4 is 16 yep 56 so Y could be greater than Y is greater than negative 14 and let's remember we have this and here and Y is less than negative 7 so we have to meet both of these constraints over here so let's draw them let's draw them on the number line so I have my number line over here and let's say negative 14 is over here negative 14 so you have negative 13 12 11 10 9 8 7 that's negative 7 and the negative 6 5 4 3 2 1 this would be 0 and then you could keep going up more positive and so we're looking for all the Y's that are less than negative 7 so let's look at this less than negative 7 so not including negative 7 so we'll do an open circle around negative 7 and less than negative 7 and if that was the only constraint we would keep going to the left but we have this other constraint and Y has to be greater than negative 14 and Y is greater than negative 14 so you make a circle around negative 14 and everything that's greater than that and if you didn't have this other constraint you would keep going but the the Y's that satisfy both of them are all of the Y's in between these are the Y's these are the Y's that are both less than negative 7 and greater than negative 14 and we can verify that things here work so let's try some values out so a value that would work well let me just do negative 10 is right here 8 9 this is negative 10 that should work so let's try it out so we'd have 3 times negative 10 3 times negative 10 plus 7 should be less than 2 times negative 10 so this is negative 30 plus 7 is negative 23 which is indeed less than negative 20 so that works and negative 10 has to work for this one as well so you have 4 times negative 10 which is negative 40 plus 8 negative 40 plus 8 should be greater than negative 48 well negative 40 plus 8 is negative 32 we're going we're going eight in the positive direction so we're getting less negative and negative thirty-two is greater than negative 48 it's less negative so this works so negative ten works now let's just verify things that shouldn't work so zero should not work zero should not work it's not in the solution set so let's try it out we fit three times zero plus 7 that would be 7 and 7 is not less than zero so it would violate this condition right over here if you put a zero over here if you put a negative 15 over here it should violate this condition right over here because it wasn't in this guy's solution set anyway hopefully you found that useful