University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
In mathematics, it can be useful limit the solution or even have multiple solutions for an inequality. For this we use acompound inequality, inequalities with multiple inequality signs. When solving compound inequalities, we use some of the same methods used in solving multi-step inequalities. The solutions to compound inequalities can be graphed on a number line, and can be expressed asintervals.
Solving a three part inequality, basically it's the same as solving a two part inequality but instead of now dealing with just one side we're dealing with another side as well. So first step is always just to get our x's by itself, subtract one over we're trying to solve for this x so just subtract 1 and instead of doing it from both sides we now have to do it from all three. So 11 minus 1, 10 less than 3x and then lastly it equal to 6 okay. Solving for x, we need to divide by negative 3, remember when we are in an inequality form if we divide by a negative we have to actually flip that sign. So 10 divided by a negative 3 we can't simplify that so we're just left with negative 10 thirds our sign switches x this sign has to switch as well 6 divided by negative 3 is negative 2 okay.
So what we actually have is x has to be between negative 10 thirds and -2, so this answer right now is in inequality form okay. We want to throw it into another form we could do our brackets and remember this is called "interval noation" we are not including negative ten thirds so we have a soft bracket here we are including -2 so we have this hard bracket here okay. We could also plot this on a number line we are going from negative ten thirds to negative 2, including negative 2 so that gets filled in not including ten thirds so that doesn't, filling an outline okay. Set bill notation again just taking the same exact thing here with a x such that bracket, x such that negative ten thirds is less than x is less than equal to -2. So solved it out and the four different ways of representing the same exact answer.
