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Inequalities with variables on both sides (with parentheses)

Sal solves the inequality 5x+7>3(x+1), draws the solution on a number line and checks a few values to verify the solution. Created by Sal Khan and Monterey Institute for Technology and Education.
Video transcript
solve for x and we have five x plus seven is greater than three times x plus one so lets just try to isolate x on one side of this inequality but before we do that lets just simplify this right hand side so we get five x plus seven is greater than, lets distribute this three, so three times x plus one is the same thing as three times x plus 3 times one so it's going to be three x plus three times one is three now if we want to put our xs on the left hand side we can subtract three x from both sides that'll get rid of this three x on the right hand side so let's do that, lets subtract three x from both sides and we get on the left hand side, five x minus 3 x is two x plus 7, is greater than 3x-3x, those cancel out, that was the whole point behind subtracting three x from both sides, is greater than 3 now we can subtract 7 from both sides to get rid of this positive 7 right over here so lets subtract 7 from both sides and we get, on the left hand side, two x plus 7 minus 7 is just 2x is greater than 3 minus 7, which is negative 4 and then lets see, we have 2x is greater than negative four if we just wanted x over here we could divide both sides by two, since 2 is a positive number we don't have to swap the inequalities so let's just divide both sides by two and we get x is greater than negative four dividied by two is negative two so the solution will look like this, draw the number line I can draw a straighter number line than that there we go still not that great, but it'll serve our purposes so say that's negative 3, negative 2, negative 1, zero one two three, x is greater than negative two it does not include negative two it is not greater than or equal to negative two so we have to exclude negative two and we exclude negative two by drawing an open circle at negative two but then all the values greater than that are valid xs that would solve, that would satisfy, this inequality so anything above it will work and let's just try let's try something that should work and let's try something else that shouldn't work so zero should work, it is greater than negative two its right over here, so lets verify that 5 times zero, plus 7 should be greater than 3 times zero plus one so this is seven, because this is just a zero seven should be greater than three right, three times one, so seven should be greater than three and it definitely is now lets try something that should not work, lets try negative three so five times negative three plus 7, lets see if its greater than three times negative 3 plus one so this is negative 15, plus 7 is negative 8 that is negative 8, lets see if that is greater than negative three plus one is negative 2 times three is negative 6 negative 8 is not greater than negative 6 negative 8 is more negative than negative 6, its less than so, its good that negative 3 didn't work because we didn't include that in our solution set so we tried something that is in our solution set and it did work and something that's not and it didn't work so we're feeling pretty good