How do you solve this problem for X: -3(x+1)-x/2+3/2<0

you distribute the -3, and add like terms...you solve it like you would if the less than sign is an equal sign

only difference is if you divide or multiply by a negative you change the sign...so in this case you would change it to a greater than sign if you divide or multiply by a negative

ok what terms do you distribute the -3 to?

just -3(x+1) since the -3 is in front of that parenthesis

ok so it would be 3x+3?

but there is a negative...so u distribute the - as well, -3x-3

so then it would look like this them -3x-3-x/2+3/2<0 ?

yup

yup

add like terms...keep all the x's on one side of the inequality and everything else on the other

ok so like this -3x+x/2<0-3+3/2?

you should be adding 3 to both sides , not subtracting

so it should be -3x-x/2<3-3/2

so it would look like this -3/2x<1.5?

how would you evaluate -3-1/2? having the x is the same thing. if it would be easier for you to get rid of the fraction, you could multiple the whole inequality by 2 to get rid of the fractions

ok i just get how to even do these and you are making me even moe confused.

oh sorry...i'm trying to explain it to you instead of just giving you the answers

on the left side of the inequality we have -3x-x/2, right?

the x in x/2 can also be written as (1/2)x. let's change that to a decimal so it's easier to work with...(1/2)x is also 0.5x. So we have -3x-0.5x which equals -3.5x.

so now you should have -3.5x<1.5

now to solve for x, you would just divide both sides by -3.5, which will change the inequality sign to >

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