Solve the inequalities in Exercises 5 to 16 for real $x$.
$$
\frac{(2 x-1)}{3} \geq \frac{(3 x-2)}{4}-\frac{(2-x)}{5}
$$

Answer

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Video Transcript

Okay, so we have 4. 50 times X minus two is less than one third times X minus six. Okay, so if we distribute here the 4. 50 what do we have? We get 4/5 acts and four. First time. Negative two. That's gonna be a negative Eat fifth. Okay, that's less than one third distributes. So we get a one third acts and then one third times negative. Six. Well, three goes into 313 goes from the negative 62 times. So that's gonna be, um, negative two. So which is so minus two? Okay, so let's get our ex terms on one side of the equation here. So we subtract one third. I gotta get 4/5 ax minus one third X, um is less than well, we add over the eight. Fifth. So we get 8/5 um, minus two. A positive positive. 8/5. And then we have the minus two over here. Okay, so what is 4/5 X minus one third x? Well, that's become denominator of Let's see. Multiply four. Fifth by three. Over three. So that becomes 12. 15 x minus. Um uh, multiply one third x by 5/5. So minus 5/15 X eyes less than while we get eight. Fifth. Multiply a negative to buy, uh, five or five. That's minus 10 fifth. Okay, now we can combine in terms here. So 12. 15 X minus 5. 15 is 12, 11, 10, 98 7/15. So, 7/15 X, um is less than 8/5 minus 15th. So that's a negative to fifth. Okay, So, um, to solve for X, we just well divide both sides by 7. 15, which is the same thing is multiplying by 15 sevens, right? 15 7th times and 15th is just one. So we get X, um, is still less than inequality is not flip because we're not dividing or multiplying by a negative. So X is going to be less than, um well, 15 sevens times negative to fifth. Okay, so what's that? 5.515 15, 3 times. So that's negative. Three times to his negative 6/7. So therefore, X is less than negative. 6/7. Okay, so X is less than, um negative 67 So what would you say? That in interpretation would be? Well, X. Give me a smallest possible X is less than negative. 67 So you go from negative infinity up to negative 6/7. Okay, so to show that on a number line right there is, by the way, so that a number line we can go, Here's our number line. Okay, let's say let's say here's zero. So, um, we go up to the biggest X. Could be is negative 6/7 and it's strictly less than so. Let's say, um, negative 67 is what a little bit native 77 would be one negative one. Um, S O negative 67 says, Let's say somewhere, let's just say it's right here. This is negative. 6/7 sevens on Ben. The better the backs is anything less than that, shall we shade to the left. All right, take care

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