### Video Transcript

Solve the inequality negative four multiplied by π¦ plus four minus 12 is less than negative 50 minus 47 minus π¦ in the set of rational numbers.

To solve this inequality, we need to isolate the variable, which is π¦. We notice that the π¦-variable currently appears on both sides of the inequality. So weβll need to collect these terms together at some point.

First though, weβll distribute the negative four on the left-hand side of the inequality over the parentheses to give negative four π¦ minus 16 minus 12 on the left-hand side. We also need to distribute the negative sign, which is representing a factor of negative one, over the parentheses on the right-hand side of the inequality. This gives negative 50 minus 47 plus π¦. Remember, itβs plus π¦ here because weβre multiplying negative π¦ by negative one.

We can then simplify the constants on each side of the inequality to give negative four π¦ minus 28 is less than negative 97 plus π¦. Next, weβll group the π¦-terms on the same side of the inequality. Subtracting π¦ from both sides eliminates the π¦-term on the right-hand side and gives negative five π¦ minus 28 is less than negative 97. The next step is to isolate the π¦-term by adding 28 to both sides of the inequality, giving negative five π¦ is less than negative 69.

The final step is to divide both sides of the inequality by negative five. But we need to be really careful here. When we multiply or divide both sides of an inequality by a negative value, this reverses the direction of the inequality. So, on the left-hand side, dividing by negative five gives π¦. On the right-hand side, we have negative 69 over negative five. And we reverse the inequality sign from less than to greater than, giving π¦ is greater than negative 69 over negative five. We can then cancel the factors of negative one in the numerator and denominator on the right-hand side to give 69 over five.

Finally, as weβre asked to solve this inequality in the set of rational numbers, we can give our answer in set notation as the set of all values of π¦ such that π¦ is a rational number and π¦ is greater than 69 over five. Itβs worth pointing out that the issue of dividing by a negative value could also be avoided if we chose to collect the π¦-terms on the other side of the inequality. If at an earlier stage we had added four π¦ to both sides instead of subtracting π¦ from both sides, we would have obtained negative 28 is less than negative 97 plus five π¦. Then, adding 97 to both sides would give 69 is less than five π¦. Dividing both sides by five would give 69 over five is less than π¦.

Itβs conventional, however, to give the solution to an inequality with the variable on the left-hand side. So we would then need to reverse the two sides of the inequality, taking care to ensure the sign still pointed towards the value of 69 over five, to give π¦ is greater than 69 over five. This is equivalent to our solution and avoided the need to divide by a negative. But it did require us to carefully reverse the two sides of the inequality at the end.