### Video Transcript

Letβs talk about solving equations with rational numbers. Before we do that, we need to talk about a few key concepts. The first one is the reciprocal; itβs also called the multiplicative inverse. The reciprocal is whatever you multiply a number by and the product becomes one. Itβs represented here by π times one over π. When you multiply those together, π over π equals one. five times one-fifth equals one. one-fifth is the reciprocal or multiplicative inverse of five.

Next step, the addition property of equality. This property states that if π equals π, then π plus π equals π plus π. Hereβs an example of using the addition property of equality to solve an equation. Iβve added five to both sides here to solve for π₯. Sometimes we might say if you do something to one side of the equation then you have to do the same thing to the other. When we say this, we are using the addition property of equality.

Our third and final key concept, the multiplication property of equality which is similar to the addition property of equality. It states if π equals π then π times π equals π times π. Hereβs our example: π divided by five equals seven. To solve for π, Iβll multiply by five on both sides of the equation, and Iβm using the multiplication property of equality to do this.

Now on to solving equations. Example one: one-half π₯ equals negative five-sixths. The goal here is to solve for π₯. And we need to do that by first using the reciprocal to get π₯ by itself. Hereβs what that looks like: I multiply both sides by two over one. Once I do that, I get π₯ equals negative ten-sixths. But this is not the simplified form of this fraction, and we always wanna keep the fractions in simplest form. The simplified form is negative five-thirds. To find this, I divided negative ten-sixths by two on the top and two on the bottom.

Next, π₯ minus one-seventeenth equals five-seventeenths. Weβre gonna solve this one with the addition property of equality. By adding one-seventeenths to both sides of the equation, our final answer becomes π₯ equals six-seventeenths.

Example three: when three-fourths is divided by π over π, the result is five-eighths. Solve for π over π. The first step is to turn this word problem into an equation. Weβve done that here by writing three-fourths divided by π over π equals five-eighths. Donβt forget! When we divide by a fraction, that means weβre multiplying by the reciprocal. So thatβs what weβre going to do; weβre going to change the division to multiplication and change the π over π to π over π. Just gonna slide the problem up so we can keep going. Weβre trying to get π over π by itself, so I multiplied by four-thirds on both sides. After multiplying by four-thirds on both sides, weβre left with π over π equals twenty over twenty-four. Again, we always want the most simple form that we can find, so I know that twenty over twenty-four can be reduced. A simplified form of that fraction is five-sixths; we divided the top and the bottom by four. So we found π over π, but our question wasnβt asking us what π over π was. Our question was asking us to solve for π over π, so we flip our fraction for the final answer, and π over π equals six-fifths.

The next question says three and seven-thirteenths divided by some number equals one. Three and seven-thirteenths divided by some number equals one; I just want you to think about this problem for a minute. What do you think should go there? How do you think we should solve this problem? If youβre still not sure, hereβs a hint: three divided by what equals one, or five divided by what equals one. Itself! three divided by three equals one and five divided by five equals one. Anything divided by itself equals one. We only had to remember that fact to answer this question. The answer here is three and seven-thirteenths because three and seven-thirteenths divided by itself equals one.

Example five: fourteen twenty-sevenths divided by some number equals one and five-sixths, so we get fourteen twenty-sevenths divided by π equals eleven-sixths. Now Iβve have made two changes: Iβve changed divided by π into multiplied by one over π. So we went from dividing by something to multiplying by its reciprocal. In order to isolate our variable, I multiplied by the reciprocal twenty-seven fourteenths on both sides of the equation. To simplify, I divided twenty-seven by three and six by three, and I got nine and two. This will help finding the simplest form of this fraction. Now weβre left with one over π equals eleven times nine over two times fourteen. When we multiply that out, we get one over π equals ninety-nine over twenty-eight. But if you look closely, youβll see that weβre not looking for one over π; weβre actually looking for π. So our final answer here has to be twenty-eight over ninety-nine.

As you solve these problems, remember the key concepts. They are your tools for solving problems like these. Your reciprocal, being π times one over π equals one. The addition property of equality, if you add something to one side, you have to add it to the other. And the same goes for multiplication, if you multiply by something on one side, you have to do the same thing on the other. And now youβre ready, so itβs your turn to try some.