### Video Transcript

Find the number that should be added to each of the numbers 32, nine, 22, and four to make them proportional.

The first thing we wanna do is go ahead and set up a rough proportion: 32 over nine and 22 over four. If these two were proportional, they would be equal. We want to add some unknown value to all four of these numbers, but we want to add the same number to all four of these numbers.

We could write an equation that says 32 plus π₯ over nine plus π₯ is equal to 22 plus π₯ over four plus π₯. And then, we need to solve for π₯. To solve for π₯, we canβt have any π₯-values in the denominator. To remove the π₯ from the denominator on the left, Iβll multiply the left side of our equation by nine plus π₯ over one. Nine plus π₯ over one times nine plus π₯ in the denominator is equal to one. The only thing left on the left side is 32 plus π₯.

However, if we multiply the left-hand side by something, we have to multiply the right-hand side by the same amount. We multiply that side by nine plus π₯ over one. The right side now says 22 plus π₯ times nine plus π₯ over four plus π₯.

We follow this same process to get four plus π₯ out of the denominator, multiplying by four plus π₯ over one on the right and the left. On the right, four plus π₯ over four plus π₯ cancels out. Our new equation looks like this: four plus π₯ times 32 plus π₯ equals 22 plus π₯ times nine plus π₯.

We need to FOIL both sides. Four times 32 equals 128. Four times π₯ equals four π₯. Then, we move to the inside π₯ times 32 equals 32 π₯ and finally π₯ times π₯ equals π₯ squared. Letβs clear this up just a little bit, copy down our π₯ squared. Four π₯ plus 32π₯ equals 36π₯ plus 128. Now, we FOIL the left side. 22 times nine equals 198, 22 times π₯ equals 22π₯, π₯ times nine equals nine π₯, and π₯ times π₯ equals π₯ squared. Well, again, rewrite this and clean it up: π₯ squared plus 22π₯ plus nine π₯ equals 31π₯ plus 198.

Iβm just gonna bring this back up to the top to give us a little bit more room to solve for π₯. I notice that both the left- and the right-hand side has π₯ squared. I can subtract π₯ squared from the left and the right and itβs cancelled out on both sides. This tells us 36π₯ plus 128 is equal to 31π₯ plus 198.

We can subtract 31π₯ from the right side and the left side. On the right side, it cancels out, leaving us with 198. 36π₯ minus 31π₯ equals five π₯, bring down the 128. Subtract 128 from both sides. On the left, it cancels out, leaving us with five π₯ is equal to 70. If we divide both sides of the equation by five, the right is left with π₯. 70 divided by five equals 14. π₯ equals 14.

And if we did everything correctly, 32 plus 14 over nine plus 14 should be equal to 22 plus 14 over four plus 14. 32 plus 14 equals 46, nine plus 14 equals 23, 22 plus 14 equals 36, four plus 14 equals 18, 46 over 23 equals two, 36 over 18 equals two, and two is equal to two.

So we found a correct solution. If you add the number 14 to all four of these terms, you make them proportional.