### Video Transcript

Charlotte wants to make dresses and suits. Each dress or suit will have the same quantity of cloth and the same number of buttons. The following inequality represents the number of dresses π· and the number of suits π that she can make with 25 square meters of cloth: five π· plus seven π is less than 25. Additionally, the following inequality represents the number of dresses π· and the number of suits π that she can make with 100 buttons: 12π· plus 18π is less than 100. Given that she has 25 square meters of cloth and 100 buttons, does she have enough cloth to make two dresses and three suits?

Letβs begin by identifying some of the key pieces of information in this question. Sheβs making dresses and suits. The first inequality weβre interested in gives us information about the number of dresses and suits she can make with 25 square meters of cloth. Itβs five π· plus seven π is less than 25. The inequality that represents the number of items she can make with 100 buttons is 12π· plus 18π is less than 100. We need to use all of this information to establish whether she will have enough cloth to make two dresses and three suits. So what weβll do is draw each of these inequalities on the coordinate plane. Letβs clear some space to do so.

Now, before we add these to the coordinate plane, there are two more inequalities that we need to construct. We know that the number of dresses that she makes and separately the number of suits π that she makes cannot be negative. We canβt have a negative number of items. This means that the number of dresses and the number of suits must individually be part of the set of natural numbers. In inequality form, we can write this as π· must be greater than or equal to zero and π must be greater than or equal to zero. Since both π· and π are nonnegative, we can consider the construction of our inequalities purely in the first quadrant as shown. Then, we can draw the lines that represent π· equals zero and π equals zero.

If we designate the π₯-axis to be π·, the number of dresses, and the π¦-axis to be π, the number of suits, then the line π equals zero is the π₯-axis and the line π· equals zero is the π¦-axis. We wonβt shade the whole region just yet. But we know since π· must be greater than or equal to zero, we need to shade the right-hand side of the line π· equals zero. Similarly, to represent the inequality π is greater than or equal to zero, we shade above the line π equals zero.

With that completed, letβs now add the first inequality to our diagram. Letβs begin by plotting the line five π· plus seven π equals 25 on the coordinate plane. We want to find the points at which it intersects the π·-axis and the π-axis. So weβll individually let π· and π be equal to zero and solve for the remaining unknown. If π· is equal to zero, our equation becomes seven π equals 25, meaning π is 25 over seven. This line therefore passes through the π-axis or the π¦-axis a little bit above 3.5. Then, if we let π be equal to zero, we get five π· equals 25 which means π· equals five. And our line passes through the π·-axis or the π₯-axis at five.

We add this line to our diagram. And at this point, we know we have a strict inequality. Five π· plus seven π is strictly less than 25. And so we add a dotted line in the place of five π· plus seven π equals 25. We need to establish which side of the line satisfies this inequality. And so we choose a point that lies either side of this line and then we substitute it into the inequality and see if it satisfies it.

Itβs always sensible where possible to choose the point zero, zero. So we let π· equals zero and π equals zero, which means that five π· plus seven π is also zero. This is indeed less than 25 as we required. So the side of the line that the point zero, zero lies on satisfies our inequality. We can therefore shade the entire region on this side of the line. Now, we actually wonβt go past the π₯- and π¦-axes because we know π is greater than zero and π· is greater than zero.

Now, we have enough to answer the question, βDoes she have enough cloth for two dresses and three suits?β To do so, weβll plot the point two, three on the coordinate plane. If it lies in the shaded region, then we know she does have enough cloth. If it doesnβt, then she does not. Well, the point two, three does not lie in our shaded region. So we can confirm she does not have enough cloth for two dresses and three suits. And in fact, we can check this by substituting π· equals two and π equals three into the inequality. When π· is two and π is three, five π· plus seven π is 31. This is greater than 25. So this number of dresses and suits does not satisfy the inequality.

So what was the point in drawing this on the coordinate plane then? Well, plotting inequalities on the coordinate plane allows us to answer problems that involve more than one inequality. So letβs draw the graph of 12π· plus 18π equals 100. The line 12π· plus 18π equals 100 looks like this. Once again, we draw a dotted line because we have a strict inequality. And if we were to substitute π· equals zero and π equals zero into our inequality, we would find it is indeed less than 100. Zero plus zero is zero, which is less than 100. So we shade the side of the line that contains the point zero, zero. We can now see that 100 buttons is enough for two dresses and three suits, since the point two, three lies on the side of the line that satisfies our inequality.

Weβre also now able to see the region that satisfies all four of our inequalities. Itβs this triangle on the lower left-hand side of our diagram. We know that any point that lies inside this triangle satisfies all four of our inequalities. So, for instance, the point one, one satisfies all four inequalities, meaning that she has enough cloth and enough buttons to make one dress and one suit. However, we have done enough to answer this question. She does not have enough cloth for two dresses and three suits.