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Video: Using Set Notation to Express Solutions of Linear Equations

Tim Burnham

We solve multistep linear equations, such as 29 − 7𝑥 = 8, and express our answers as solution sets, such as {3}, rather than simply stating that 𝑥 = 3, and so on. We also explain and use the notation for the set of integers and the set of natural numbers.

08:36

Video Transcript

In this video, we’ll be solving multi-step linear equations. But we won’t be giving our answers in the form 𝑥 equals five or 𝑥 equals minus three over two and so on. We’ll be using set notation.

Let’s see an example. Find the solution set of the equation five 𝑥 minus three equals twenty-seven.

So straight of, I’m gonna add three to each side of my equation, just to leave myself with the 𝑥 term on its own on the left. Adding three to the left-hand side, just five 𝑥 minus three plus three, just leaves us with five 𝑥. And on the right-hand side, twenty-seven plus three is thirty. Now I want to know what one 𝑥 is equal to. So I’m just gonna divide both sides by five. And a fifth of five 𝑥 is just one 𝑥, and a fifth of thirty is six. So the value of 𝑥 that satisfies the equation is six. But the question said: Find the solution set of the equation. So we have to present that in set notation. Now there’s only one value of 𝑥 that satisfies that equation. So our solution set only contains one element and that is six. So the answer is six.

Next then. Given that 𝑥 is an element of the set of natural numbers, find the solution set of the equation three 𝑥 plus five equals fifteen. So first, we’re just gonna go ahead and solve this equation like we normally would do. So we wanna get 𝑥 on its own on the left-hand side. So we’re gonna subtract five from both sides, which leaves us with just three 𝑥 on the left and ten on the right. And then we’re gonna divide both sides by three. So that gives us 𝑥 on the left-hand side and ten over three on the right-hand side. Or, three and a third, if you prefer to write it like that. So the value of 𝑥 that satisfies that equation is three and a third. But the question says that 𝑥 is an element of the set of natural numbers and they don’t include fractional numbers. They just are counting numbers one, two, three, four, and five, and so on. So if that’s the only value of 𝑥 that satisfies the equation, but 𝑥 has to be a natural number, then that means there are no solutions in the solution set of the equation three 𝑥 plus five equals fifteen. So we have to say that the solution set is empty. It’s the null set.

Now we’ve got to determine the solution set of twenty-nine minus seven 𝑥 equals eight, using the substitution set negative four, zero, three and ten. So what this question boils down to is, picking which of these numbers in this substitution set satisfy the equation twenty-nine minus seven 𝑥 is equal to eight. So we’ve gotta try them all out in turn and then make a list of which ones satisfy that equation. So doing them one by one, when 𝑥 is equal to negative four, the left-hand side of the equation is twenty-nine minus seven times negative four. And seven times negative four is negative twenty-eight. So it’s twenty-nine take away negative twenty-eight. So that’s twenty-nine plus twenty-eight, which is equal to fifty-seven. So the left-hand side is equal to fifty-seven, but we wanted it to equal eight. And fifty-seven obviously isn’t equal to eight. So 𝑥 equals minus four is not a solution in our solution set.

Okay. Let’s try 𝑥 equals zero. So the next one is twenty-nine minus seven times zero. Well, seven times zero is zero. So that becomes twenty-nine minus zero, which is just twenty-nine. But we wanted the answer to be eight not twenty-nine. So again, zero is not in the solution set.

Let’s try 𝑥 equals three. The left-hand side becomes twenty-nine minus seven times three. And seven times three is twenty-one. So that’s twenty-nine minus twenty-one, which is equal to eight. Ah! Good news! That is the solution we’re looking for, so it looks like the value 𝑥 equals three is in our solution set. Let’s go on and try the last one, 𝑥 equals ten. Substituting in 𝑥 equals ten, so that equation gives us twenty-nine minus seven times ten on left-hand side, which is twenty-nine minus seventy, which is negative forty-one. But negative forty-one isn’t eight, which was the answer we were looking for. So ten is not in our solution set.

So the only element of the substitution set which gave us an answer, which was correct according to that equation, was 𝑥 equals three. So the solution set only contains one element and that is the number three.

Okay. Let’s look at this question then. Find the solution set of four 𝑥 minus three equals two 𝑥 plus seven using the substitution set negative five, zero, two, five. So again, we’ve got to try all those different values for 𝑥 and just check that the left-hand side equals the right-hand side and that it satisfies that equation.

Right. When 𝑥 is equal to negative five, the left-hand side becomes four times negative five minus three. And four times negative five is negative twenty. So that’s negative twenty minus three. And negative twenty minus three is negative twenty-three. Okay. Let’s try 𝑥 equals minus five on the right-hand side. Substituting negative five into two 𝑥 plus seven, that’s two times negative five plus seven. Two times negative five is negative ten. So that becomes negative ten plus seven, which is negative three. And negative twenty-three on the left-hand side, negative three on the right-hand side, they’re not equal. So it looks like 𝑥 equals negative five is not in the solution set. Let’s move on to the next one 𝑥 equals zero. Substituting 𝑥 equals zero into the left-hand side, four 𝑥 minus three becomes four times zero minus three. Four times zero is obviously zero. So that becomes zero minus three, which is just negative three. And on the right-hand side, two 𝑥 plus seven becomes two times zero plus seven, when 𝑥 is zero. Two times zero is zero. So that’s zero plus seven, which is just seven. So when 𝑥 is zero, the left-hand side becomes negative three and the right-hand side becomes seven; and they’re not equal. So it looks like 𝑥 equals zero isn’t in our solution set either.

Okay. Let’s try 𝑥 equals two. When 𝑥 equals two, four 𝑥 minus three becomes four times two minus three. Now four times two is eight. So that’s eight minus three, which is five. And the right-hand side becomes two times two plus seven, which is eleven. So when 𝑥 equals two, the left-hand side is five and the right-hand side is eleven. So that’s not equal. So that doesn’t satisfy the equation. So 𝑥 equals two is not in our solution set. We just got one more chance to get it right with 𝑥 equals five. And then the left-hand side becomes four times five minus three, which is seventeen. And the right-hand side becomes two times five plus seven, which is seventeen. So now, the left-hand side is seventeen and the right-hand side is seventeen; they are equal. So 𝑥 equals five is in our solution set and the solution set consists of just one element, the number five.

Lastly then, let’s look at this question. Find the solution set of 𝑥 over five plus two equals negative one in ℤ, the set of integers. So that’s positive and negative integers.

So first, let’s just go through and solve this and see what value of 𝑥 we get, and we’ll see whether it’s an integer or not. 𝑥 over five plus two equals negative one. So if I subtract two from each side of that equation, then I’ve just got a term involving 𝑥 on the left-hand side. So 𝑥 over five plus two minus two is just 𝑥 over five. And if I subtract two from the right-hand side, negative one, take away another two is negative three. Now I can multiply both sides by five just to leave me with one whole 𝑥. Five lots of a fifth of 𝑥 is one whole 𝑥. And five lots of negative three are negative fifteen. So 𝑥 is equal to negative fifteen.

Now that is an integer negative fifteen. So it is in the set of integers. So it will be in the solution set. So there we go, the solution set consists of just one element negative fifteen and it is an integer.