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In this video, we’ll be solving multi-step linear equations.
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But we won’t be giving our answers in the form 𝑥 equals five or 𝑥 equals minus three over two and so on.
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We’ll be using set notation.
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Let’s see an example.
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Find the solution set of the equation five 𝑥 minus three equals twenty-seven.
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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.
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Adding three to the left-hand side, just five 𝑥 minus three plus three, just leaves us with five 𝑥.
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And on the right-hand side, twenty-seven plus three is thirty.
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Now I want to know what one 𝑥 is equal to.
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So I’m just gonna divide both sides by five.
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And a fifth of five 𝑥 is just one 𝑥, and a fifth of thirty is six.
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So the value of 𝑥 that satisfies the equation is six.
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But the question said: Find the solution set of the equation.
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So we have to present that in set notation.
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Now there’s only one value of 𝑥 that satisfies that equation.
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So our solution set only contains one element and that is six.
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So the answer is six.
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Next then.
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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.
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So first, we’re just gonna go ahead and solve this equation like we normally would do.
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So we wanna get 𝑥 on its own on the left-hand side.
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So we’re gonna subtract five from both sides, which leaves us with just three 𝑥 on the left and ten on the right.
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And then we’re gonna divide both sides by three.
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So that gives us 𝑥 on the left-hand side and ten over three on the right-hand side.
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Or, three and a third, if you prefer to write it like that.
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So the value of 𝑥 that satisfies that equation is three and a third.
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But the question says that 𝑥 is an element of the set of natural numbers and they don’t include fractional numbers.
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They just are counting numbers one, two, three, four, and five, and so on.
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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.
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So we have to say that the solution set is empty.
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It’s the null set.
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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.
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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.
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So we’ve gotta try them all out in turn and then make a list of which ones satisfy that equation.
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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.
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And seven times negative four is negative twenty-eight.
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So it’s twenty-nine take away negative twenty-eight.
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So that’s twenty-nine plus twenty-eight, which is equal to fifty-seven.
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So the left-hand side is equal to fifty-seven, but we wanted it to equal eight.
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And fifty-seven obviously isn’t equal to eight.
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So 𝑥 equals minus four is not a solution in our solution set.
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Okay.
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Let’s try 𝑥 equals zero.
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So the next one is twenty-nine minus seven times zero.
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Well, seven times zero is zero.
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So that becomes twenty-nine minus zero, which is just twenty-nine.
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But we wanted the answer to be eight not twenty-nine.
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So again, zero is not in the solution set.
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Let’s try 𝑥 equals three.
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The left-hand side becomes twenty-nine minus seven times three.
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And seven times three is twenty-one.
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So that’s twenty-nine minus twenty-one, which is equal to eight.
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Ah!
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Good news!
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That is the solution we’re looking for, so it looks like the value 𝑥 equals three is in our solution set.
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Let’s go on and try the last one, 𝑥 equals ten.
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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.
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But negative forty-one isn’t eight, which was the answer we were looking for.
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So ten is not in our solution set.
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So the only element of the substitution set which gave us an answer, which was correct according to that equation, was 𝑥 equals three.
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So the solution set only contains one element and that is the number three.
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Okay.
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Let’s look at this question then.
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Find the solution set of four 𝑥 minus three equals two 𝑥 plus seven using the substitution set negative five, zero, two, five.
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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.
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Right.
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When 𝑥 is equal to negative five, the left-hand side becomes four times negative five minus three.
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And four times negative five is negative twenty.
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So that’s negative twenty minus three.
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And negative twenty minus three is negative twenty-three.
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Okay.
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Let’s try 𝑥 equals minus five on the right-hand side.
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Substituting negative five into two 𝑥 plus seven, that’s two times negative five plus seven.
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Two times negative five is negative ten.
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So that becomes negative ten plus seven, which is negative three.
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And negative twenty-three on the left-hand side, negative three on the right-hand side, they’re not equal.
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So it looks like 𝑥 equals negative five is not in the solution set.
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Let’s move on to the next one 𝑥 equals zero.
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Substituting 𝑥 equals zero into the left-hand side, four 𝑥 minus three becomes four times zero minus three.
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Four times zero is obviously zero.
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So that becomes zero minus three, which is just negative three.
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And on the right-hand side, two 𝑥 plus seven becomes two times zero plus seven, when 𝑥 is zero.
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Two times zero is zero.
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So that’s zero plus seven, which is just seven.
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So when 𝑥 is zero, the left-hand side becomes negative three and the right-hand side becomes seven; and they’re not equal.
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So it looks like 𝑥 equals zero isn’t in our solution set either.
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Okay.
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Let’s try 𝑥 equals two.
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When 𝑥 equals two, four 𝑥 minus three becomes four times two minus three.
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Now four times two is eight.
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So that’s eight minus three, which is five.
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And the right-hand side becomes two times two plus seven, which is eleven.
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So when 𝑥 equals two, the left-hand side is five and the right-hand side is eleven.
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So that’s not equal.
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So that doesn’t satisfy the equation.
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So 𝑥 equals two is not in our solution set.
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We just got one more chance to get it right with 𝑥 equals five.
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And then the left-hand side becomes four times five minus three, which is seventeen.
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And the right-hand side becomes two times five plus seven, which is seventeen.
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So now, the left-hand side is seventeen and the right-hand side is seventeen; they are equal.
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So 𝑥 equals five is in our solution set and the solution set consists of just one element, the number five.
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Lastly then, let’s look at this question.
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Find the solution set of 𝑥 over five plus two equals negative one in ℤ, the set of integers.
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So that’s positive and negative integers.
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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.
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𝑥 over five plus two equals negative one.
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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.
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So 𝑥 over five plus two minus two is just 𝑥 over five.
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And if I subtract two from the right-hand side, negative one, take away another two is negative three.
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Now I can multiply both sides by five just to leave me with one whole 𝑥.
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Five lots of a fifth of 𝑥 is one whole 𝑥.
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And five lots of negative three are negative fifteen.
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So 𝑥 is equal to negative fifteen.
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Now that is an integer negative fifteen.
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So it is in the set of integers.
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So it will be in the solution set.
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So there we go, the solution set consists of just one element negative fifteen and it is an integer.