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Solve the simultaneous equations 𝑥 over six plus 𝑦 over six equals one and 𝑥 over six plus 𝑦 over nine equals 17 over two.
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We’re looking for the point for 𝑥 and 𝑦 that would make both of these statements true.
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Now, when I notice these fractions, the first thing that I think to do will be to get rid of the fractions before we try to solve.
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We’re trying to get rid of six, nine, and two in the denominator.
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And that means we need to multiply through by their least common multiple.
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Six, nine, and two are all divisible by 18.
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So, let’s multiply both of these equations through by 18.
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18 times 𝑥 over six equals 18𝑥 over six.
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And 18 divided by six equals three.
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So, we now have three 𝑥.
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The same thing is true for 𝑦.
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18 over six 𝑦 equals three 𝑦.
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And 18 times one equals 18.
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And then for the second equation, 18 times 𝑥 over six will be three 𝑥, just like in the first equation.
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18 times 𝑦 over nine, it will be 18 over nine 𝑦, which equals two 𝑦.
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18 divided by nine equals two.
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And then we multiply 18 by 17 over two.
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18 times 17 over two can be reduced.
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18 divided by two is nine.
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Now we need to multiply nine times 17, which equals 153.
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We can label our equations equation one and equation two.
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We want to subtract equation two from equation one.
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When we do that, we’ll say three 𝑥 minus three 𝑥 equals zero.
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Three 𝑦 minus two 𝑦 equals 𝑦.
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Zero plus 𝑦 just equals 𝑦.
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Be careful here.
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We have 18 and 153.
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And we’re subtracting.
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18 minus 153 equals negative 135.
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Our 𝑦 equals negative 135.
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We can take equation one and we can plug 𝑦 in.
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We can plug in negative 135 for 𝑦.
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However, there’s one thing we can do to make this process a little bit simpler.
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All of our coefficients are divisible by three.
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And so, I want to divide every part of equation one by three.
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Three 𝑥 divided by three equals 𝑥.
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Three 𝑦 divided by three equals 𝑦.
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18 divided by three equals six.
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And at this point, I’m gonna plug in negative 135 for 𝑦.
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𝑥 plus negative 135 equals six.
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To solve for 𝑥, we add 135 to both sides.
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And 𝑥 equals six plus 135.
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𝑥 equals 141.
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These lines cross their simultaneous equations at the point 𝑥 equals [141] and 𝑦 equals negative 135.
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I wanna address what would happen if you hadn’t gotten rid of those fractions all the way at the beginning.
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What would have happened if we hadn’t multiplied through by 18?
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We’ll start with the same two equations.
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I see that there’s 𝑥 over six in both of these equations.
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And that makes me think I could substitute one equation in to the other.
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So, I subtract 𝑦 over six from both sides of the first equation, which gives me a new equation.
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𝑥 over six equals one minus 𝑦 over six.
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This is our new equation one.
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And we’re gonna substitute what we found for 𝑥 into equation two.
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We know that 𝑥 over six equals one minus 𝑦 over six.
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And that means, in place of 𝑥 over six in the second equation, we’ll write one minus 𝑦 over six.
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At this point, we’ll try and combine like terms, negative 𝑦 over six plus 𝑦 over nine.
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We can’t add fractions that don’t have a common denominator.
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We need to give them a new common denominator of 18.
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Negative 𝑦 over six is equal to negative three over 18.
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Positive 𝑦 over nine equals two 𝑦 over 18.
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From there, we can add these two fractions together by adding their numerators.
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Negative three 𝑦 plus two 𝑦 equals negative 𝑦.
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And the denominator didn’t change.
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It stays 18.
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To solve for 𝑦, we need 𝑦 by itself.
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So, we subtract one from both sides of the equation.
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Negative 𝑦 over 18 equals 17 over two minus one.
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One is the same thing as two over two.
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So, I’ll rewrite that.
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Then, these fractions have a common denominator.
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And I can say 17 minus two equals 15 and the denominator stays the same, two.
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Then, we multiply both sides of the equation by 18.
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Negative 𝑦 equals.
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We can simplify 18 divided by two equals nine.
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Nine times 15 equals 135.
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Negative 𝑦 equals 135.
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And that means positive 𝑦 equals negative 135.
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We’re still not finished, though, because we only know our 𝑦-value.
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We’ll have to plug this 𝑦 value back into equation one.
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Here, we have this.
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𝑥 over six equals one minus negative 135 over six.
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Pay attention here.
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We’re subtracting a negative.
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So, we can make that addition.
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And in place of the whole number one, we can write the fraction six over six.
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And then we can combine these two numerators.
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Six plus 135 equals 141.
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The denominator didn’t change.
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It stays six.
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If we multiplied through by six, the sixes would cancel out.
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And we’d be left with 𝑥 equals 141.
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Both of these methods work.
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However, I usually find that when we’re working with fractions, there are more chances for error.
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And it’s often easier to deal with the fractions in the first step.
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One final thing we can do is plug these solutions in and see if we get true statements.
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Is 141 over six plus negative 135 over six equals to one?
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When we add those two fractions, we get six over six.
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Six over six is equal to one.
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The next one, 141 over six plus negative 135 over nine, is equal to 17 over two.
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We wanna know if that is true.
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To check this, I’m just gonna plug it into the calculator.
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141 divided by six equals 23 and a half.
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Negative 135 divided by nine equals negative 15.
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17 divided by two equals eight and a half.
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And 23 and a half minus 15 equals eight and a half.
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And our solution is 𝑥 equals 141.
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𝑦 equals negative 135.