Question Video: Solving a System of Linear Equations in Two Unknowns with Fraction Coefficients | Nagwa Question Video: Solving a System of Linear Equations in Two Unknowns with Fraction Coefficients | Nagwa

# Question Video: Solving a System of Linear Equations in Two Unknowns with Fraction Coefficients Mathematics • Third Year of Preparatory School

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Solve the simultaneous equations (𝑥/6) + (𝑦/6) = 1 and (𝑥/6) + (𝑦/9) = 17/2.

08:09

### Video Transcript

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.

We’re looking for the point for 𝑥 and 𝑦 that would make both of these statements true. 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. We’re trying to get rid of six, nine, and two in the denominator. And that means we need to multiply through by their least common multiple. Six, nine, and two are all divisible by 18. So, let’s multiply both of these equations through by 18. 18 times 𝑥 over six equals 18𝑥 over six. And 18 divided by six equals three. So, we now have three 𝑥. The same thing is true for 𝑦. 18 over six 𝑦 equals three 𝑦. And 18 times one equals 18. And then for the second equation, 18 times 𝑥 over six will be three 𝑥, just like in the first equation. 18 times 𝑦 over nine, it will be 18 over nine 𝑦, which equals two 𝑦. 18 divided by nine equals two. And then we multiply 18 by 17 over two. 18 times 17 over two can be reduced. 18 divided by two is nine. Now we need to multiply nine times 17, which equals 153.

We can label our equations equation one and equation two. We want to subtract equation two from equation one. When we do that, we’ll say three 𝑥 minus three 𝑥 equals zero. Three 𝑦 minus two 𝑦 equals 𝑦. Zero plus 𝑦 just equals 𝑦.

Be careful here. We have 18 and 153. And we’re subtracting. 18 minus 153 equals negative 135. Our 𝑦 equals negative 135. We can take equation one and we can plug 𝑦 in. We can plug in negative 135 for 𝑦. However, there’s one thing we can do to make this process a little bit simpler. All of our coefficients are divisible by three. And so, I want to divide every part of equation one by three. Three 𝑥 divided by three equals 𝑥. Three 𝑦 divided by three equals 𝑦. 18 divided by three equals six. And at this point, I’m gonna plug in negative 135 for 𝑦. 𝑥 plus negative 135 equals six. To solve for 𝑥, we add 135 to both sides. And 𝑥 equals six plus 135. 𝑥 equals 141.

These lines cross their simultaneous equations at the point 𝑥 equals [141] and 𝑦 equals negative 135.

I wanna address what would happen if you hadn’t gotten rid of those fractions all the way at the beginning. What would have happened if we hadn’t multiplied through by 18? We’ll start with the same two equations. I see that there’s 𝑥 over six in both of these equations. And that makes me think I could substitute one equation in to the other. So, I subtract 𝑦 over six from both sides of the first equation, which gives me a new equation. 𝑥 over six equals one minus 𝑦 over six. This is our new equation one. And we’re gonna substitute what we found for 𝑥 into equation two. We know that 𝑥 over six equals one minus 𝑦 over six. And that means, in place of 𝑥 over six in the second equation, we’ll write one minus 𝑦 over six.

At this point, we’ll try and combine like terms, negative 𝑦 over six plus 𝑦 over nine. We can’t add fractions that don’t have a common denominator. We need to give them a new common denominator of 18. Negative 𝑦 over six is equal to negative three over 18. Positive 𝑦 over nine equals two 𝑦 over 18. From there, we can add these two fractions together by adding their numerators. Negative three 𝑦 plus two 𝑦 equals negative 𝑦. And the denominator didn’t change. It stays 18.

To solve for 𝑦, we need 𝑦 by itself. So, we subtract one from both sides of the equation. Negative 𝑦 over 18 equals 17 over two minus one. One is the same thing as two over two. So, I’ll rewrite that. Then, these fractions have a common denominator. And I can say 17 minus two equals 15 and the denominator stays the same, two. Then, we multiply both sides of the equation by 18. Negative 𝑦 equals. We can simplify 18 divided by two equals nine. Nine times 15 equals 135. Negative 𝑦 equals 135. And that means positive 𝑦 equals negative 135.

We’re still not finished, though, because we only know our 𝑦-value. We’ll have to plug this 𝑦 value back into equation one. Here, we have this. 𝑥 over six equals one minus negative 135 over six. Pay attention here. We’re subtracting a negative. So, we can make that addition. And in place of the whole number one, we can write the fraction six over six. And then we can combine these two numerators. Six plus 135 equals 141. The denominator didn’t change. It stays six. If we multiplied through by six, the sixes would cancel out. And we’d be left with 𝑥 equals 141.

Both of these methods work. However, I usually find that when we’re working with fractions, there are more chances for error. And it’s often easier to deal with the fractions in the first step. One final thing we can do is plug these solutions in and see if we get true statements. Is 141 over six plus negative 135 over six equals to one?

When we add those two fractions, we get six over six. Six over six is equal to one. The next one, 141 over six plus negative 135 over nine, is equal to 17 over two. We wanna know if that is true. To check this, I’m just gonna plug it into the calculator. 141 divided by six equals 23 and a half. Negative 135 divided by nine equals negative 15. 17 divided by two equals eight and a half. And 23 and a half minus 15 equals eight and a half. And our solution is 𝑥 equals 141. 𝑦 equals negative 135.

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