Can you solve the given system? | (Harvard University) |

[00:00:00]Welcome to pre-math. In this video, we have got this interesting problem that involves a system of three equations with three variables, as you can see over here. Our very first equation is a + b + c = 216.

[00:00:19]Our second equation is a + c = 3 * b. And finally, our third equation is b + c = 0.5 * a. And now our task is to find the value of a, b, and c.

[00:00:43]Please don't forget to give a thumbs up and subscribe. Let's go ahead and get started. And here's our very first step.

[00:00:49]I'm going to label the very first equation as our equation number one, the second one as our equation number two, and then finally number three. And here's our next step. Let's focus on this equation one.

[00:01:06]And here I have copied down equation one over here. And now I'm going to rearrange. I'm going to put a and c together. So, therefore, I can write a + c and then + b = 216.

[00:01:23]And now let's focus on a + c. And from equation two, a + c = 3 * b. So, I'm going to replace that one with 3 * b over here. So, therefore, I can write 3 * b + b = 216.

[00:01:44]And now let's combine the like terms.

[00:01:46]So, 4 * b = 216.

[00:01:51]And now we are going to divide both sides by 4 to isolate b. So, this 4 and 4 is gone. So, therefore, our lowercase b value turns out to be 50 4. So, thus we figured out our b value.

[00:02:08]And now our task is to find the value of a and c.

[00:02:14]And here's our next step. Let's focus on this equation three. And here's our equation three. b + c = 0.5 * a. And now we know that 0.5 could be written as 1 / 2. So, therefore, this equation is going to become b + c = 2 a / 2. And now I'm going to remove this fraction by multiplying by 2 across the board. And here we can see two and two cancels out. So, therefore, two times b + 2 * c = lowercase a. And now we know that b value is 54. So, therefore, I'm going to substitute this b value 54 over here. So, therefore, this is going to become 2 * 54 + 2 * c = 2 a. In other words, 108 + 2 * c = a. And now I'm going to move this 2 * c on the other side. So, therefore, this equation is going to become a 2 * c = 2 108.

[00:03:34]And I'm going to label this one as our equation number four.

[00:03:40]And here's our next step. Let's focus on this equation two.

[00:03:44]And here's our equation two. a + c = 2 3 * b.

[00:03:49]And we know our b value is 54. So, I'm going to replace that one with 54 over here. So, therefore we can write A + C = 3 * 50 4. So, therefore A + C is going to be equal to 162.

[00:04:13]I'm going to label this one as our equation number five. And here's our next step. Let's focus on equation five and four.

[00:04:23]And here I have copied them down. And now we are going to subtract equation four from equation five. So, I'm going to change the sign. This positive become negative, negative become positive, and positive become negative once again. And now we can see positive A and negative A they are gone. So, C + 2 C is going to give us 3 * C = 54.

[00:04:50]And now we are going to divide both sides by three twice. Let's see. This three and three is gone. So, therefore our C value turns out to be 18.

[00:05:02]So, thus our C value turns out to be 18.

[00:05:05]And here's our final step. Let's focus on this equation five. And here we can see our C value is 18. So, I'm going to replace that one with 18 over here. So, therefore equation five is going to become A + 18 = 162.

[00:05:28]And I'm going to subtract 18 from both sides to isolate A. And here we can see 18 and negative 18 they are gone. So, therefore our lower case A value turns out to be 144.

[00:05:48]So, this our lower case A value turns out to be 144 as well. So, this after all the calculations and manipulations, our solution turns out to be our A value is 144, our B value is 54, and finally our C value is 18. And that's our final answer.

[00:06:11]Thanks for watching and please don't forget to subscribe to my channel for more exciting videos. Bye.