Grade 8 MATH Term 2 Week 5: Solving Problems on Linear Equations One Variable | MATATAG 3rd Quarter — Transcript

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1. 0:07[bell] Hello everyone, welcome math easy
2. 0:10YouTube channel where you can find free
3. 0:12Tagalog mathematics tutorial. I am your
4. 0:15teacher Van and today on solving word
5. 0:18problems involving linear equation in
6. 0:20one variable. But before we begin, let
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18. 0:53Okay. So without further ado, let us now
19. 0:55begin with our lesson. Okay. At the end
20. 0:57of this video, you will be able to solve
21. 0:59problems like for example number
22. 1:02problems, geometry problems and money
23. 1:04problems involving linear equations in
24. 1:07one variable. But before that let's have
25. 1:09first a review okay on linear equation
26. 1:12in one variable.
27. 1:14linear equation in one variable. It is
28. 1:17an equation that can be written in the
29. 1:19form ax + b is equal to c where a, b and
30. 1:24c are real numbers and a should not be
31. 1:27equal to zero. Okay. And take note is
32. 1:30lang po variable and the degree is just
33. 1:33one or the exponent is just one.
34. 1:36Examples are 3x + 5 is equal to 11 7 x -
35. 1:414 is equal to 10. We also have here x +
36. 1:442 is equal to 6. You might be wondering
37. 1:46a
38. 1:49is just 1. So
39. 1:53numerical coefficient here we also have
40. 1:56here 5x= -20
41. 2:03actually is zero. Okay.
42. 2:08So we only have here ax is equal to c.
43. 2:11So which is 5x is equal to -20. Now if
44. 2:15you want to remember if you want to
45. 2:17review our lesson on this one you can
46. 2:19actually watch again linear equation in
47. 2:21one variable in our previous upload.
48. 2:26Okay. So in our lesson for today
49. 2:30in our previous video on solving linear
50. 2:33equations in one variable using the
51. 2:35different properties. So apply in real
52. 2:38life like for example solving uh number
53. 2:41problems, geometry problems, uh money
54. 2:44problems and others. We are going to
55. 2:46show you some examples later on. Okay.
56. 2:49So let's begin with this one. A word
57. 2:52problem involving a linear equation is a
58. 2:54real life situation that can be
59. 2:56translated into an equation with one
60. 2:58unknown. So unknown linear equation in
61. 3:02one variable. So steps in solving a word
62. 3:05problem. Let us uh remember the
63. 3:07following steps. First is you
64. 3:10understanding the problem. So in in
65. 3:12order for us to understand the problem,
66. 3:14we have to read the uh problem
67. 3:17carefully. Identify unknown given and so
68. 3:21on. Next one is to represent the unknown
69. 3:24with a variable. So we are going to
70. 3:26choose a variable
71. 3:29to represent the quantities. And here we
72. 3:32are going to translate the condition
73. 3:34into an equation. So if formulate
74. 3:38equation
75. 3:40solve variable. So yes you fourth stage
76. 3:44or fourth step is to solve the equation
77. 3:47using
78. 3:49methods which is properties of equality.
79. 3:53And then number five of course is to
80. 3:55check okay double check
81. 3:58and then we will interpret the answer.
82. 4:04word problem. Okay.
83. 4:10At the end of the solution.
84. 4:13Okay. Now let's proceed to our first
85. 4:15example. We have here the sum of two
86. 4:18consecutive integers is 27. Find the
87. 4:21numbers. So this is an example of a
88. 4:24number problem. So step we have to
89. 4:27reread the problem and identify the
90. 4:29unknown quantity. So here we know that
91. 4:32the unknown is numbers that are
92. 4:36consecutive an numbers integers. Okay.
93. 4:39So we are going to look for two
94. 4:42consecutive integers. So in our solution
95. 4:44first we have to uh represent the
96. 4:47quantity that is step two to choose a
97. 4:50variable to represent the unknown. So we
98. 4:53will be using x. So let the first number
99. 4:56be equal to X.
100. 4:59Letters but in our lesson for today we
101. 5:01will just be using common which is
102. 5:04variable X. And what else? Second
103. 5:07number. The next number will be X + one.
104. 5:11Bit X + one because they are
105. 5:13consecutive.
106. 5:16And since we are dealing with integers
107. 5:19first integer
108. 5:21one to have the second integer. All
109. 5:24right. So this is step two. Step three
110. 5:27is to formulate now the equation. So
111. 5:29equation
112. 5:31using this. Let's go back to the problem
113. 5:34is the sum of two consecutive integers
114. 5:36is 27. Meaning if this is the first
115. 5:39integer x and then this is the second
116. 5:41integer x + one we just need to combine
117. 5:45or add. So so that lang sum i 27. So the
118. 5:50equation will be the first number x plus
119. 5:53the second number which is x + 1 and
120. 5:56then it will be equal to 27. Y equation
121. 6:01and we are now ready for step four which
122. 6:03is to solve. Okay,
123. 6:06first number.
124. 6:09So in our equation, we can actually
125. 6:11simplify this out since uh plus you can
126. 6:15just remove the parenthesis here. So we
127. 6:18have here x + x + 1 is equal to 27. Now
128. 6:22in order for us to get the value of x
129. 6:24here, we can first simplify
130. 6:28combine similar terms. Yes, we can add
131. 6:31these two similar terms. x + x will be
132. 6:342x okay + 1 is equal to 27. Now our goal
133. 6:39is to separate isolate xang
134. 6:43equation variable and then the rest of
135. 6:46these numbers
136. 6:48right side of the equation. So
137. 6:52we can remove one here using what
138. 6:55property? Yes, we will be using the
139. 6:58subtraction property of equality by just
140. 7:00uh subtracting itself. So 1 - 1
141. 7:06then of course
142. 7:09side.
143. 7:11So we have here 27 - 1 then it will now
144. 7:14be 26. So 2x so we have 2x is equal to
145. 7:1926. Now let's rewrite it here. Okay. Now
146. 7:25number which is 2. The numerical
147. 7:27coefficient here which is 2
148. 7:30cx. Okay. Yes we are going to use the
149. 7:33division property of equality which is
150. 7:36to divide two by itself
151. 7:39one
152. 7:42and then divide left side. Divide two
153. 7:46right side. So 26 / 2 will be 13. And
154. 7:50our answer is x is equal to 13. Well,
155. 7:53this is just for the first number. Now,
156. 7:56how about the second number? So, let's
157. 7:58solve for the second number. Since our
158. 8:00uh representation is x + 1,
159. 8:04so let us just substitute our x is 13.
160. 8:121 + 1. So, 14. Okay. And step five is to
161. 8:19check
162. 8:22since 13 and 14 are consecutive
163. 8:25integers. Well, this is satisfied. Next
164. 8:27one is the sum should be 27. Add 13
165. 8:3327. Yes. Okay. Uh in fact, if we're
166. 8:36going to substitute it here in the
167. 8:38original equation, it will be the same
168. 8:41in solving linear equation in our
169. 8:44previous lesson. So in our conclusion or
170. 8:47our interpretation we can say that the
171. 8:50first number is uh 13 and then the
172. 8:53second number is 14.
173. 8:57Okay let's have example number two.
174. 8:59Let's read the difference between two
175. 9:01number is 45. The larger number is five
176. 9:04less than three times the smaller
177. 9:06number. Find the two numbers. So
178. 9:10the lang numbers. Okay. And then the
179. 9:12first statement says the difference
180. 9:15between two numbers is 45. We can use
181. 9:17this in the equation. The second
182. 9:20statement says larger number is five
183. 9:22less than 3 * the smaller number. We can
184. 9:24use this in the representation. So
185. 9:28representation
186. 9:29let x be equal to the smaller number.
187. 9:33Okay. So we have here x smaller number.
188. 9:39So let's go back to the problem. The
189. 9:42difference between two numbers is 45
190. 9:48equation. The larger number is five less
191. 9:51than three times the smaller number. So
192. 9:53second statement. Okay.
193. 10:00So five less than three times the
194. 10:03smaller number. Let us translate this
195. 10:06algebraic statement into uh algebraic
196. 10:10expression. So pano po it will be 3x - 5
197. 10:14bucket 3x is you 3 * the smaller number.
198. 10:18Okay and then minus 5 less than okay so
199. 10:22it will now be the larger number 3x - 5.
200. 10:26Next one is the equation. Okay let's go
201. 10:29back to the problem again. First
202. 10:31statement the difference between two
203. 10:34numbers is 45.
204. 10:37So
205. 10:40number difference is 45. So number
206. 10:473US
207. 10:50number which is x
208. 10:5445. Okay. Now let's simplify by uh uh
209. 10:58simplify. Let us solve by simplifying
210. 11:00the equation. So solve for the smaller
211. 11:02number. So
212. 11:073x - 5 - x is equal to 45. We can now
213. 11:11simplify by combining similar terms or
214. 11:14operating you similar terms. 3x - x we
215. 11:18can subtract 3x
216. 11:212x. So 2x - 5 is now equal to 45. Now
217. 11:25let us remove uh -5 by adding ano
218. 11:31y positive 5. So let us add positive.
219. 11:36And this is what we call the addition
220. 11:38property of equality. So
221. 11:43plus therefore 2x is equal now to 50
222. 11:47because 45 + 5 is 50. Now using our
223. 11:51division property of equality divide
224. 11:54ulit s 2x by itself which is 2 2 / 2 and
225. 11:58then sabil 50 / 2. So we have now here x
226. 12:02is equal to 25. You po y smaller smaller
227. 12:08number which is x. Now let us solve for
228. 12:10the larger number which is 3x - 5. So
229. 12:13using our value of x is a substitute
230. 12:16lang
231. 12:18representation
232. 12:193x - 5. So let us change x into 25. So
233. 12:23gaw 25 x 3 * 25 will be 75 - 5. And then
234. 12:3075 - 5 will be equal now to 70. Ea po
235. 12:34larger number. Now let's double check.
236. 12:36Okay.
237. 12:38Subtract 70 - 25 the answer will be 45.
238. 12:43Yes, correct. 70 - 25 is 45. Therefore,
239. 12:48the smaller number is 25 and the larger
240. 12:50number is 70.
241. 13:07>> [music]
242. 13:25>> Okay, let's have example number three.
243. 13:27This time this is all about a geometry
244. 13:29problem. Okay, so let me read. A 96 m
245. 13:33long fence is used to enclose a
246. 13:35rectangular playground. The length of
247. 13:37the playground is 8 m more than its
248. 13:40width. Find the length and the width of
249. 13:42the playground. So
250. 13:45length and then width playground shape
251. 13:49playground. It's a rectangle. So length
252. 13:53and width rectangular playground. First
253. 13:56let's have the representation. Let will
254. 13:59let x be the variable. So x let's go
255. 14:03back to the problem first. So
256. 14:07statement
257. 14:09compare
258. 14:11quantities.
259. 14:13Okay. So length and then width. So first
260. 14:16sentence a 96 m long fence is used to
261. 14:18enclose a rectangular playground. We
262. 14:20cannot use this. Maybe this will be used
263. 14:22in the equation. Next statement the
264. 14:24length of the playground is 8 m more
265. 14:26than its width. Yes comparison
266. 14:30for our representation. So we have here
267. 14:35length. Okay. Of course
268. 14:38be the width of the playground. Now it
269. 14:42says here
270. 14:458 m. Okay. So how do we represent the
271. 14:48length? We can use x + 8. Okay. This
272. 14:51will be equal to the length of the
273. 14:53playground. Now next step will be the
274. 14:55equation. Okay, first sentence. A 96 m
275. 14:59long fence is used to enclose a
276. 15:02rectangular playground.
277. 15:05Enclosed
278. 15:11rectangometry
279. 15:19is it the area or the perimeter? Okay,
280. 15:23correct. It is the perimeter. Now ano
281. 15:25formula perimeter it's P is equal to 2 L
282. 15:29which is the length plus 2 W which is
283. 15:32the width. Okay. So
284. 15:35equation and in our first sentence
285. 15:38perimeter it's 96 m. Okay. Soap
286. 15:44playground 96 m long fence. Okay. So our
287. 15:48perimeter is 96 and let us substitute
288. 15:52you length which is x + 8 representation
289. 15:57and then you width which is x. Okay. So
290. 16:01po equation 96 is equal to 2 * the
291. 16:04quantity of x + 8 + 2 * x. Okay. We can
292. 16:09simplify this by distributing
293. 16:15numbers parenthesis. So we have here 2 *
294. 16:18x
295. 16:202x and then 2 * 8 16. Okay. 2 * x lang
296. 16:26po yan kaying 2x. Okay. Now we are going
297. 16:29to solve for the width first value x.
298. 16:33Okay. Representation x is the width of
299. 16:35the playground. So combine equation. Yes
300. 16:40we can combine 2x and then 2x 4x. Okay.
301. 16:45+ 16. Now take note if you notice right
302. 16:49side variable which is x. So
303. 16:55later on
304. 16:58isolate variable xant
305. 17:03what are we going to do?
306. 17:06We can actually subtract itself. Okay.
307. 17:10Subtraction property of equality.
308. 17:14we can just add the additive inverse
309. 17:16which is 16. So 16 - 16 will be 0. Okay.
310. 17:25So we have here 96 - 16 it will now be
311. 17:2980. Okay. Right is 4x. Now we have here
312. 17:3480 is equal to 4x.
313. 17:364. We can just divide it by itself. So
314. 17:39gig 4 / 4 80 / 4 using the uh division
315. 17:44property of equality. Now 80 divided by
316. 17:474 will be 20. Okay. So x is equal to 20
317. 17:51symmetric property
318. 17:54left side x and then s right side 20. x
319. 17:58will now be equal to 20. Okay. This is
320. 18:01our first
321. 18:04quantity
322. 18:05width. Okay. Now for the second quantity
323. 18:08which is the length we just use the
324. 18:10representation x + 8 and then substitute
325. 18:13s 20 x mag 20 + 8 therefore the length
326. 18:18will be 28. Okay. So we can double check
327. 18:22these answers by just substituting it in
328. 18:25the original equation. Original equation
329. 18:29equals then our answer will be uh
330. 18:33correct. Okay. So yes, therefore the
331. 18:35width is 20 m and the length is 28 m.
332. 18:42Okay, next example another geometry
333. 18:45problem. Okay, so let me read the
334. 18:46supplement of an angle measures 40° more
335. 18:49than twice its complement. Find the
336. 18:51measure of the angle. So first
337. 18:56measurement angle. So
338. 18:59it's just the measure of the angle given
339. 19:02we have here 40°
340. 19:07it's 40° more than twice its complement.
341. 19:10So take note of the word complement
342. 19:13supplement. So you have to remember the
343. 19:15supplement and complement in our uh
344. 19:18angles. Okay angle pairs. So una let's
345. 19:22have the representation.
346. 19:24Okay let x be the measure of unknown
347. 19:26angles. So
348. 19:29and then let 90 - x be the measure of
349. 19:32its complement bucket 90 - x. Okay,
350. 19:35remember
351. 19:36complmentary angles if they add if you
352. 19:39add uh two angles together and they uh
353. 19:42they they form 90° angle
354. 19:46complmentary angles. So reverse process
355. 19:53complement. So 90 - x will be the
356. 19:55complement. Next one, how about the
357. 19:58supplement? So it is 180 minus x. Okay,
358. 20:02take note supplementary angles are two
359. 20:04angles that when you add together
360. 20:07180°.
361. 20:08So reverse
362. 20:12is angle you will also get the
363. 20:14supplement. All right. Now what will be
364. 20:17the equation? Okay. First statement. The
365. 20:20supplement of an angle measures 40°
366. 20:24more than twice its complement. So
367. 20:29supplement is equal to 40 plus 2 * the
368. 20:33complement. Okay. So let us substitute
369. 20:36the values. Supplement is 180 - x. So
370. 20:41180 - x you complmentary angle is 90 -
371. 20:45xan representation. So therefore
372. 20:49equation we can simplify this first
373. 20:53distribute 2 * the quantity 90 - x. So
374. 20:58we have here 2 * 90 it will be 180 and
375. 21:02then 2 * -x will be uh minus or -2x
376. 21:07uh 40 + 180 add so it will be 220. Okay,
377. 21:13next one we have here
378. 21:16-2x
379. 21:18and then 180 220
380. 21:28-2x. Okay,
381. 21:32left side. Okay, so using our addition
382. 21:34property of equality, we can remove -2x
383. 21:38adding the additive inverse. So zero po
384. 21:41yan. So therefore left side
385. 21:452x okay plus 2x therefore we can now
386. 21:48combine these two terms -x + 2x it will
387. 21:51be positive x okay and then we have here
388. 21:55180
389. 21:58by removing 180 here on the left side
390. 22:01using subtraction property
391. 22:04180
392. 22:07c x. Okay. So therefore we also subtract
393. 22:11220 by 180. So miging x now will be
394. 22:15equal to uh 40. Okay. Therefore the
395. 22:18measure of the angle is 40°.
396. 22:24Okay. For example number five. This will
397. 22:26be all about money problem. So let me
398. 22:28read the problem. Anna and beya have 450
399. 22:31pesos in total. If Anna has 90 pesos
400. 22:34more than Beya, how much does each have?
401. 22:36Okay. So here we can represent first the
402. 22:40uh variable. Okay, we can represent the
403. 22:42unknown. So we have here ano bait
404. 22:47quantity.
405. 22:52Okay.
406. 22:53So we have here Anna or so let x be
407. 22:58basal.
408. 23:01Next
409. 23:03peri it is more than bay money 90 pesos
410. 23:07more than bas money. So it will be x +
411. 23:1090. So poi Anna. Now for the equation we
412. 23:14have x plus the quantity x + 90 is equal
413. 23:18to 450. Why? Basing on our first uh
414. 23:20sentence. Anna and ba have 40 450 pesos
415. 23:24in total.
416. 23:28It will be 450. Okay, so Bea's money is
417. 23:30X, Anna's money is X + 90. We can now
418. 23:35remove the parenthesis here.
419. 23:382X + 90. Okay,
420. 23:41parenthesis and then we add X + X it
421. 23:43will be 2X. Okay, + 90 is equal to 450.
422. 23:48Now let us solve for BAS money first
423. 23:50which is you variable X. Okay. Now here
424. 23:542X + 90
425. 23:58left side. So - 90 using subtraction
426. 24:02property and then sabang side 450 - 90
427. 24:07pesos. So 360 okay 2x to remove 2 we can
428. 24:13just divide both side by 2 and then
429. 24:16therefore x will now be equal to 360 / 2
430. 24:20it is 180. Okay. Now to solve for Anna's
431. 24:25money we can just substitute okay here
432. 24:28in the representation x + 90 x will now
433. 24:31be 180. So 180 + 90 it will be 270.
434. 24:37So if we're going to check this add
435. 24:40first sentence
436. 24:43450
437. 24:46180 + 270 yes it will be 450.
438. 24:50Okay. Now, therefore, Bea has 180 pesos
439. 24:54and then Anna has 270
440. 24:57pesos.
441. 25:01Okay, let's have example number six.
442. 25:03Another money problem. Okay, let me
443. 25:05read. Lara keeps her savings in a jar
444. 25:08containing 50 pesos and 100 peso bills.
445. 25:11The total amount of money is 4,250.
446. 25:15The number of 50 peso bills is five more
447. 25:18than twice the number of 100 peso bills.
448. 25:21How many bills of each kind does have
449. 25:2750 pesos and then 100 peso bills. Okay.
450. 25:32And then total it will be 4,250.
451. 25:37Okay. So in representing the unknown
452. 25:39okay take note
453. 25:42statement or sentence compare
454. 25:45quantities. So I think this is the third
455. 25:48sentence. The number of 50 peso bill is
456. 25:51five more than twice the number of 100
457. 25:54peso bill. So ano madam quantity is it
458. 25:57the 100 peso bill or the 50 peso bill?
459. 26:01Okay it will be the 100 peso bill. So
460. 26:06as x represent as x. Okay. Now for the
461. 26:1250 peso bill. Okay. Take note. Okay.
462. 26:15Five more than twice the number of 100
463. 26:18peso. So how are we going to represent
464. 26:20this verbal statement? It will now be 2x
465. 26:23+ 5. Okay. Twice. So 2x and then five
466. 26:27more than plus 5. Okay. 2x + 5. It will
467. 26:30be the number of 50 peso bills. Now for
468. 26:34the equation going back to the uh first
469. 26:36two sentences
470. 26:38the total amount of money is 4,250
471. 26:47bills and then it will be 4,250
472. 26:51but take note okay multiply
473. 26:54100 uh times the number of 100 peso bill
474. 26:59is 100x okay and then you uh 50 peso
475. 27:04bill is 2x + 5. Okay.
476. 27:10Numbers of bills. Okay. Bill 100 peso
477. 27:13bill 50 peso billip
478. 27:16number nil. Okay.
479. 27:19Equation. It is now equal to 4,250.
480. 27:24Okay. So again we can distribute this 50
481. 27:27* 2x and then 50 * 5. Okay.
482. 27:32We have 50 * 2x it's 100x and then 50 *
483. 27:365 it will be 250
484. 27:39okay equals 4,250
485. 27:42okay sol number of 100 peso bill so y x
486. 27:47okay now we can simplify this by adding
487. 27:50common terms or similar terms so 100x +
488. 27:54100x will be 200x okay plus 250 it's
489. 27:59equal to 42 250. Now we can remove 250
490. 28:04here. Okay. By subtracting itself 250.
491. 28:07So magig
492. 28:09zero saba naman we can subtract it again
493. 28:12by 250.
494. 28:14Okay. Therefore 4,000 na poa. And then
495. 28:18here we can divide 200x by 200 both
496. 28:21sides. Okay. Therefore our x will be 20.
497. 28:26Okay. 4,000 divided by 200 is 20. Ano po
498. 28:29y 20. po number of 100 peso bills. Now
499. 28:34for the number of 50 peso bills, it is
500. 28:362x + 5. So substitute lang
501. 28:40representation or expression 2x + 5
502. 28:4520 x. So 2 * 20 and then it will be uh 2
503. 28:50* 20 is 40 + 5 it will be 45. So the
504. 28:54number of 50 peso bill is 45. So
505. 28:57therefore, Lara has 20 pieces of 100
506. 29:00peso bills and then 45 pieces of 50
507. 29:05bills in her jar.
508. 29:09Okay, another common type of uh problem
509. 29:12on linear equation in one variable is
510. 29:14the age problem. Okay, let's have this
511. 29:16example. The age of Carla is six less
512. 29:19than twice the age of Diane. The sum of
513. 29:22their ages is 48. Find their ages. Okay.
514. 29:25So, let's represent their age. Let X be
515. 29:28the age of Dian. Okay.
516. 29:33X. And then for uh the age of Carla, we
517. 29:38can use the first statement
518. 29:416 less than twice the age of Dian. So
519. 29:44using this verbal statement 6 less than
520. 29:46twice the age of Dian which is X. We can
521. 29:49have 2X - X uh - 6 as its algebraic
522. 29:53expression. to age of Carerala. Okay.
523. 29:57Next, using the second sentence, the sum
524. 30:00of their ages is 48. We can now
525. 30:02formulate the equation. So, add to it,
526. 30:06it will be 48. So, x + 2x - 6, it will
527. 30:10be 48. Okay. We can remove the
528. 30:12parentheses here. It will be uh x + 2x
529. 30:173x. Okay. So, 3x - 6 is equal to 48.
530. 30:22Okay. By removing -6 here we can use the
531. 30:26addition property of equality. -6 plus
532. 30:29it's additive inverse positive 6 is
533. 30:32equal to 0. And then 48 magg
534. 30:376. So magig 3x is now equal to 54. Okay.
535. 30:42We can now divide 3x and 54 by 3. There
536. 30:46you go. E1 i x is equal to 18. Okay. po
537. 30:52age of Dian. Okay. Now, how about the
538. 30:55age of Carla? So, using 2x - 6, we can
539. 30:59substitute the value of x is 18. So, 2 *
540. 31:0318 will be 36 - 6, it will now be 30.
541. 31:08Okay? So, the age of Carla is 30.
542. 31:10Therefore, Carla is 30 years old and
543. 31:13Dian is 18 years old.
544. 31:19Okay, for our last example, let's have a
545. 31:21problem on motion. Okay, so motion, it
546. 31:25talks about the speed. So factors speed
547. 31:29or rate or distance and then time. So
548. 31:32let me read the problem first. A bus
549. 31:35leaves the terminal at 50 km per hour. 3
550. 31:39hours later, a van leaves the same
551. 31:41terminal at 70 km per hour following the
552. 31:44same route. How long will it take for
553. 31:47the van to overtake the bus? So,
554. 31:53how long will it take for the van to
555. 31:55overtake the bus?
556. 31:58And take note,
557. 32:01travel, it's the bus. Okay. Travel with
558. 32:04a speed of 50 km per hour and then po
559. 32:09van after 3 hours ano po speed it's 70
560. 32:12km hour. So
561. 32:16overtake van.
562. 32:20Okay. So let's have the solution. Okay.
563. 32:23First, okay, let us represent. Okay.
564. 32:26Represent unknown.
565. 32:30Time or because [clears throat] uh we
566. 32:32are looking for the time we will be
567. 32:34using letter t to represent the time.
568. 32:37But here time
569. 32:40it's the time in hours the van travels
570. 32:43before overtaking the bus. Okay. Why? In
571. 32:47order for us to know the time overtake
572. 32:54time.
573. 32:59Okay.
574. 33:02travel time. So in order to do that we
575. 33:06have to remember that the distance when
576. 33:09it comes to speed is equal to rate
577. 33:21to rate time time. So ano ba time bus
578. 33:24moa so time bus since 3 hours it will be
579. 33:28t + 3. So rate ano speed it's 50. So
580. 33:33again distance is equal to rate time
581. 33:35time. So 50 * time which is t + 3 bus
582. 33:41equal
583. 33:43rate van which is 70 kilm. Okay.
584. 33:50So it is unknown. Okay. Now this is will
585. 33:53this will be our equation. This time let
586. 33:56us distribute itong 50 * the quantity of
587. 34:00t + 3. So 50 * t is 50t and then naman
588. 34:0450 * 3 is 150. Okay. Next is
589. 34:10similar terms. Okay. Either way 50 right
590. 34:14side or 70
591. 34:17left side. Okay. Soil
592. 34:20is c 50t. So we can remove 50t subtract
593. 34:24it by itself. So zero po yan. So 70t
594. 34:28let's subtract it also by 50t. Okay
595. 34:3250.
596. 34:34Next c 150. Next
597. 34:3870 t minus 50 it will be 20 t. Okay.
598. 34:43Next we can remove 20 by dividing
599. 34:46itself. 20 divided by 20 para 1 150
600. 34:51divided by 20
601. 34:547.5. Okay
602. 34:57it will be 7.5 hours. Okay.
603. 35:027.5 hours
604. 35:11equal distance. So overtake it will be
605. 35:16more than 7.5. So therefore the van will
606. 35:19catch up to the bus after 7 hours and 30
607. 35:23minutes of travel. So
608. 35:27we can now say that the van overtook the
609. 35:30bus. Okay. Final answer.
610. 35:34Okay. For you to practice, I have here
611. 35:36some exercises. You can do this on your
612. 35:38own. Uh this is similar to our examples
613. 35:41a while ago. So
614. 35:47and then you can just do this and then
615. 35:49later you can go back to the the video
616. 35:51and then play
617. 35:53solution.
618. 35:59Okay, I hope you are now ready. I will
619. 36:01just show you the solution
620. 36:04check on your own. So, we have here the
621. 36:06solution for number one. Okay, next the
622. 36:09solution for number two. You can just
623. 36:12pause this video anytime you want. Okay.
624. 36:14And then next the solution for number
625. 36:17three.
626. 36:18And then number four. And then last is
627. 36:22number five. So I hope
628. 36:28okay if you want further exercises for
629. 36:30you to master our topic you can actually
630. 36:32answer our activity here and then uh po
631. 36:37second part number 5 to 8. So eight
632. 36:39items poto. Now if you want to check
633. 36:41this you can check this with your
634. 36:43teachers later on especially if they are
635. 36:45using our powerpoint presentations. But
636. 36:48for now I have to say bye. Thank you for
637. 36:51listening. I hope
638. 36:54examples. Okay, and see you on our next
639. 36:58video. Bye-bye and God bless.
640. 37:03Please click the subscribe button to
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642. 37:11Lessons
643. 37:15[music]
644. 37:20>> [music]
645. 37:21>> God bless and see you on the next video.
646. 37:24Bye-bye.