WEBVTT
00:00:00.420 --> 00:00:11.950
Find π΄ and π΅ such that four π₯ minus two over π₯ plus three times π₯ minus two equals π΄ over π₯ plus three plus π΅ over π₯ minus two.
00:00:12.890 --> 00:00:20.620
We have this equation that we want to be true for all values of π₯, and we need to find π΄ and π΅ to make this so.
00:00:21.640 --> 00:00:27.060
Letβs get rid of all these fractions by multiplying up by π₯ plus three times π₯ minus two.
00:00:28.130 --> 00:00:43.440
So on the left-hand side, you have four π₯ minus two, and on the right-hand side, we have π΄ times π₯ plus three times π₯ minus two over π₯ plus three plus π΅ times π₯ plus three times π₯ minus two over π₯ minus two.
00:00:44.550 --> 00:00:47.250
And we can see that thereβs some cancelation that will happen.
00:00:47.660 --> 00:01:00.000
The factor of π₯ plus three in the numerator of the first fraction cancels with the denominator, and the same is true in the second fraction, where the factor of π₯ minus two cancels with the denominator.
00:01:01.210 --> 00:01:27.040
Tidying up a little, we still have four π₯ minus two on the left-hand side and we have π΄ times π₯ minus two plus π΅ times π₯ plus three on the right-hand side, and we can expand the brackets on the right-hand sides to get π΄π₯ minus two π΄ plus π΅π₯ plus three π΅, and we can combine like terms to get π΄ plus π΅ times π₯ plus negative two π΄ plus three π΅.
00:01:28.210 --> 00:01:34.770
The reason that we needed to combine the like terms on the right-hand side was to allow us to compare coefficients.
00:01:35.610 --> 00:01:49.000
On the left-hand side, the coefficient of π₯ is four, and on the right-hand side, the coefficient of π₯ is π΄ plus π΅, and these coefficients must be equal, so four must be equal to π΄ plus π΅.
00:01:49.940 --> 00:01:58.710
For a very similar reason, the constant term on the left, negative two, must be equal to the constant term on the right, negative two π΄ plus three π΅.
00:01:59.570 --> 00:02:06.570
Having gone through all this algebra, weβre left with two linear equations in the variables π΄ and π΅.
00:02:07.630 --> 00:02:13.960
And to finish, we just need to solve these equations simultaneously to obtain the values of π΄ and π΅.
00:02:14.940 --> 00:02:18.120
There are a number of ways of solving simultaneous equations.
00:02:18.900 --> 00:02:39.720
We could rearrange the first equation to get π΅ in terms of π΄ and then substitute this into our second equation to get an equation in π΄ alone, which we can then solve in the normal way, multiplying out the brackets, combining like terms, and rearranging to find that π΄ is 14 over five.
00:02:40.770 --> 00:02:52.670
Given the value of π΄, we can substitute this into our expression for π΅ in terms of π΄ to get that π΅ is four minus 14 over five, which is six over five.
00:02:53.280 --> 00:03:06.320
There are of course other methods for solving these simultaneous equations, for example, elimination or matrix methods, but all of them will give the same answer that π΄ is 14 over five and π΅ is six over five.
00:03:07.510 --> 00:03:25.330
In the context of our question, this means that, for all π₯, four π₯ minus two over π₯ plus three times π₯ minus two is equal to 14 over five over π₯ plus three plus six over five over π₯ minus two.