Kataki 'oku fu'u general ho'o fiema'u pea 'oku 'ikai keu fu'u mahino'i ho'o 'uhinga. Kaikehe teu fai pe ha fakaofiofi ki ho'o fiema'u. Let's have a look at Solving simultaneous equations first: Example: solve the following equations simultaneously. 3x + 2y = 8 x + y = 3 In this case, teke lava pe 'o solova 'aki 'ae Substitution pe koe Elimination method: 1. Using Substitution Method: Ui e first equation koe Eq1, pea second one koe Eq2. Sai, rearrange Eq2 ke hoko koe: y = -x + 3 Substitute y in Eq2 into y in Eq1. 'o pehe ni: 3x + 2(-x+3) = 8, expand pea simplify then solve for x 3x + -2x + 6 = 8 x + 6 = 8 x = 2. Now find y since x is found to be 2. Teke lava pe 'o ngaue'aki ha taha pe 'oe ongo equation ka teu fili 'ae Eq2 he 'oku matangofua ange ia ke ngaue'aki. x + y = 3 , substitute in x = 2 2 + y = 3 y = 1. solution (2, 1) or x = 2, y = 1 2. Elimination Method: 3x + 2y = 8 x + y = 3 Sai teu liunga 3 'ae Eq2 ke tatau e value 'o x he ongo equation kae lava keu eliminate 'a x, Fakatokanga'i ange teu lava pe 'o eliminate 'a y kapau teu liunga 2 'ae Eq2. Anyway, as a matter of choice, I want to eliminate x 3(Eq2) = 3(x + y = 3) , note that you have to multiply each term by 3 = 3x + 3y = 9 , I will call this new equation Eq3 Now, koe ongo equation 'e pehe ni: 3x + 2y = 8 , Eq1 3x + 3y = 9 , Eq3 Take away Eq1 from Eq3 to avoid having negative numbers but of course you can still take away Eq3 from Eq1 if you are good with directed numbers. 3x + 3y = 9 - 3x + 2y = 8 y = 1, now find x Using Eq2, you can quickly find x x + y = 3 x = 2 Solution (2, 1) or x = 2, y = 1 Note, the answers for both methods are the same. Sai, ta hoko atu ki ho'o fehu'i fekau'aki moe algebraic sequence, are you talking of Arithmetic sequence or Geometric Sequence? 'Oku fiema'u ia kete 'ilo pe koe ha e type of sequence kae lava kete formulate ha ongo equations pea te toki solve simultaneously. Hange ko eni: Let's use an Arithmetic Sequence as an example: Problem: The second term of an Arithmetic sequence is 8 and the ninth term is 29, find the first term, a and the common difference, d We can formulate our two equations as: Using the general formula for Arithmetic Sequence: tn = a + (n-1)d We know, n=2 and n=9, tn = 8 and tn=29, in our scenario. Substitute these values in: 8 = a + (2-1)d, ... Eq1 29 = a + (8-1)d, ... Eq2 We get: 8 = a + d, ... Eq1 29 = a + 7d, ...Eq2 Use the Elimination method to eliminate a in this case (it is easier to eliminate a) Take away Eq1 from Eq2 29 = a + 8d - 8 = a + d 21 = 7d, solve for d d = 3 Since d = 3, we can find a Using 8 = a + d 8 = a + 3 a = 5 so for this sequence, the first term, a = 5, the common difference, d = 3 Just for fun, we can generate the sequence as: 5, 8, 11, 14, 17, 20, 23, 26, 29,...