how do we find k1 and k2 in the following example: 16= 2k1 + 3k2 & 25= 5k1+4k2
+130516 16= 2k1 + 3k2
25= 5k1 + 4k2
Multiply the first equation by 4 and the second equation by -3.....this will eliminate k2.....so we have
64 = 8k1 + 12k2
-75 = -15k1 - 12k2 add these together
-11 = -7k1 divide by -7 on both sides
11/7 = k1 substitute this into either equation to find k2 ...I'll use the first one
16 = 2(11/7) + 3k2
16= 22/7 + 3k2 subtract 22/7 from borh sides
16 - 22/7 = 3k2
[112 - 22]/7 = 3k2
90/7 = 3k2 divide by 3 on both sides
90/21 = 30/7 = k2
So...our solution is (k1, k2) = (11/7, 30/7 )
+130516 16= 2k1 + 3k2
25= 5k1 + 4k2
Multiply the first equation by 4 and the second equation by -3.....this will eliminate k2.....so we have
64 = 8k1 + 12k2
-75 = -15k1 - 12k2 add these together
-11 = -7k1 divide by -7 on both sides
11/7 = k1 substitute this into either equation to find k2 ...I'll use the first one
16 = 2(11/7) + 3k2
16= 22/7 + 3k2 subtract 22/7 from borh sides
16 - 22/7 = 3k2
[112 - 22]/7 = 3k2
90/7 = 3k2 divide by 3 on both sides
90/21 = 30/7 = k2
So...our solution is (k1, k2) = (11/7, 30/7 )
