Answer:
Original number of sixes = 6
Original number of nines = 10
Explanation:
We are told in the question that:
A certain number of sixes and nines is added to give a sum of 126
Let us represent originally
the number of sixes = a
the number of nines = b
Hence:
6 × a + 9 × b = 126
6a + 9b = 126.....Equation 1
If the number of sixes and nines is interchanged, the new sum is 114.
For this second part, because it is interchanged,
Let us represent
the number of sixes = b
the number of nines = a
6 × b + 9 × a = 114
6b + 9a = 114.......Equation 2
9a + 6b = 114 .......Equation 2
6a + 9b = 126.....Equation 1
9a + 6b = 114 .......Equation 2
We solve using Elimination method
Multiply Equation 1 by the coefficient of a in Equation 2
Multiply Equation 2 by the coefficient of a in Equation 1
6a + 9b = 126.....Equation 1 × 9
9a + 6b = 114 .......Equation 2 × 6
54a + 81b = 1134 ........ Equation 3
54a + 36b = 684.........Equation 4
Subtract Equation 4 from Equation 3
= 45b = 450
divide both sides by b
45b/45 = 450/45
b = 10
Therefore, since the original the number of nines = b,
Original number of nines = 10
Also, to find the original number of sixes = a
We substitute 10 for b in Equation 1
6a + 9b = 126.....Equation 1
6a + 9 × 10 = 126
6a + 90 = 126
6a = 126 - 90
6a = 36
a = 36/6
a = 6
Therefore, the original number of sixes is 6