Question

The sum of three numbers is 15. If the second number is subtracted from the sum of first and third numbers, then we get 5. When the third number is subtracted from the sum of twice the first number and the second number, we get 4. Find the three numbers.

Answer 1

Let the three numbers be x, y and z.
According to the given conditions,
x + y + z = 15
x + z – y = 5 i.e. x – y + z = 5
2x + y – z = 4

D = `|(1, 1, 1),(1, -1, 1),(2, 1, -1)|`

= 1(1 – 1) – 1(–1 – 2) + 1(1 + 2)
= 1(0) – 1(– 3) + 1(3)
= 0 + 3 + 3
= 6 ≠ 0

D_x = `|(15, 1, 1),(5, -1, 1),(4, 1, -1)|`

= 15(1 – 1) – 1(– 5 – 4) + 1(5 + 4)
= 15(0) – 1(– 9) + 1(9)
= 0 + 9 + 9
= 18

D_y = `|(1, 15, -1),(1, 5, 1),(2, 4, -1)|`

= 1(– 5 – 4) – 15(– 1 – 2) + 1(4 – 10)
= 1(– 9) – 15(– 3) + 1(– 6)
= – 9 + 45 – 6
= 30

D_z = `|(1, 1, 15),(1, -1, 5),(2, 1, 4)|`

= 1(– 4 – 5) – 1(4 – 10) + 15(1 + 2)
= 1(– 9) – 1(– 6) + 15(3)
= –9 + 6 + 45
= 42
By Cramer's Rule,

x = `"D"_x/"D" = 18/6` = 3,

y = `"D"_y/"D" = 30/6` = 5,

z = `"D"_x/"D" = 42/6` = 7

∴ The three numbers are 3, 5 and 7.