if 2x+3y-5z=18, 3x+7y+z=29,x+y+3z=17 find xy+yz+zx
Answers
2x + 3y - 5z = 18 ----(1)
3x + 2y + z = 29 ----(2)
x + y + 3z = 17 ----(3)
Multiply equation (3) by 2 and then subtracting equation (1) from (3)
⇒ (x + y + 3z) × 2 - (2x + 3y - 5z) = 17 × 2 - 18
⇒ 2x + 2y + 6z - 2x - 3y + 5z = 34 - 18
⇒ 11z - y = 16 ----(4)
Multiply equation (1) by 3 and (2) by 2 and then Subtracting equation (1) from (2)
⇒ (3x + 2y + z) × 2 - (2x + 3y - 5z) × 3 = 29 × 2 - 18 × 3
⇒ 6x + 4y + 2z - 6x - 9y + 15z = 58 - 54
⇒ 17z - 5y = 4 ----(5)
Multiply equation (4) by 5 and Subtracting equation (4) from (5)
⇒ 17z - 5y - (11z - y) × 5 = 4 - 16 × 5
⇒ 17z - 5y - 55z + 5y = 4 - 80
⇒ z = 2
By putting the z = 2 in equation (4)
⇒ 11 × 2 - y = 16
⇒ y = 6
By putting the value of y = 6 and z = 2 in equation (1)
⇒ 2x + 18 - 10 = 18
⇒ x = 5
∴ xy + yz + zx = 6 × 5 + 6 × 2 + 5 × 2 = 52
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