Let a = 693, b = 567 and c = 441
Now first we will find HCF of 693 and 567 by using Euclid’s division algorithm as under
693 = 567 x 1 + 126
567 = 126 x 4 + 63
126 = 63 x 2 + 0
Hence, HCF of 693 and 567 is 63
Now we will find HCF of third number i.e., 441 with 63 So by Euclid’s division alogorithm for 441 and 63
441 = 63 x 7+0
=> HCF of 441 and 63 is 63.
Hence, HCF of 441, 567 and 693 is 63.
Answer:
7
Step-by-step explanation:
7x + 68 = 2x + 103
Subtract 2x from each side
5x + 68 = 103
Subtract 68 from each side
5x = 35
divide 5 from 35 and you get X = 7
Answer:
D
Step-by-step explanation:
Because I know
You are given two equations, solve for one variable in one of the equations. Say you solved for x in the second equation. Then, plug in that value of x in the x of the first equation. Solve this (first) equation for y (as it should become apparent) and you'll get a number value. Plug in this numerical value of y into the y of the second equation. Solve for x in the second equation. And there you have it: (x, y)