The sum of three consecutive even numbers is one hundred sixty-two. What is the smallest of the three numbers?
2 answers:
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To solve this, the problem is solvable by elimination. So you see how the 5y and -5y cancel out? You basically add those two and you add the -7x and 4x and 10 and -10 leaving you with -3x=0. Dividing by each side gives you x=0. Then, plug 0 in for x in -7x+5y=10. Giving you
-7(0)+5y=10.
0+5y=10
5y=10
Divide by 5 on each side giving you y=2. So your answer is (0,2).
-7(0)+5y=10.
0+5y=10
5y=10
Divide by 5 on each side giving you y=2. So your answer is (0,2).
If you have 1275 ÷ 4, you write it as shown, with 1275 on the inside and 4 on the outside. then you go digit by digit and find how many times 4 goes into the number. it goes into 1 zero times, so an imaginary 0 is put in the first spot above the 1. next, 4 goes into 12 three times, so 3 is written above the 2. After that, subtract 4×3 from 12, which leaves 0, and bring down the 7. 4 goes into 7 once with a remainder of 3, so a 1 is put up top and the 5 brought down next to the three. 4 goes into 35 eight times with a remainder of 3,so 8 is written up top. the 3 is the remainder, written as R3. so your answer is 1274÷4= 318 R3.
Hope this helps.
Hope this helps.