Question

there are three numbers such that the second is twice the first and the third is 6 more than the second. Their sum is 41. Find these numbers. Please explain with steps.

Answer 1

Answer:

The third number is 3 times the second.

Therefore four times the second number plus a number that is 8 less equals 67.

So 67 divided by 4 is 16 point something. But we know straight away that the answer is not 16 because 4 times 16 is 64 and that leaves us only 3 (67–64) to find the first number, knowing that the first number is 8 less than the second number.

Try 15. 15 times 4 equals 60 add on 7 (15–8), total equals 67.

Correct, so we have 7, 15 and 45.

Alternatively you could go: 4 times the second number plus the second number less 8 equals 67.

Therefore five times the second number is 75. Therefore the second number is 15 and the other two are therefore 7 and 45

Answer 2

$\huge\sf\underline{\pink{A}\orange{N}\blue{S}\red{W}\green{E}\purple{R}}$

Let the number be x,y,and z

$y=2x\: z=6+2x$

According to Question

$x+y+z=41$

$x+2x+6+2x=41$

$5x+6=41$

$5x=41-6$

$x=\frac{35}{5}$

$x=7$

$y=2×7=14$

$z=6+14=20$