One number is 11 more than twice another number. if the sum of the numbers is twice their difference, find the numbers.

So, we need to make some equations to find this.

Let x be "one number"

x = (2y) +11

x+y = 2(x-y)

Now we can substitute

2y+11 +y = 2(2y+11)-2y

3y+11 = 4y+22-2y

3y+11 = 2y+22

y+11 = 22

y= 11

So the "Other number" is equal to 11.

Now, we can solve for x

x = 2y+11

x = 2(11) + 11

x = 22+11

x = 33

So, the numbers are 33 and 11

step-by-step explanation:

5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10