Can somebody solve this math problem for me ?

Question

There are three consecutive integers. Ten times the sum of the second and third numbers is one thousand,two hundred more seven times the first number. What are the integer?

Please show the steps

Answer 1

x,y,z
y=x+1 z=x+2

10(y+z)=1200+7x
substitute....
10(x+1+x+2)=1200+7x
20x+30=1200+7x
13x=1170
90,91,92

Answer 2

Let the integers to be (x , x+1 , x+2 )
10 (x+1 + x + 2) = 1200 + 7 ( x)
10 ( 2x + 3 ) = 1200 =7 x
20x + 30 = 1200 + 7x
20x - 7x = 1200 - 30
13 x = 1170
x = 1170 / 13
x = 90
the integers are : 90 ,91 , 92

Answer 3

Hello,
x
x+1
x+2

10(x+1 + x+2) = 7x + 1200

10(2x + 3) = 7x + 1200

20x + 30 = 7x + 1200

13x = 1170

x = 90
x+1 = 91
x+2 = 92

Hope this helps!!

Answer 4

14 is 30% of what numer? write as 14 = .3X, X is what you want. Divide the two aspects via .3, or 14/.3 = X = 40 six.667 what's two% of 40? you want .02*40 = 0.8 sammy sosa had 193 hits, his batting typical became into 0.320 (32%) this is the full of at-bats (32%) what proportion at-bats did he have? this potential 193/ B = .32, the place B is the variety at-bat. pass multiply. 193 = .32 B, divide via .32 or B = 193/.32 = 603.a hundred twenty five or 603.