Respuesta :
Answer:
10, 12, 14, 16
Step-by-step explanation:
Given: Four consecutive even integers such that seven times the first exceeds their sum by 18.
Lets assume the first number be "x".
As it is even integers
∴ Second consecutive number will be [tex](x+2)[/tex]
Third consecutive number will be [tex](x+4)[/tex]
Fourth consecutive number will be [tex](x+6)[/tex]
Now, as given seven times the first exceeds their sum by 18.
∴ [tex][x+(x+2)+(x+4)+(x+6)]+18 = 7x[/tex]
solving the equation to find the number.
⇒ [tex][x+(x+2)+(x+4)+(x+6)]+18 = 7x[/tex]
Opening parenthesis
[tex]x+x+2+x+4+x+6+18 = 7x[/tex]
⇒ [tex]4x+30= 7x[/tex]
subtracting both side by 4x
⇒ [tex]30= 3x[/tex]
Dividing both side by 3
∴ x= 10.
Hence, subtituting the value x to find four consecutive even integers.
First number is 10
Second consecutive number will be [tex](10+2)[/tex] = 12
Third consecutive number will be [tex](10+4)[/tex]= 14
Fourth consecutive number will be [tex](10+6)[/tex] = 16
∴ Four consecutive even integers are 10,12,14,16.
