even integers differ by 2, so the three numbers are x, x+2 and x+4. Now you are told that
x(x+2) = 11(x+4)-8
Now solve for x.
Solve the following algebraically using one variable: Find three consecutive even integers such
that the product of the smallest integer and the middle integer is eight less than eleven times the larger integer.
2 answers
let n be the smallest integer
n+2 the middle
n+4 the larger
n*(n+2)+8=11(n+4)
solve for n. I appears to be a quadratic equation
n^2-9n-36=0
(n-12)(n+3)=0
n=12, or n=-3
reject the n=-3 as it is not an even integer.
n+2 the middle
n+4 the larger
n*(n+2)+8=11(n+4)
solve for n. I appears to be a quadratic equation
n^2-9n-36=0
(n-12)(n+3)=0
n=12, or n=-3
reject the n=-3 as it is not an even integer.