Answer:
The numbers are 10 and 11.
Step-by-step explanation:
Let [tex]x[/tex] and [tex]x+1[/tex] be consecutive positive integers.
When the problem says that the square of the first decreased by 67 this means [tex]x^2-67[/tex] and this is equal to three times the second [tex]3(x+1)[/tex].
So [tex]x^2-67=3(x+1)[/tex] will be our equation.
Next, we solve the equation:
[tex]x^2-67=3x+3\\\\x^2-70=3x\\\\x^2-3x-70=0[/tex]
Solve by factoring
[tex]x^2-3x-70= \left(x^2+7x\right)+\left(-10x-70\right)\\\\x^2-3x-70=x\left(x+7\right)-10\left(x+7\right)\\\\x^2-3x-70=\left(x+7\right)\left(x-10\right)=0[/tex]
Using the Zero Factor Principle: If ab = 0, then either a = 0 or b = 0, or both a and b are 0.
[tex]x+7=0\\x=-7[/tex]
And
[tex]x-10=0\\x=10[/tex]
Because the numbers need to be positive integers, we only take x = 10 as a valid solution.
The numbers are 10 and 11.
