the product of two consecutive even integersis 34 less than 7 times their sum.find the two integers.
let the first integer be x
then the second even integer is x+2
x(x+2) = 7(x + x+2) - 34
x^2 + 2x = 14x + 14 - 34
x^2 - 12x + 20 = 0
(x-10)(x-2) = 0
x = 10 or x = 2
the numbers are 10 and 12 or the numbers are 2 and 4
check:
case 1: numbers are 10 and 12
product = 120
7 times their sum = 154
is 120 less than 154 by 34 ? YES
case 2: numbers are 2 and 4
product = 8
7 times their sum = 42
is 8 less than 42 by 34 ? YES!
So we have two answers:
the integers could be 10 and 12
or
the integers could be 2 and 4
To find the two consecutive even integers, let's first assign variables to represent the integers. Let's call the first even integer "x" and the second even integer "x + 2" since consecutive even integers have a difference of 2.
Given that the product of the two consecutive even integers is 34 less than 7 times their sum, we can write an equation to represent this relationship:
(x)(x + 2) = 7(x + x + 2) - 34
Now, let's solve this equation step by step:
1. Expand the expressions on both sides of the equation:
x^2 + 2x = 7(2x + 2) - 34
2. Distribute 7 to the terms inside the parentheses:
x^2 + 2x = 14x + 14 - 34
3. Simplify the right side of the equation:
x^2 + 2x = 14x - 20
4. Move all the terms to one side of the equation to form a quadratic equation:
x^2 + 2x - 14x + 20 = 0
5. Combine like terms:
x^2 - 12x + 20 = 0
6. Factor the quadratic equation:
(x - 2)(x - 10) = 0
7. Set each factor equal to zero and solve for x:
x - 2 = 0 --> x = 2
x - 10 = 0 --> x = 10
Therefore, the two consecutive even integers are 2 and 4, or 10 and 12.