We are tasked with finding three consecutive even integers, such that the product of the first and third, minus the second, is one more than 10 times the third integer.

Let's define the variables:

Let the first even integer be $x$.

• The second consecutive even integer will then be $x + 2$.
• The third consecutive even integer will be $x + 4$.

According to the problem:

$(x \cdot (x + 4)) - (x + 2) = 1 + 10 \cdot (x + 4)$ This equation reflects the given condition. Now let's solve it step by step.

1. Expand both sides:
   • Left side: $x(x + 4) - (x + 2) = x^2 + 4x - x - 2 = x^2 + 3x - 2$
   • Right side: $1 + 10(x + 4) = 1 + 10x + 40 = 10x + 41$

Thus, the equation becomes: $x^2 + 3x - 2 = 10x + 41$

2. Move all terms to one side: $x^2 + 3x - 2 - 10x - 41 = 0$ $x^2 - 7x - 43 = 0$

3. Solve the quadratic equation: We can solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For $x^2 - 7x - 43 = 0$, the coefficients are $a = 1$, $b = -7$, and $c = -43$. Plugging these into the quadratic formula:

$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(-43)}}{2(1)}$ $x = \frac{7 \pm \sqrt{49 + 172}}{2}$ $x = \frac{7 \pm \sqrt{221}}{2}$

Since $\sqrt{221}$ is not a perfect square, this quadratic doesn't have integer solutions, which suggests there may be a mistake or misunderstanding in the problem setup.

Would you like me to explore alternative methods, or did you intend for the problem to involve non-integer values? Let me know if you'd like further clarification or assistance.

Related Questions:

1. How do you determine consecutive integers from a word problem?
2. What are the common ways to solve quadratic equations?
3. Can quadratic equations always have real number solutions?
4. What happens if the discriminant in the quadratic formula is negative?
5. How do you interpret problems involving integer constraints in algebra?

Tip:

When solving word problems, always double-check the setup and make sure the defined variables correspond exactly to the described scenario!