### Instruction: Find all values of $x$ that satisfy
\[5x - 1 < (x + 1)^2 < 7x - 3.\].### Response: To find the values of $x$ that satisfy the inequality, I need to compare the two expressions on the left and right of the middle term.
I can start by simplifying the inequality by expanding the square and combining like terms.
This gives me $5x - 1 < x^2 + 2x + 1 < 7x - 3$.
Next, I can split the inequality into two parts: $5x - 1 < x^2 + 2x + 1$ and $x^2 + 2x + 1 < 7x - 3$.
I can solve each part separately by moving everything to one side and factoring.
For the first part, I get $x^2 - 3x + 2 > 0$, which factors as $(x - 2)(x - 1) > 0$.
For the second part, I get $x^2 - 5x + 4 < 0$, which factors as $(x - 4)(x - 1) < 0$.
Now I can use a sign chart to determine when each factor is positive or negative.
For the first part, I see that $(x - 2)(x - 1) > 0$ when $x < 1$ or $x > 2$.
For the second part, I see that $(x - 4)(x - 1) < 0$ when $1 < x < 4$.
To satisfy both parts, I need to find the intersection of these two intervals.
This means that the values of $x$ that satisfy the original inequality are those that satisfy $2 < x < 4$.