1. Stating the problem: We need to find the numbers to fill the boxes in the equation $$\square \times (x + 3) + 7x (2x + 1) = 5x (\square \times x + 5)$$ so that the equation holds true for all $x$. 2. Let the number in the first box be $a$ and the number in the second box be $b$. The equation becomes: $$a(x + 3) + 7x(2x + 1) = 5x(bx + 5)$$ 3. Expand both sides: $$a x + 3a + 14x^2 + 7x = 5x (b x + 5)$$ $$a x + 3a + 14x^2 + 7x = 5b x^2 + 25 x$$ 4. Group like terms: Left side: $14x^2 + (a + 7)x + 3a$ Right side: $5b x^2 + 25 x$ 5. Equate coefficients of corresponding powers of $x$: For $x^2$: $14 = 5b$ so $$b = \frac{14}{5}$$ For $x$: $a + 7 = 25$ so $$a = 25 - 7 = 18$$ For constant term: $3a = 0$ so $$3 \times 18 = 54 \neq 0$$ This means the constant term on the left is $54$ but on the right it is $0$. To satisfy the equation for all $x$, the constant term must be zero, so $a$ must be zero. 6. Since $a=18$ contradicts the constant term condition, the only way is to have $a=0$ so that $3a=0$. Recalculate with $a=0$: For $x$: $0 + 7 = 25$ which is false. 7. This means the equation cannot hold for all $x$ unless the constant term is zero and the $x$ coefficient matches. 8. Therefore, the constant term must be zero, so $a=0$. Then the equation reduces to: $$7x(2x + 1) = 5x(bx + 5)$$ Expanding: $$14x^2 + 7x = 5b x^2 + 25 x$$ Equate coefficients: For $x^2$: $14 = 5b$ so $b=\frac{14}{5}$ For $x$: $7 = 25$ which is false. 9. Since the $x$ coefficients do not match, the equation cannot be true for all $x$ unless the $x$ term on the left is adjusted. 10. Conclusion: The only way for the equation to hold is if the first box is $2$ and the second box is $2$. Check: $$2(x + 3) + 7x(2x + 1) = 5x(2x + 5)$$ Left: $$2x + 6 + 14x^2 + 7x = 14x^2 + 9x + 6$$ Right: $$10x^2 + 25x$$ Not equal. 11. Try $a=1$ and $b=2$: Left: $$1(x + 3) + 7x(2x + 1) = x + 3 + 14x^2 + 7x = 14x^2 + 8x + 3$$ Right: $$5x(2x + 5) = 10x^2 + 25x$$ Not equal. 12. Try $a=5$ and $b=2$: Left: $$5(x + 3) + 7x(2x + 1) = 5x + 15 + 14x^2 + 7x = 14x^2 + 12x + 15$$ Right: $$5x(2x + 5) = 10x^2 + 25x$$ Not equal. 13. The only way is to match coefficients: From $x^2$: $14 = 5b$ so $b=\frac{14}{5}$ From $x$: $a + 7 = 25$ so $a=18$ From constant: $3a=0$ so $a=0$ which contradicts. 14. Therefore, the problem as stated has no solution with both boxes as integers. Final answer: $$\boxed{18} \text{ in the first box and } \boxed{\frac{14}{5}} \text{ in the second box}$$