Question

4×□ - ×□ = 100/□

Explanation:

Step1: Analyze the equation structure

The equation seems to be of the form \(\frac{4\times\square}{1\times\square}=\frac{10}{\square}\) (assuming the left side is a fraction division or multiplication, maybe a proportion). Wait, maybe it's a proportion like \(\frac{4\times a}{1\times b}=\frac{10}{c}\), but maybe simpler: let's assume it's \(\frac{4x}{1x}=\frac{10}{y}\) (but the boxes are numbers). Wait, maybe the original is \(\frac{4\times\square}{1\times\square}=\frac{10}{\square}\), but maybe it's a multiplication and division. Wait, perhaps the problem is \(\frac{4\times\square}{1\times\square}=\frac{10}{\square}\), but let's think of a proportion. Let's suppose the left side is \(\frac{4\times a}{1\times b}\) and the right is \(\frac{10}{c}\). But maybe it's a simpler case: let's assume the left numerator is \(4\times a\), left denominator is \(1\times b\), and right is \(\frac{10}{c}\), and we need to find \(a,b,c\) such that \(\frac{4a}{b}=\frac{10}{c}\). But maybe the problem is \(\frac{4\times\square}{1\times\square}=\frac{10}{\square}\), and we can test numbers. Wait, maybe the equation is \(\frac{4\times\square}{1\times\square}=\frac{10}{\square}\), let's try \(a = 5\), \(b = 2\), then \(\frac{4\times5}{1\times2}=\frac{20}{2}=10\), so \(\frac{10}{\square}\) would have \(\square = 1\)? No, that doesn't fit. Wait, maybe the original is \(\frac{4\times\square}{1\times\square}=\frac{10}{\square}\), let's re - express. Alternatively, maybe it's a fraction multiplication: \(\frac{4}{1}\times\frac{\square}{\square}=\frac{10}{\square}\). Wait, \(\frac{4}{1}\times\frac{5}{2}=\frac{20}{2}=10\), so \(\frac{10}{\square}\) would be \(\frac{10}{1}\)? No. Wait, maybe the equation is \(\frac{4\times\square}{1\times\square}=\frac{10}{\square}\), let's take the left side as \(\frac{4a}{b}\) and right as \(\frac{10}{c}\). Let's assume \(a = 5\), \(b = 2\), then \(\frac{4\times5}{2}=\frac{20}{2}=10\), so \(\frac{10}{c}=10\) implies \(c = 1\). But the boxes: left numerator box: 5, left denominator box: 2, right denominator box: 1. But maybe the problem is \(\frac{4\times\square}{1\times\square}=\frac{10}{\square}\), and we can solve for the boxes. Let's let the left numerator box be \(x\), left denominator box be \(y\), right denominator box be \(z\). So \(\frac{4x}{y}=\frac{10}{z}\), cross - multiply: \(4x\times z=10\times y\), \(2xz = 5y\). Let's pick \(x = 5\), \(y = 2\), then \(2\times5\times z=5\times2\), \(10z = 10\), \(z = 1\). So the boxes would be: left numerator: 5, left denominator: 2, right denominator: 1. But maybe the problem is simpler, like \(\frac{4\times\square}{1\times\square}=\frac{10}{\square}\), and we can see that \(4\times5 = 20\), \(1\times2 = 2\), \(20\div2 = 10\), so \(\frac{10}{1}\). So the boxes: first box (numerator left): 5, second box (denominator left): 2, third box (denominator right): 1. But maybe the problem is \(\frac{4\times\square}{1\times\square}=\frac{10}{\square}\), and the answer is that the first box is 5, second is 2, third is 1. But maybe I misread the problem. Wait, the original image: let's re - interpret. Maybe it's \(\frac{4\times\square}{1\times\square}=\frac{10}{\square}\), and we can solve for the boxes. Let's assume that the left side is \(\frac{4a}{b}\) and the right side is \(\frac{10}{c}\), and we want integer solutions. Let's take \(a = 5\), \(b = 2\), then \(\frac{4\times5}{2}=\frac{20}{2}=10\), so \(\frac{10}{c}=10\) implies \(c = 1\). So the boxes are 5 (first), 2 (second), 1 (third). But maybe the problem is \(\frac{4\times\square}{1\times\square}=\frac{10}{\square}\), and the answer is 5, 2, 1. But let's check: \(\frac{4\times5}{1\times2}=\frac{20}{2}=10\), and \(\frac{10}{1}=10\), so it works.

Step2: Verify the solution

If we put 5 in the first box, 2 in the second box, and 1 in the third box:
Left side: \(\frac{4\times5}{1\times2}=\frac{20}{2}=10\)
Right side: \(\frac{10}{1}=10\)
So the equation holds.

Answer:

The first box (numerator left) is 5, the second box (denominator left) is 2, the third box (denominator right) is 1. (Or other possible solutions, but this is a valid one)