Question

use the hanger balance to model and solve the equation $3(g + 1) = 9$. what does $3(g + 1)$ represent? 3 groups of $g + 1$ how many 1s go on the right side of the hanger balance? blank box

Explanation:

Step1: Analyze the equation

The equation is \(3(g + 1)=9\). On the left - hand side, we have \(3\) groups of \((g + 1)\). To find the number of \(1\)s on the right - hand side, we first solve the equation for the total number of units represented by the right - hand side.

Step2: Solve the equation for the right - hand side value

We know that the left - hand side \(3(g + 1)\) is equal to the right - hand side (let the number of \(1\)s be \(n\), so the right - hand side is \(n\times1=n\)). From the equation \(3(g + 1)=9\), we can also think in terms of the hanger balance. The left - hand side has \(3\) sets of \((g + 1)\), and the right - hand side has a total value of \(9\) (since \(3(g + 1) = 9\)). But we can also use the fact that if we consider the left - hand side as \(3\) groups of \((g + 1)\), and we want to find the number of \(1\)s on the right, we can solve the equation \(3(g + 1)=9\) for the total number of \(1\)s. Divide both sides of the equation \(3(g + 1)=9\) by \(3\), we get \(g + 1=\frac{9}{3}=3\). But another way: the left - hand side has \(3\) groups, and the right - hand side is equal to the left - hand side. The left - hand side's non - \(g\) part: each \((g + 1)\) has one \(1\), and there are \(3\) groups, so the number of \(1\)s from the constant part is \(3\times1 = 3\)? Wait, no. Wait, the equation is \(3(g + 1)=9\). The right - hand side is \(9\) units of \(1\)? No, wait, the question is "How many 1s go on the right side of the hanger balance?". Since the left - hand side is \(3(g + 1)\) and the equation is \(3(g + 1)=9\), the right - hand side has a total value of \(9\), but if we are talking about the number of \(1\) - unit weights, since each \(1\) - unit weight has a value of \(1\), the number of \(1\)s is \(9\)? Wait, no, that can't be. Wait, let's re - examine the hanger. The left - hand side has 3 \(g\)s and 3 \(1\)s (because there are 3 groups of \((g + 1)\), so 3 \(g\)s and 3 \(1\)s). The equation is \(3(g + 1)=9\), so the right - hand side should balance the left - hand side. So the total value of the left - hand side is \(9\), so the right - hand side should have 9 \(1\)s? Wait, no, maybe I misread. Wait, the question is "How many 1s go on the right side of the hanger balance?". Let's solve the equation \(3(g + 1)=9\). Divide both sides by 3: \(g + 1 = 3\). But the hanger balance: the left has 3 \(g\)s and 3 \(1\)s. The right should have a total weight equal to the left. The left's total weight is \(3(g + 1)=9\), so the right should have 9 \(1\)s? No, that doesn't make sense. Wait, maybe the hanger is set up such that the left has 3 groups of \((g + 1)\) and the right has some number of \(1\)s. Since \(3(g + 1)=9\), the number of \(1\)s on the right is 9? But that seems too much. Wait, no, let's look at the equation again. \(3(g + 1)=9\). If we expand \(3(g + 1)\), we get \(3g+3\). So the left - hand side has \(3g\) and \(3\) (from the \(3\times1\)). The right - hand side is equal to the left - hand side, so if we consider the non - \(g\) part, the \(3\) is 3 \(1\)s, but the total left - hand side is \(3g + 3\), and it's equal to the right - hand side. Wait, the equation is \(3(g + 1)=9\), so the right - hand side is 9, which is 9 \(1\)s. But that seems incorrect. Wait, maybe the hanger is balanced when the left (3 groups of \(g + 1\)) equals the right (number of \(1\)s). So \(3(g + 1)=n\times1\), and we know that \(3(g + 1)=9\), so \(n = 9\)? No, that can't be. Wait, maybe I made a mistake. Wait, let's solve the equation \(3(g + 1)=9\). Step 1: Divide both sides by 3: \(g + 1=3\). Step 2: Subtract 1 from both sides: \(g = 2\). Now, if we substitute \(g = 2\) into the left - hand side, \(3(2 + 1)=3\times3 = 9\). So the right - hand side must have a total value of 9, which is 9 \(1\)s? But the hanger on the left has 3 \(g\)s (each \(g = 2\), so \(3\times2 = 6\)) and 3 \(1\)s ( \(3\times1=3\)), and \(6 + 3=9\), so the right - hand side should have 9 \(1\)s. But that seems like a lot. Wait, maybe the question is simpler. The left - hand side has 3 groups of \((g + 1)\), and the equation is \(3(g + 1)=9\). So the number of \(1\)s on the right is 9? But that seems wrong. Wait, no, maybe the hanger is such that the left has 3 \(g\)s and 3 \(1\)s, and the right has to balance it. So the total weight on the left is \(3g+3\), and the right is some number of \(1\)s, say \(n\). So \(3g + 3=n\). But we also know that \(3(g + 1)=9\), so \(n = 9\). So the number of \(1\)s is 9.

Answer:

9