Let's complete the equation shown by this balance scale. The left side has two identical boxes, each with a weight of s + 3. The right side has two weights of seven, which gives us a total weight of 14. Because the scale is balanced, the two sides are equal. We can write an equation to represent this. Since there are two identical groups of s + 3, we can write this as 2 * that quantity. So, we'll fill in the first blank with a two for the two groups. The expression inside the group is s plus three. So we'll fill in the second blank with three. Our equation is 2 * s + 3 = 14.
Now what does a single group of s+ 3 balance? The first equation shows that two groups balance a total of 14. This means one group must balance half of that amount. We can divide both sides of our equation by two. This shows us that one box of s + 3 is equal to a weight of 7. Our simplified equation is now s + 3 = 7.
To find the solution for s, we just need to get s by itself. We can do this by subtracting three from both sides of the equation. On the right side, 7 - 3 is 4.
This means the solution is s= 4.
Thinking about the problem in terms of groups is a helpful strategy. When you see multiple identical groups on one side, simplifying the problem by finding the value of a single group can make the solution much clearer.
Let's look at another problem using the same idea. Here the left side has a single weight of 60. The right side has three identical boxes each labeled t + 4. This means 60 is equal to three groups of t + 4. We can write this as 60 = 3 * t + 4.
To make the equation more simple, let's figure out what a single group of t + 4 would balance. We have a total weight of 60 that is split evenly among three groups. So, we'll divide 60 by 3, which gives us 20. This means that a weight of 20 is equal to t + 4.
Now, we have a much simpler equation to solve. 20 = t + 4. To find the final value of t, we just need to subtract four from both sides of the equation. 20 - 4 gives us 16. So our final solution is 16= t.
When we see these problems as being made of groups, if we find what balances just one group, the solution becomes much clearer. This is the same process as dividing both sides of the original equation by the number of groups to simplify it.
