Let's solve this equation. The balance scale shows that two squares and a three balance a weight of 13, which represents the equation 2 s + 3 = 13. To find the value of s, we need to isolate s on one side. So, let's subtract three from each side.
This leaves us with 2 s on the left. On the right side, we have 10. Our new equation is 2 s= 10.
When solving an equation, you often have options for your first step. One choice often leads to a solution more easily.
Let's look at a different problem. 3 * the quantity C + 10 equals 300. One way to start is to use the distributive property.
3 * C is 3 C and 3 * 10 is 30. The resulting equation is 3 C + 30 = 300.
Let's try a different first move with the same equation. Instead of distributing, we can divide both sides by three. On the left side, dividing by 3 leaves us with C + 10. On the right side, 300 / 3 is 100. Our new equation is C + 10 = 100.
Comparing the two results, C + 10= 100 is easier to solve than 3 C + 30 = 300.
Let's look at a different equation.
Which move leads to 7x = 63? If we subtract 77 from both sides, that would give us 7x = 63. Now, what if we wanted to get to the equation x + 11 = 20. Look at the original equation again. Both terms on the left, 7x and 77 are divisible by 7. We can factor out the 7, which gives us 7 * the quantity x + 11.
We can then divide both sides by 7 to get x + 11 on the left and 20 on the right. Equations can often be solved in more than one way. The key is to look at the structure of an equation before you make a move. Considering your options can help you choose the path that makes the problem simpler and faster to solve.
