Dividing Grouped Expressions to Solve Equations

Here we want to figure out what balances C + 1.

The first scale shows three boxes of C++ 1 balanced by a weight of 33. This represents the equation 3 * the group C + 1 = 33.

To find the value of one group, we divide 33 into three equal parts.

33 / 3 = 11. So C + 1 equ= 11.

When an equation has multiple identical groups, you can use division to simplify.

By dividing both sides by three, we isolate the group C + 1.

This simplifies the equation to C + 1 = 11.

Here's a new problem. 11 * b minus 2 = 77.

We're asked to simplify by isolating the group b minus 2. We have 11 groups of b minus 2. So to find what one group equals, we divide both sides by 11.

77 / 11 = 7.

The expression b minus 2 = 7.

Now with d minus 2= 7, we can solve for b. We need a number that when you subtract two gives 7. To find it, we add 2 to both sides. This isolates b. 7 + 2 = 9. The solution is b = 9. Let's solve 100 * z + 1 = 500.

First simplify by dividing both sides by 100. This isolates the group z + 1.

500 / 100 = 5, giving us Z + 1 = 5. To find Z, subtract 1 from both sides. 5 - 1 = 4. So, Z = 4.

Let's solve 44 = 4 * a - 5. We isolate a - 5 by dividing both sides by 4.

44 / 4 = 11. Our simplified equation is 11 = a - 5. To solve for a, we add 5 to both sides.

11 + 5 = 16. So, a = 16.

Thinking of 4 * a - 5 as four groups of a - 5. Lets us simplify the problem first. By dividing both sides by the number of groups, we create a simpler one-step equation. This is an efficient strategy for solving equations in this format.