Algebra I · Elise's Elected Education

Imagine a backpack with some books inside. You don't know how many. We'll call that number x (it just means "the unknown").

You also have 3 books sitting on the desk next to the backpack.

Altogether, you count 10 books.

So: backpack books (x) + 3 desk books = 10 total.

x + 3 = 10

Question: how many books are in the backpack?

Easy way to figure it out — if there are 10 total and 3 are on the desk, the backpack must have 10 minus 3 = 7.

That's the whole game. We're just figuring out how to undo what's been done to x, until x is alone.

Picture a literal balance scale. On the left side: a box with x marbles inside, plus 3 loose marbles. On the right side: 10 marbles.

[ 📦x + 🔵🔵🔵 ] ⚖️ [ 🔵🔵🔵🔵🔵🔵🔵🔵🔵🔵 ]

left side = right side

The scale is balanced — both sides weigh the same. We want to know what's in the box.

If we take 3 marbles off the left side, we have to take 3 marbles off the right side too — otherwise the scale tips.

[ 📦x ] ⚖️ [ 🔵🔵🔵🔵🔵🔵🔵 ]

still balanced

Now the box on the left is alone. And the right has 7 marbles. So the box must have 7 marbles in it.

x = 7. That's it.

To solve x + 3 = 10 — get x alone on one side. Here's the process every time:

x + 3 = 10

Step 1. Look at what's "stuck" to x. Here: +3.

Step 2. Do the opposite of that to both sides. Opposite of +3 is -3.

x + 3 - 3 = 10 - 3

Step 3. Simplify. +3 - 3 on the left cancels. 10 - 3 on the right = 7.

x = 7

Step 4. Check it. Put 7 back in for x: 7 + 3 = 10. ✓ True. We're done.

Same recipe works whether the number is small, big, decimal, negative — doesn't matter. Undo what's stuck to x, on both sides.

Solving Linear Equations