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Before diving headfirst into geometrical proofs, it's a good idea to revisit algebra. We've already learned how to solve equations for a variable. Now we'll do algebra in the format of the two-column proof.

Sample Problem

Show that if 3x – 7 = 5, then x = 4.

Here, our given statement is 3x – 7 = 5, and we are asked to prove x = 4.

StatementsReasons
1. 3x – 7 = 5Given
2. 3x – 7 + 7 = 5 + 7Addition of 7 to equation (1)
3. 3x + 0 = 5 + 7Substitution of –7 + 7 = 0 into (2)
4. 3x = 5 + 7Substitution of 3x + 0 = 3x into (3)
5. 3x = 12Substitution of 5 + 7 = 12 into (4)
6. 3x3 = 123Dividing equation (5) by 3
7. x = 123Substitution of 3x3 = x into (6)
8. x = 4Substitution of 123 = 4 into (7)

Is there such a thing as being too descriptive? Yep, and that was it (since over half the proof was devoted to telling the reader how to do arithmetic). We will typically take numerical computation for granted, and write proofs like this:

StatementsReasons
1. 3x – 7 = 5Given
2. 3x = 12Add 7 to both sides of equation (1)
3. x = 4Divide equation (2) by 3

See? That proof looks a lot like how we'd write it in algebra. The only difference is that you give reasons as you go, convincing the readers (like your math teacher) that you know what you're doing. You got this.

Sample Problem

Show that if 5(x + 12) = 30 and x + y = 100, then y = 106.

This time, our two given statements are 5(x + 12) = 30 and x + y = 100. We are supposed to prove y = 106. Here we go.

StatementsReasons
1. 5(x + 12) = 30Given
2. x + y = 100Given
3. 5x + 60 = 30Distributive property (1)
4. 5x = -30Subtract 60 from both sides of (3)
5. x = –6Divide both sides of (4) by 5
6. -6 + y = 100Substitute x = -6 into (2)
7. y = 106Add 6 to (6)

As you can see, there are lots of ways of phrasing your reasons. The important part is that you justify at each step with why your statement is true. Of course, if your "reader" prefers it to be written a certain way, it's probably a good idea to follow his or her suggestions. Just saying.

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