FrontierMath
Algebra I/Linear Equations/The Properties of Algebra
Algebra I Regents in 22 days
Algebra I · Lesson 1

The Properties of Algebra

The rules that govern Algebra.


Consider the equation 3x=123x=12. How would you solve for xx?

An algebra student may divide both sides of the equation by 33 to get x=4x=4, but why must you divide both sides of the equation? Why not just the left-hand side or the right-hand side of the equation?

The answer isn't intuition. It's a set of rules that govern what you're allowed to do to an equation without breaking it. These rules are the Properties of Algebra and once you know them, every algebraic move you make has a reason behind it.

Let's look at the following equation:

2x+4=102x + 4 = 10

A common first instinct is to subtract 4 from the left side.

2x=102x = 10     \implies x=5x = 5

This looks reasonable — but let's check by substituting x=5x = 5 back into the original equation:

2(5)+4=14102(5) + 4 = 14 \neq 10

Something went wrong. We got x=5x = 5, but x=5x = 5 doesn't even satisfy the equation we started with!

The mistake: we subtracted 44 from the left side without subtracting it from the right side! The moment we did that, we changed the equation into a different one — and solved the wrong problem.

This is exactly what the Properties of Algebra are designed to prevent.

Now that we have seen what happens when we break the rules, let's see what the rules actually are:

The Addition Property of Equality

Assume you have an equation like

x9=1x-9=1

To solve for xx, you must add 9 to both sides of the equation. Think of a balance scale — if both sides weigh the same, adding the same weight to each side keeps it balanced. Let's finish solving for xx:

x9=1+9=+9x=10\begin{aligned} x - 9 &= 1 \\[10pt] +9 &= +9 \\[10pt] x &= 10 \end{aligned}

A common practice in Algebra is plugging our final solution for xx back into our orginal equation to check its validity. Lets plug back our soluion for x back into the equation:

(10)1=?99=9\begin{aligned} (10) - 1 &\stackrel{?}{=} 9 \\[10pt] 9 &= 9 \checkmark \end{aligned}

x=10x = 10 checks out. Let's move on to our second rule...

The Subtraction Property of Equality

This rule is very similar to the Addition Property of Equality, only with subtraction instead of addition being used to isolate for the variable. Let's see it in action:

Assume you have the following equation

x+10=20x + 10 = 20

Notice how we isolate for x below.

x+10=2010=10x=10\begin{aligned} x + 10 &= 20 \\[10pt] -10 &= -10 \\[10pt] x &= 10 \end{aligned}

Let's plug back in just to check:

(10)+10=?2020=20\begin{aligned} (10) + 10 &\stackrel{?}{=} 20 \\[10pt] 20 &= 20 \checkmark \end{aligned}

Now to our next rule...

The Division Property of Equality

Given an equation like 3x=93x = 9, you must divide both sides of the equation to solve for xx. See below:

3x=93x3=93x=3\begin{aligned} 3x &= 9 \\[10pt] \frac{3x}{3} &= \frac{9}{3} \\[10pt] x &= 3 \end{aligned}

Let's verify quickly:

3(3)=?9    9=93(3) \overset{?}{=} 9 \implies 9 = 9 \checkmark

Very simalerly, we have...

The Multiplication Property of Equality

We must use the Multiplication Property of Equality when fractions are involved in our equation. For instance,

x2=4\frac{x}{2} = 4

If you multiply both sides by the denominator, in this case 22, the denominator would be eliminated and you would have solved for xx.

x2=42x2=242x2=24x=8\begin{aligned} \frac{x}{2} &= 4 \\[10pt] 2 \cdot \frac{x}{2} &= 2 \cdot 4 \\[10pt] \cancel{2} \cdot \frac{x}{\cancel{2}} &= 2 \cdot 4 \\[10pt] x &= 8 \end{aligned}

We can also use this same property with questions like

25x=10\frac{2}{5}x = 10
where you must solve both sides by multiplying by the reciprocal of 2/5, which is 5/2. Let's multiply by the reciprocal and solve for xx:

25x=105225x=10525225x=1052x=1052    x=25\begin{aligned} \frac{2}{5}x &= 10 \\[10pt] \frac{5}{2} \cdot \frac{2}{5}x &= 10 \cdot \frac{5}{2} \\[10pt] \frac{\cancel{5}}{\cancel{2}} \cdot \frac{\cancel{2}}{\cancel{5}}x &= 10 \cdot \frac{5}{2} \\[10pt] x &= 10 \cdot \frac{5}{2} \\[10pt] \implies x &= 25 \end{aligned}

Notice how when we multiply by the reciprocal on the Left Hand side, everything cancels out aside from xx. Also, remember to multiply both sides by the reciprocal! Onto our next property...

The Distributative Property

Let's say you have an equation like 4(2+5)=x4(2+5) = x . You may remember from your past math classes that must solve the contents inside the parenteses first becasuse of P.E.M.D.A.S./B.O.D.M.A.S. You certienly can do that here: 4(2+5)=x    4(7)=x    x=284(2+5) = x \implies 4(7) = x \implies x = 28 , but there is another way of solving for x. You can, instead, distribute the 4 across each term inside the parentheses — multiplying it by both the 2 and the 5 separately:

4(2+5)=x42+45=x8+20=xx=28\begin{aligned} 4(2+5) &= x \\[10pt] 4\cdot2 + 4\cdot5 &= x \\[10pt] 8 + 20 &= x \\[10pt] x &= 28 \end{aligned}

Same answer, different path. In general, the Distributive Property states:

a(b+c)=ab+aca(b + c) = ab + ac

This becomes especially useful when you can't simplify what's inside the parentheses first. For example, consider:

3(x+4)=213(x + 4) = 21

You can't add xx and 44 together since they're unlike terms, so you must distribute:

3(x+4)=213x+12=2112=123x=9x=3\begin{aligned} 3(x+4) &= 21 \\[10pt] 3x + 12 &= 21 \\[10pt] -12 &= -12 \\[10pt] 3x &= 9 \\[10pt] x &= 3 \end{aligned}

Notice how we used the Subtraction Property and Division Property of Equality after distributing — the properties work together.