Solving equations

Solving equations is the most fundamental concept of algebra. Solving an equation means finding out the value of a variable. Sometimes, this is as simple as combining like terms to a solution, but often we must "isolate" the variable - or get it alone.

Equations in their simplest form edit

Consider the following simple equation:
$x=3+7$

If you combine like terms by adding 3 + 7, which equals 10, then you get:

$x=10$
You have just solved for x.

Order of Operations, Basic Laws of Algebra and Algebraic Properties of Equality edit

Many equations are more complex than this one, and we have to do more than combine like terms. In order to solve them, we need to know something about the Basic Laws of Algebra and the Algebraic Properties of Equality, which help us understand how numbers in equations work. We also need to know the proper Order of Operations, because doing steps in the wrong order can give us the wrong answer. These work together to help solve equations. Let's take an example of each in action.

Order of Operations edit

The Order of Operations is simply the sequence we use in a math problem to reach an answer. If you do a sequence out of order, you'll get the wrong result. There are only a few steps and easily memorized. We always begin working within any parentheses or brackets in a problem before moving to the problem in general. Remembering that, we then simplify equations in the following order, going from left to right in the problem:

Raise exponents/extract roots
Multiply/divide
Add/subtract

$(6+2^{3})-2^{2}*4-(2-\scriptstyle {\sqrt {16}}+{\tfrac {10}{5}})=x$

First, inside the parentheses, we raise exponents and extract roots.

$(6+8)-2^{2}*4-(2-4+{\tfrac {10}{5}})=x$

Next, inside the parentheses, we multiply and divide.

$(6+8)-2^{2}*4-(2-4+2)=x$

Then, we add and subtract inside the parentheses.

$(14)-2^{2}*4-(0)=x$

Now, outside the parentheses, we raise exponents and extract roots.

$(14)-4*4-(0)=x$

We multiply and divide.

$(14)-16-(0)=x$

We add and subtract.

$-2=x$

The Algebraic Properties of Equality edit

The good thing about equations is that you can add to, subtract from, multiply or divide one side by whatever you like as long as you do exactly the same thing to the other. To help you understand this concept, think of a scale. You want the scale to remain equal - the two sides to balance - so whatever you do to one side, you must do to the other to keep balance. Doing this allows you to gradually get the variable alone.

$x-3=7$

This one is still pretty simple. To isolate x, we must add 3 to both sides of the equation. (The Algebraic Properties of Equality come in handy here.)

$x-3+3=7+3$

We combine like terms to find the value of x.

$x=10$

The Basic Laws of Algebra edit

The Basic Laws of Algebra tell us a few things about adding and multiplying numbers. The arrangement and grouping of addition and multiplication don't matter. $1+2=2+1$ , $(1*2)*3=1*(2*3)$ . They also tell us that adding numbers and multiplying them gives you the same answer as multiplying numbers and then adding them. If $2(3+4)=2(7)=14$ , then $2(3)+2(4)=6+8=14$ . This final point is called the "Distributive Law", and it comes up in algebra quite a lot.

$2(x+4)=6$

To get our variable alone, we first need to get it out of the parentheses. We use the Distributive Law.

$2(x)+2(4)=6$
$2x+8=6$

Now we'll use the Algebraic Properties of Equality to get rid of the 8, combining like terms as we do so.

$2x+8-8=6-8$
$2x=-2$

Now we'll use the Algebraic Properties of Equality again to get rid of the 2.

${\tfrac {2x}{2}}={\tfrac {-2}{2}}$
$x=-1$

Checking your work edit

How can we be sure this is right? We can go back to the original problem and plug in our answer. If the math works, our solution can be said to be true.

We've already solved this one:
$2(x+4)=6$
Our answer was $x=-1$
$2(-1+4)=6$
$2(3)=6$
$6=6$
Our answer checks out.

Problems with multiple variables edit

Suppose we are given the following problem and told to "solve for $b$ ". We can't, not completely, because we don't know what $a$ is. But we can isolate $b$ .

$8(a-b)+2(a+3b+b)=6a+2(a+2^{2})+b$

First we do any math we can within the parentheses. In this case, we can combine like terms (by adding them) and raise an exponent.

$8(a-b)+2(a+3b+b)=6a+2(a+4)+b$
$8(a-b)+2(a+4b)=6a+2(a+4)+b$

Next we do the multiplication by using the Distributive Law.

$8a-8b+2a+8b=6a+2a+8+b$

This leaves us with more adding and subtracting like terms.

$10a=8a+8+b$

We'll use the Algebraic Properties of Equality to get the $b$ alone.

$10a-8a=8a-8a+8+b$
$2a=8+b$
$2a-8=8-8+b$
$2a-8=b$

This is as far as we can go, so how can we check this? We can "plug in" a value for either of the variables and solve the other. Since we've isolated for $b$ already, let's pick a value for $a$ . Let's say $a$ stands for 4. We substitute it in the problem.
$2(4)-8=b$
$8-8=b$
$b=0$

Once we have a possible solution set, we can check our work by going back to the original problem and plugging in both numbers:
$8(4-0)+2(4+3(0)+0)=6(4)+2(4+4)+0$
$8(4)+2(4)=6(4)+2(8)$
$32+8=24+16$
$40=40$

This is true. This does not mean, of course, that $b=0$ in the problem $2a-8=b$ . $b$ could equal 0, but only if $a$ equals 4. Without knowing the value of $a$ , we cannot solve for $b$ further than $2a-8=b$