Solving equations with equivalent steps

When rearranging equations we use equivalent steps, which do not change whether the equation is true. The point of equivalent steps is to bring the equation into a simpler form from which we can work out the solution.

What is an equation

First, a short reminder of what an equation is and how we will write it. If we have two functions f(x) and g(x), we can set them equal and get an equation:

$f (x) = g (x)$

Here we call f(x) the left-hand side of the equation and g(x) the right-hand side. In place of the abstract functions f(x) and g(x) we can imagine particular functions, for example f(x) = 3x and g(x) = x + 2. Then the equation would look like this:

$3 x = x + 2$

The solution of the equation is then a set (say) S such that for every x in S, the equation is still true after we substitute x into it. So the solution of our example equation would be the set

$S = {1} .$

Substituting into the equation, we get

$3 \cdot 1 = 1 + 2$

An equivalent step is one that does not change whether the equation is true. So a step is equivalent if we turn the function f(x) into a function f₁(x), do the same to g(x) to get g₁(x), and the equality

$f_{1} (x) = g_{1} (x)$

still holds.

The first and simplest equivalent step for equations (not inequalities!) is swapping the left-hand and right-hand sides. It is probably obvious that if

$f (x) = g (x),$

then the swapped equation is true as well:

$g (x) = f (x) .$

I cover equations in more detail in the article What is an equation.

Adding

The simplest example is adding an expression to both sides of the equation. We always apply an equivalent step to both sides of the equation. So let's take this equation:

$x - 3 = 1.$

We can solve this equation by adding three to both sides. When rearranging equations we usually write the step in brackets at the end of the line where we started it, like this

$\begin{array}{rcl} x - 3 & = & 1 (+ 3) \\ 3 + x - 3 & = & 1 + 3 \\ x & = & 4 \end{array}$

In general, if we again have two functions f(x) and g(x), we can add a third function h(x) to both sides without changing whether the equation is true.

$\begin{array}{rcl} f (x) & = & g (x) (+ h (x)) \\ f (x) + h (x) & = & g (x) + h (x) \end{array}$

As you can see, we don't have to add just a number – we can add a function too:

$x = 2 x + 6$

By convention we want the variables on the left, which we get by adding −2x to the equation, in other words by subtracting 2x.

$\begin{array}{rcl} x & = & 2 x + 6 (- 2 x) \\ x - 2 x & = & 2 x - 2 x + 6 \\ - x & = & 6 \\ x & = & - 6 \end{array}$

Of course we can also add fractions if that is useful.

$x - \frac{1}{2} = 471$

We add one half and we have the result:

$\begin{array}{rcl} x - \frac{1}{2} & = & 471 (+ \frac{1}{2}) \\ x - \frac{1}{2} + \frac{1}{2} & = & 471 + \frac{1}{2} \\ x & = & 471.5 \end{array}$

Adding is the classic way to “move” an expression from one side of an equation to the other. If you have the equation

$3 = 2 x + 17,$

how do you get the term with the variable, that is 2x, onto the left-hand side? You simply add −2x to the whole equation. That gives you zero on the right-hand side instead of 2x, because 2x − 2x = 0. And on the other side we get −2x:

$\begin{array}{rcl} 3 & = & 2 x + 17 (- 2 x) \\ 3 - 2 x & = & 2 x - 2 x + 17 \\ 3 - 2 x & = & 17 \end{array}$

Multiplying

We can also multiply an equation by a function. The only condition is that the function must not be zero. We cannot multiply an equation by zero, or by any function that can be simplified to zero, for example f(x) = x − x. So if we have the equation

$3 x = 7,$

we can multiply it by two:

$\begin{array}{rcl} 3 x & = & 7 (\cdot 2) \\ 2 \cdot 3 x & = & 2 \cdot 7 \\ 6 x & = & 14 \end{array}$

That probably wasn't very useful. Multiplying is used, for example, when you need to get rid of fractions:

$3 = \frac{2}{5 x}$

If we multiply this equation by the denominator of the fraction on the right-hand side, the expression 5x cancels nicely thanks to the properties of fractions.

$\begin{array}{rcl} 3 & = & \frac{2}{5 x} (\cdot 5 x) \\ 5 x \cdot 3 & = & 5 x \cdot \frac{2}{5 x} \end{array}$

Now we multiply the fraction by 5x, which puts 5x into the numerator, and then we can cancel 5x because it appears in both the denominator and the numerator.

$\begin{array}{rcl} 15 x & = & \frac{5 x \cdot 2}{5 x} \\ 15 x & = & 2 \end{array}$

At the start of this section I said that we must not multiply by zero. Now we multiplied by 5x – but this expression contains a variable, and if we put zero in for it, the whole expression is zero and we are multiplying by zero after all. What do we do?

The answer is to state the conditions under which we may do this step. Here we have to say that we may multiply by 5x only if x is not zero. But now let's look at the equation we multiplied. The expression 5x is already in it, in the denominator of a fraction. What does that mean? We cannot divide by zero, so the domain of the function $\frac{2}{5 x}$ already rules out x being zero.

So once more – the equation contains a fraction with 5x in the denominator. Because we cannot divide by zero, the equation only makes sense if x is not zero. That is why we can safely multiply the whole equation by 5x: the necessary conditions are already taken care of by the domain of the function in the denominator.

Multiplying by more complicated expressions

Here is an important point: if we multiply an equation by an expression, we must always multiply the whole right-hand side and the whole left-hand side. As an example, take this equation:

$x + 2 = x - 4$

If we wanted to multiply it by two, we would have to multiply whole sides, like this:

$\begin{array}{rcl} 2 \cdot (x + 2) & = & 2 \cdot (x - 4) \\ 2 x + 4 & = & 2 x - 8 \end{array}$

It would be a mistake to leave out the brackets, because we would get a wrong result:

$\begin{array}{rcl} 2 \cdot x + 2 & = & 2 \cdot x - 4 \\ 2 x + 2 & = & 2 x - 4 \end{array}$

You can see that the result is different from the previous case. Similarly, if you multiply by a more complicated expression, you must multiply the whole equation. As an example, take the equation

$2 x + 1 = \frac{2}{x + 3} .$

We multiply this equation by x + 3 to get rid of the variable in the denominator on the right-hand side.

$\begin{array}{rcl} 2 x + 1 & = & \frac{2}{x + 3} (\cdot (x + 3)) \\ (x + 3) (2 x + 1) & = & (x + 3) \frac{2}{x + 3} \\ (x + 3) (2 x + 1) & = & \frac{2 (x + 3)}{x + 3} \end{array}$

We now have the fraction in a form where we can cancel it nicely by x + 3:

$\begin{array}{rcl} (x + 3) (2 x + 1) & = & \frac{2 (x + 3)}{x + 3} \\ (x + 3) (2 x + 1) & = & 2 \end{array}$

Now we can expand the brackets on the left-hand side and finally subtract two to “move” the two to the left-hand side, which gives zero on the right-hand side.

$\begin{array}{rcl} (x + 3) (2 x + 1) & = & 2 \\ 2 x^{2} + x + 6 x + 3 & = & 2 \\ 2 x^{2} + 7 x + 3 & = & 2 (- 2) \\ 2 x^{2} + 7 x + 1 & = & 0 \end{array}$

Dividing

Just as we can multiply an equation, we can also divide it. The restriction is the same as for multiplying – we must not divide by zero. Dividing can easily be turned into multiplying. If we want to divide an equation by an expression w, it is the same as multiplying the equation by the fraction 1/w. Dividing is often used in examples like this one:

$12 x = 7$

This is the kind of equation where there is not much more we can do than divide the whole equation by twelve, which gives us the value of x.

$\begin{array}{rcl} 12 x & = & 7 (\div 12) \\ x & = & \frac{7}{12} \end{array}$

We would get the same result if we multiplied the equation by 1/12:

$\begin{array}{rcl} 12 x & = & 7 (\cdot \frac{1}{12}) \\ 12 x \cdot \frac{1}{12} & = & 7 \cdot \frac{1}{12} \\ \frac{12 x}{12} & = & \frac{7}{12} \\ x & = & \frac{7}{12} \end{array}$

Raising to a power

When rearranging equations we can also use powers, but with some restrictions. First, let's refresh our knowledge of powers. Is this equality true?

$\sqrt{x^{2}} = x$

If you square a number and then take the square root, do you get the original number back? This is a tricky question, and the answer is: not always. Imagine putting in a negative number for x, for example minus three:

$\begin{array}{rcl} \sqrt{(- 3)^{2}} & = & - 3 \\ \sqrt{9} & = & - 3 \\ 3 & \neq & - 3 \end{array}$

As you can see, the equation suddenly doesn't hold, because when we square a negative number we get a positive number. But if we then take the square root of a positive number, we again get a positive number. When would this equality hold? When we know we are working only with positive, or rather non-negative, numbers. So we may square an equation when we know we are working with positive numbers.

For example, in geometry we often solve equations containing the lengths of line segments, and then we can square without worry, because a length cannot be negative. Even when squaring or taking roots, the rule is that we must apply it to the whole equation. For example:

$\sqrt{x + 2} = x + 4$

Suppose we are working only with positive numbers, so we may square the whole equation:

$\begin{array}{rcl} \sqrt{x + 2} & = & x + 4 (^{2}) \\ {(\sqrt{x + 2})}^{2} & = & (x + 4)^{2} \\ x + 2 & = & x^{2} + 8 x + 16 \\ x^{2} + 7 x + 14 & = & 0 \end{array}$

We got a quadratic equation, which we can solve using the methods described in that article.

How would you solve an equation that you need to square, but you can't be sure you are working only with positive numbers? You simply square it and then check the answers, because squaring can give you answers that are not solutions of the equation. For example:

$\sqrt{x} = x - 2$

Now we square the equation:

$\begin{array}{rcl} \sqrt{x} & = & x - 2 (^{2}) \\ {(\sqrt{x})}^{2} & = & (x - 2)^{2} \\ x & = & x^{2} - 4 x + 4 \\ x^{2} - 5 x + 4 & = & 0 \end{array}$

Again we get a quadratic equation, so we solve it using a suitable technique, for example factorising (the exact method is in the linked article):

$(x - 1) (x - 4) = 0$

We can see that one solution is 1 and the other is 4. But we know we did a step that may not give an equivalent equation, so we still have to check. We substitute these solutions into the original equation and find out which of them is a solution of the original equation. First we substitute one:

$\begin{array}{rcl} \sqrt{x} & = & x - 2 \\ \sqrt{1} & = & 1 - 2 \\ 1 & \neq & - 1 \end{array}$

The check shows that one is not a solution of the original equation. What about four?

$\begin{array}{rcl} \sqrt{x} & = & x - 2 \\ \sqrt{4} & = & 4 - 2 \\ 2 & = & 2 \end{array}$

Four is a solution of the original equation, so we have found the answer.

Chapters: Solving equations with equivalent steps, Solving inequalities with equivalent steps

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