Equivalent modifications of equations

When modifying equations, we use equivalent modifications, which are characterized by the fact that they do not change the validity of the equation. The point of equivalent adjustments is to get the equation into some simpler form from which we can already calculate the result of the equation.

What is an equation

First, a brief discussion of what an equation is and how we will write it. If we have two functions f(x) and g(x), then we can put them into an equation and get the equation:

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

In this case, we call f(x) the left side of the equation and g(x) the right side. We can think of the abstract functions f(x) and g(x) as concrete functions, for example f(x) = 3x and g(x) = x + 2. Then the equation would look like this:

$$3x=x+2$$

The solution to the equation is then the set (for example) S, where for every x in the set S, the given equation still holds when substituted into the equation. So the solution of the example equation would be the set

$$S=\left\{1\right\}.$$

After substituting into the equation, we get

$$3\cdot1=1+2$$

An equivalent modification of the equation is then a modification that does not change the validity of the equation. So an equivalent modification would be a modification where we get the function f₁(x) from the function f(x) and the same modification would get g₁(x) from the function g(x) and yet the equality would still hold

$$f_1(x)=g_1(x).$$

The first and simplest equivalent adjustment of equations (not inequalities!) is to swap the left and right sides. It is probably obvious that if the

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

then swapping them will also be valid

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

I discuss equations more in my article What is an equation.

Addition

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

$$x-3=1.$$

We can solve this equation by adding the number three to both sides. When adjusting equations, we usually write the adjustment after the slash in the line where we started the adjustment, as follows

$$\begin{eqnarray} x-3&=&1\quad /+3\\ 3+x-3&=&1+3\\ x&=&4 \end{eqnarray}$$

In general, we can write down that if we again have two functions f(x) and g(x), we can add a third function h(x) to them and not change the validity of the equation.

$$\begin{eqnarray} f(x)&=&g(x)\quad/+h(x)\\ f(x)+h(x)&=&g(x)+h(x) \end{eqnarray}$$

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

$$x=2x+6$$

By habit, we want to have the variables on the left, which we achieve by adding −2x, or subtracting 2x to the equation.

$$\begin{eqnarray} x&=&2x+6\quad/-2x\\ x-2x&=&2x-2x+6\\ -x&=&6\\ x&=&-6 \end{eqnarray}$$

Of course, we can also add fractions, if that's convenient.

$$x-\frac12=471$$

Add one half and you have the result:

$$\begin{eqnarray} x-\frac12&=&471\quad/+\frac12\\ x-\frac12+\frac12&=&471+\frac12\\ x&=&471{,}5 \end{eqnarray}$$

Adding to an equation is classically used when you want to "flip" an expression from one side of the equation to the other side of the equation. If you have an equation

$$3=2x+17,$$

then how do you get the expression with the variable, i.e. 2x, to the left side of the equation? You simply add −2x to the whole equation. This gives you zero on the right side of the equation instead of 2x because 2x − 2x = 0. And on the other side you get −2x:

$$\begin{eqnarray} 3&=&2x+17\quad/-2x\\ 3-2x&=&2x-2x+17\\ 3-2x&=&17 \end{eqnarray}$$

Multiplying

We can also multiply the equation by some function. The only condition is that the function must not be zero. We can't multiply the equation by zero, nor can we multiply it by some function that can be modified to zero, such as f(x) = x − x. So if we have an equation

$$3x=7,$$

we can multiply it by two:

$$\begin{eqnarray} 3x&=&7\quad/\cdot2\\ 2\cdot3x&=&2\cdot7\\ 6x&=&14 \end{eqnarray}$$

This probably doesn't make much sense, multiplication is used for example when you need to get rid of fractions:

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

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

$$\begin{eqnarray} 3&=&\frac{2}{5x}\quad/\cdot5x\\ 5x\cdot3&=&5x\cdot\frac{2}{5x}\\ \end{eqnarray}$$

Now we multiply the fraction 5x, which puts 5x in the numerator of the fraction, and then we can truncate the expression 5x because we have it in both the denominator and the numerator.

$$\begin{eqnarray} 15x&=&\frac{5x\cdot2}{5x}\\ 15x&=&2\\ \end{eqnarray}$$

I said at the beginning of this section that we must not multiply by zero. Now we've multiplied by the expression 5x - except that this expression contains a variable, and if we add zero to this variable, the whole variable comes out zero as well, and lo and behold, we're already multiplying by zero. What to do?

The way to solve this is to specify the conditions under which we can make this adjustment. So at this point, we have to write that we can only multiply 5x under the condition that x is different from zero. But now let's take another look at the equation we multiplied by. The expression 5x is already there, and it's in the denominator of the fraction. What does that mean? We can't divide by zero, so the very definitional scope of the function 5x precludes x from being equal to zero.

So once again - the equation contains a fraction with the expression 5x in the denominator. Since we can't divide by zero, the equation only makes sense under the assumption that x is different from zero. Therefore, we can easily multiply the entire equation by 5x, because the necessary conditions are already taken care of by the definitional domain of the function in the denominator.

Multiplication by more complex expressions

Now an important note on the fact that when we multiply an equation by an expression, we must always multiply the entire right-hand side and the entire left-hand side. For an example, let's take this equation:

$$x+2=x-4$$

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

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

It would be a mistake not to put the parentheses in because we would get the wrong result:

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

You can see that the result is different from the previous case. Similarly, if you multiply by a more complex expression, you have to multiply the whole equation. For an example, let's take the equation

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

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

$$\begin{eqnarray} 2x+1&=&\frac{2}{x+3}\quad /\cdot(x+3)\\ (x+3)(2x+1)&=&(x+3)\frac{2}{x+3}\\ (x+3)(2x+1)&=&\frac{2(x+3)}{x+3} \end{eqnarray}$$

We have a fraction in a form where we can nicely truncate it with the expression x + 3:

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

Now we can still multiply the left side of the equation and finally subtract two to "move" two to the left side, giving us zero on the right side.

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

Dividing

Just as we can multiply an equation, we can also divide it. With the same restriction as in the case of multiplication - we must not divide by zero. In fact, we can simply turn division into multiplication. If we want to divide an equation by an expression w, it will be the same as if we multiplied the equation by a fraction 1/w. Division is often used in examples such as this one:

$$12x=7$$

This is the type of equation that we can't do much with, we can only divide the whole equation by twelve to get the value x.

$$\begin{eqnarray} 12x&=&7\quad/:12\\ x&=&\frac{7}{12} \end{eqnarray}$$

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

$$\begin{eqnarray} 12x&=&7\quad/\cdot\frac{1}{12}\\ 12x\cdot\frac{1}{12}&=&7\cdot\frac{1}{12}\\ \frac{12x}{12}&=&\frac{7}{12}\\ x&=&\frac{7}{12} \end{eqnarray}$$

Amplification

We can also use powers when modifying equations, but with some limitations. First, let's refresh our knowledge of powers. Does this equality hold?

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

If you multiply a number squared and then perform a square root, do you get back the original number? This is a tricky question, and the answer is not always. Imagine that you put a negative number after x, for example, minus three:

$$\begin{eqnarray} \sqrt{(-3)^2}&=&-3\\ \sqrt{9}&=&-3\\ 3&\ne&-3 \end{eqnarray}$$

As you can see, the equation suddenly doesn't hold because when you square the negative number, you get a positive number. But if we subtract back the positive number, we get a positive number again. When would this equality hold? In the case where we know that we are only working with positive, or non-negative, numbers. So we can afford to make the equation stronger if we know we are working with positive numbers.

For example, in geometry we often calculate some equations that contain the lengths of some line segments, and at this point we can safely multiply because the length itself cannot be negative. Even when multiplying or subtracting, the rule is that we must multiply the whole equation. So an example:

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

Suppose we are only in the positive numbers, so we can afford to multiply the whole equation squared:

$$\begin{eqnarray} \sqrt{x+2}&=&x+4\quad/^2\\ \left(\sqrt{x+2}\right)^2&=&(x+4)^2\\ x+2&=&x^2+8x+16\\ x^2+7x+14&=&0 \end{eqnarray}$$

We have a quadratic equation, which we can solve by the means described in the article.

How would you solve an equation where you need to multiply to the square, but you can't be sure you're only in the positive numbers? You simply multiply and then do the test, because by multiplying you can get results that are not the correct results of the equation. So an example:

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

We will now multiply the equation squared:

$$\begin{eqnarray} \sqrt{x}&=&x-2\quad/^2\\ \left(\sqrt{x}\right)^2&=&(x-2)^2\\ x&=&x^2-4x+4\\ x^2-5x+4&=&0 \end{eqnarray}$$

Again, we get a quadratic equation, so we solve it using some appropriate technique, such as decomposition (see the linked article for the exact procedure):

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

We see that one result is 1 and the other is 4. However, we know that we have made a non-equivalent adjustment, so we still have to do a test. We plug these solutions into the original equation and see which result is a valid result of the original equation. First we plug in one:

$$\begin{eqnarray} \sqrt{x}&=&x-2\\ \sqrt{1}&=&1-2\\ 1&\ne&-1 \end{eqnarray}$$

By checking, we have found that one is not a solution to the original equation. What about the four?

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

Four is the solution to the original equation, we found the result.