Linear equations

A linear equation is an equation that we can rearrange into the form ax + b = 0, where a ≠ 0. A concrete example might look like this: 2x + 4 = 0. The solution of this equation is the number −2, which you can probably work out quite logically in your head. With slightly bigger numbers that kind of reasoning gets harder, so we need a more systematic method.

The form of a linear equation

More precisely, it is an equation with one unknown x that can be rearranged, using equivalent steps, into the form ax + b = 0, where a ≠ 0. The unknown is not squared, square-rooted or anything like that. An equation with the unknown in the denominator of a fraction can only be rearranged into this form on the condition that the denominator is not zero, so we deal with those separately: see linear equations with the unknown in the denominator. Take a look at some examples of linear equations:

$$\begin{array}{rcl}2x+7& =& 0\\ x-2& =& 74\\ (x+3)-(7x\cdot 3)& =& -12x\\ \frac{x}{2}+\frac{3x}{72}& =& x\cdot \frac{45}{8}+1\end{array}$$

As you can see, a linear equation can look all sorts of ways. To solve them nicely, we first need to bring them all into one common form:

$$ax+b=0,$$

where x is the unknown and a and b are any real numbers, called the coefficients of the equation. The number a must not be zero, that is, a ≠ 0. We call ax the linear term and b the constant term.

We cannot solve a linear equation without equivalent steps, that is, steps that do not change the solution of the equation. Without them you cannot solve any more complicated equation. We can get an equation into the form ax + b = 0 either by simplifying the expressions in the equation or by using equivalent steps on the equation.

Let’s go back to the examples above. The equation 2x + 7 = 0 is already in this form, so we can see the coefficients straight away: a = 2 and b = 7. The second equation, x − 2 = 74, is not so easy. We use an equivalent step and add −74 to both sides (in other words, subtract 74 from both sides) and get x − 76 = 0. This is now in the form ax + b = 0, with coefficients a = 1 and b = −76.

The third equation is still a long way from this form, but we can deal with it too. First we “make” a zero on the right-hand side: we add the expression 12x to both sides. This turns the right-hand side into zero, because (−12x) + 12x = 0. The whole equation now looks like this:

$$(x+3)-(7x\cdot 3)+12x=0$$

All that remains is to remove the brackets and add up the terms with x.

$$\begin{array}{rcl}(x+3)-(7x\cdot 3)+12x& =& 0\\ x+3-21x+12x& =& 0\\ -8x+3& =& 0\end{array}$$

This linear equation is in the required form, and its coefficients are a = −8 and b = 3.

In the last equation we have to get rid of the fractions. So we multiply the whole equation by 72, which is divisible by all the denominators:

$$72\cdot \frac{x}{2}+72\cdot \frac{3x}{72}=72x\cdot \frac{45}{8}+72\cdot 1$$

Now we cancel the fractions:

$$36x+3x=9x\cdot 45+72$$

On the left-hand side we add, on the right-hand side we multiply:

$$39x=405x+72$$

One last step: we subtract 405x and 72 from both sides, and the equation is in the required form:

$$-366x-72=0$$

So the coefficients are a = −366 and b = −72.

How to solve a linear equation

Once a linear equation is in the form ax + b = 0, it is easy to solve. Let’s look at an example: 3x − 18 = 0. What do we need to subtract 18 from to get zero? Well, from 18 itself. So for the expression 3x − 18 to equal zero, the expression 3x must equal 18, because then we get 18 − 18 = 0. When does 3x equal 18? Exactly when x = 6, because 3 · 6 = 18.

How do we turn this into a general method? First we need to find out what the linear term ax has to equal. We do that by moving the constant term to the right-hand side, that is, by adding −b to both sides. This gives us ax + b − b = −b, which is the same as ax = −b. In the previous example we would get 3x − 18 + 18 = 18, so 3x = 18. Don’t forget that b was equal to b = −18, so when we add −b we are adding −(−18), which is +18.

Now we have the equation in the form ax = −b. How do we get x on its own? In the example we were looking for the x that gives 18 when multiplied by three. What we actually did was divide: $18÷3=6$. So we took the expression on the right-hand side (−b) and divided it by a, which was three in our example. From ax = −b we therefore get

$$x=\frac{-b}{a}=-\frac{b}{a}$$

Here is the whole method applied to the first example:

$$\begin{array}{rcl}3x-18& =& 0(+18)\\ 3x& =& 18(÷3)\\ x& =& 6\end{array}$$

Geometric meaning

If we rearrange a linear equation into the form ax + b = 0, the left-hand side is a linear function. The graph of a linear function is a straight line. Since the equation has the form ax + b = 0, its solutions are all the points where the line (the graph of the linear function) crosses the x-axis (which is also the graph of the function f(x) = 0, that is, the right-hand side).

For example, we found that the solution of 3x − 18 = 0 is x = 6. What does that mean? It means that the graph of the function 3x − 18 crosses the x-axis at x = 6.

Let’s try to solve the equation 5x + 2 = 2x − 7 graphically. How can we go about it? We can rearrange the equation into the form ax + b = 0: add −2x to get 3x + 2 = −7, and then add 7 to get 3x + 9 = 0. Now we draw the graph of the linear function y = 3x + 9 and see where it crosses the x-axis.

The second way is not to rearrange the equation at all, but to draw the graphs of the functions on both sides of the equation. That is, we draw the graphs of f(x) = 5x + 2 and g(x) = 2x − 7.

These two lines meet at a single point, labelled A. What is the x-coordinate of this point A? Once again it is x = −3. The graphical solution of 5x + 2 = 2x − 7 is therefore the x-coordinate of the point where the graphs of the left-hand and right-hand sides intersect.

Worked examples

Example 1: Let’s solve the equation 7x − 14 = 0. This one is easy: the equation is already in the form ax + b = 0, so we just apply the method:

$$\begin{array}{rcl}7x-14& =& 0(+14)\\ 7x& =& 14(÷7)\\ x& =& 2\end{array}$$

Example 2: Solve the equation 2x + 10 = 0. Again the equation is in the form ax + b = 0, but watch out: b is positive, so after rearranging you get a negative number on the right-hand side:

$$\begin{array}{rcl}2x+10& =& 0(-10)\\ 2x& =& -10(÷2)\\ x& =& -5\end{array}$$

Example 3: Solve the equation 15 − 3x = 0. The equation is almost in the form ax + b = 0; we only need to put the unknown x first. Then the equation reads −3x + 15 = 0. Watch out again: there is a minus in front of the linear term, which we have not seen before. The steps are exactly the same, though:

$$\begin{array}{rcl}-3x+15& =& 0(-15)\\ -3x& =& -15(÷(-3))\\ x& =& 5\end{array}$$

We can also solve this example a slightly different, longer way. Instead of dividing the equation straight away by −3, we can first multiply it by −1, which gets rid of both minus signs. Then we can divide by three:

$$\begin{array}{rcl}-3x+15& =& 0(-15)\\ -3x& =& -15(\cdot (-1))\\ 3x& =& 15(÷3)\\ x& =& 5\end{array}$$

Example 4: Solve the linear equation

$$3+\frac{x}{4}=\frac{x}{2}$$

This equation is a little harder because it contains fractions. First we need to get rid of them. The best way is to multiply the whole equation by the least common multiple (LCM) of all the denominators. We have two denominators, 4 and 2, and their LCM is 4. If you don’t feel like working that out, you can multiply the whole equation by the product of the denominators, 4 · 2 = 8, but that is not the recommended method, because it makes life harder for you a step later. If we multiply the equation by four, we get rid of the fractions by cancelling them down:

$$\begin{array}{rcl}3+\frac{x}{4}& =& \frac{x}{2}(\cdot 4)\\ 4\cdot 3+4\cdot \frac{x}{4}& =& 4\cdot \frac{x}{2}\\ 12+\frac{4x}{4}& =& \frac{4x}{2}\\ 12+x& =& 2x\end{array}$$

Now we put the equation into the form ax + b = 0: x − 12 = 0, and from this we can see that x = 12.

Example 5 (word problem): Take two brothers, Oliver and Harry. Harry went straight into work after school, while Oliver went on to university, where he studied for five years. Harry earned a wage of £2,500 every month. Although Oliver studied Leisure Studies with a specialism in lying on a sun lounger, he managed to get a job in the civil service, where he earned £3,500 a month. How long will it take Oliver to earn the same total amount of money as Harry? To be clear: in five years Harry earned 5 · 12 · 2,500 pounds, while Oliver has nothing after five years. After how many months will these amounts be equal?

From the question we can build a linear equation as follows. The variable x will stand for the number of months we are looking for, with x = 0 meaning zero months since Oliver graduated.

First we build a function giving Oliver’s earnings in terms of the number of months. If we are interested in Oliver’s earnings after six months, we simply work out that 3,500 · 6 = 21,000 pounds. In general, after x months Oliver has earned 3,500 · x pounds in total.

For Harry it is similar: 2,500 · x, but we must also add the amount he earned while Oliver was studying. Oliver studied for five years, so during his studies Harry earned 5 · 12 · 2,500 pounds. The whole expression for Harry is therefore 5 · 12 · 2,500 + 2,500 · x.

We put these expressions into an equation, because we want to know for which x the earnings are equal:

$$5\cdot 12\cdot \mathrm{2,500}+\mathrm{2,500}\cdot x=\mathrm{3,500}\cdot x$$

We rearrange the equation by moving all the x terms to one side:

$$1000x=5\cdot 12\cdot \mathrm{2,500}$$

And finally we divide by 1000:

$$x=5\cdot 12\cdot 2.5=150$$

After 150 months they will have earned the same amount. During Oliver’s studies Harry earned a handsome 5 · 12 · 2,500 = 150,000 pounds, and in the next 150 months he earned 2,500 · 150 = 375,000, which makes 150,000 + 375,000 = 525,000 pounds in total. Oliver worked only for those 150 months, and in that time he earned 3,500 · 150 = 525,000 pounds. We can see that after five years of study and another 150 months, the two really have earned the same.

More examples

Solve the following linear equation:

$$2\cdot (x-7)=6$$

First we expand the bracket:

$$2x-14=6$$

Now we keep the term with the unknown on the left-hand side and move everything else to the right-hand side:

$$2x=20$$

… and finally we divide by two.

$$x=10$$

Solve the equation

$$3\cdot (3-x)+5\cdot (x-2)=0$$

Again we first expand the brackets:

$$9-3x+5x-10=0$$

We collect like terms, then keep the terms with the unknown on the left-hand side and move the rest to the right-hand side:

$$\begin{array}{rcl}9-3x+5x-10& =& 0\\ 2x-1& =& 0\\ 2x& =& 1\end{array}$$

Finally we divide the whole equation by two:

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