4.2 - Linear Equations

A linear equation is an equation in which the terms containing the unknown are all of first degree. Assuming that the unknown is x, this means that the equation does not contain powers of x such as x 2 or x 3 or functions of x such as sin(x). A linear equation has the form:
a x + b = c x + d,
where a, b, c and d are constants; or it can be put into this form by distributing. Any term containing x is called a linear term and any term not containing x is called a constant term. Here are some examples of linear equations: We already know how to solve the first two of these equations because the unknown appears only once (in fact the second equation is already in solved form). However the last three have the unknown appearing more than once and these are the linear equations that we are interested in learning how to solve here. We now present several techniques for solving them.


Technique 1: Collecting like terms


If there are like terms on one side of the equation then combine them. (All the linear terms in the equation are considered to be alike, and all the constant terms are considered to be alike.) Combining the linear terms turns an equation in which the unknown appears more than once into one in which the unknown appears only once, and we have already seen how to solve that type.



Example: Solve this equation for x:
x + 5 x = −16.
Solution: Combine the like terms on the left side of the equation:
x = −16.
Notice that this gives an equation in which x appears just once. Now divide both sides by 8:
x = −2.
This is the solution. Check it by substituting back into the original equation. This gives −16 = −16, so the solution checks out.




Suppose that there are like terms, but on opposite sides of the equation. Then choose one of the terms and subtract (or add) it to both sides of the equation. This makes that term disappear from one side of the equation and makes its opposite or negative appear on the other side of the equation. This is called “transposing the term”.

Continue transposing terms until you have collected all the linear terms on one side of the equation and all the constant terms on the other side of the equation. With all the like terms now on the same side of the equation, they can be combined.



Example: Solve this equation for x:
x + 22 = −5 x + 6.
Solution: There are linear terms on both sides of the equation. Transpose the −5 x term to the left-hand-side by adding 5 x to both sides of the equation:
x + 22 + 5 x = −5 x + 6 + 5 x x + 22 = 6.
Notice how −5 x has disappeared from the right-hand-side and +5 x has appeared on the left-hand-side. There are constant terms on both sides of the equation so transpose the 22 term to the right-hand-side by subtracting 22 from both sides:
x + 22 − 22 = 6 − 22 x = −16.
Notice how +22 has disappeared from the left-hand-side and −22 has appeared on the right-hand-side.

At this point the collecting of like terms is complete: All of the linear terms containing x have been collected on the left-hand-side of the equation and all the constant terms have been collected on the right-hand-side of the equation and x now appears just once. Divide both sides by 8 and we have the solution:
x = −2.
Check it by substituting back into the original equation. This gives 16 = 16, so the solution checks out.




To collect like linear terms that have literal coefficients, simply factor x out of them. For example, to collect a x + b x − c x, factor like this:
x (a + bc).


Example: Solve this literal equation for x:
a x + b = c x + d.
Solution: We perform exactly the same steps as in the previous example. Transpose the c x term to the left-hand-side by subtracting c x from both sides of the equation:
a x + b c x = c x + d c x a xc x + b = d.
Now transpose the b term to the right-hand-side by subtracting b from both sides:
a xc x + b b = d b a xc x = db.
In the previous example we combined the x terms like this:
x + 5 x(3 + 5) x → 8 x.
The blue factoring step was not shown but it was implied. We do the same thing here to collect the x terms: we factor out the common factor of x on the left-hand-side. This puts the equation in the form:
x (ac) = db
At this point the collecting of like terms is complete. All the terms containing x have been collected on the left-hand-side of the equation and all the constant terms have been collected on the right-hand-side of the equation and x now appears just once. Divide both sides by ac and we have the solution:

Technique 2: Distributing


Suppose that a linear equation has a linear term and a constant term together in brackets. Then in order to collect linear terms on one side of the equation and constant terms on the other side, we must use the distributive law to remove the brackets.



Example: Solve this equation for x:

2 (19 − 7 x) = 5 x.
Solution: Because the 19 term and the 7 x term are together inside brackets there is no way to transpose one without the other. Therefore we must distribute on the left-hand-side. This gives:
38 − 14 x = 5 x.
Now we can transpose the 14 x term to the right-hand-side by adding 14 x to both sides. This gives:
38 = 19 x.
(Note that we could have transposed either the 14 x term to the right-hand-side or the 5 x term to the left-hand-side. We chose to transpose the 14 x term because (1) it results in one less algebra step to get the solution, and (2) it gives us a positive term, namely 19 x, on the right, rather than a negative term, −19 x, on the left.)

Now the unknown appears just once. Divide both sides by 19. This gives:
2 = x,
which means that the solution is x = 2. Substituting the solution back into the original equation gives 10 = 10, so it checks out.


Note: Just because a linear equation has brackets does not necessarily mean that you should distribute. Here are two examples where you shouldn’t:

Technique 3: Clearing fractions, decimal coefficients and large coefficients


Equations are generally easier to solve if they don’t contain fractions or decimal coefficients or large coefficients. To clear an equation of fractions, carry out the following steps:
  • Look at the denominators of all the fractions and find their least common multiple (LCM) (this is also called the least common denominator of the fractions).

  • Multiply both sides of the equation by the LCM.

  • Distribute the LCM over both sides of the equation.


Example: Solve this equation for x:
Solution: First we will clear the fractions. Here are the required steps: The fractions are now cleared and we can carry on as in previous examples. We will collect linear terms on the left-hand-side and constant terms on the right-hand-side. Transpose the 6 x term to the left-hand-side by subtracting 6 x from both sides. This gives:
−2 x − 3 = 7.
Add 3 to both sides. This gives:
−2 x = 10.
Divide both sides by −2. This gives the solution:
x = −5.
Check it by substituting it back into the original equation. This gives −23 / 6 = −23 / 6, so the solution checks out.




To clear an equation of decimals, carry out the following steps:
  • Look at the coefficients of all the terms that have decimal points and find the lowest power of 10 that they can all be multiplied by to turn them all into integers. (By power of 10 we mean 10, 100, 1000, etc.)

  • Multiply both sides of the equation by this power of 10.

  • Distribute this power of 10 over both sides of the equation.


Example: Solve this equation for x:
7.2 x − 16 = 8.48
Solution: First we will clear the decimals. Here are the required steps: The decimals are now cleared and we can carry on as in previous examples. Add 1600 to both sides. This gives:
720 x = 2448.
Divide both sides by 720. This gives the solution:
x = 3.4.
Note that generally if an equation contains decimals then you should give the solution in decimal form as well. Check the solution by substituting it back into the original equation. This gives 8.48 = 8.48, so the solution checks out.




To clear an equation of large coefficients, carry out the following steps:
  • Find the largest numerical factor that can be factored out of both sides of the equation.

  • Factor the factor out of both sides of the equation.

  • Divide both sides of the equation by the factor.


Example: Solve this equation for x:
150 x = 100 x + 250.
Solution: First we will clear the equation of large coefficients. Here are the steps: Now we can carry on as in previous examples. Subtracting 2 x from both sides of the equation gives the solution:
x = 5.
Check the solution by substituting it back into the original equation. This gives 750 = 750, so the solution checks out.




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A flowchart for solving simple equations

Here is a flowchart that shows the procedure for solving all of the basic types of equation that we have seen in this chapter. The red area at the top refers to equations of the type a · b = 0, the green area on the left refers to equations in which the unknown occurs just once, and the blue area on the right refers to linear equations. In later chapters we will encounter more complicated types of equations and then we will add new sections to this flowchart where the ??? circle is.

To use the flowchart, start at the “Startellipse and follow the arrows until you reach the “equation solved, check solution” ellipse. At a diamond you are asked a question and you proceed in one of two directions depending on your answer. At a rectangle you perform some operation. Click on any diamond or rectangle to see details or an example.