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:
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. 
3 x + 5 x = −16.Solution: Combine the like terms on the left side of the equation:
8 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. 
3 x + 22 = −5 x + 6.Solution: There are linear terms on both sides of the equation. Transpose the −5 x term to the lefthandside by adding 5 x to both sides of the equation:
3 x + 22 + 5 x = −5 x + 6 + 5 x 8 x + 22 = 6.Notice how −5 x has disappeared from the righthandside and +5 x has appeared on the lefthandside. There are constant terms on both sides of the equation so transpose the 22 term to the righthandside by subtracting 22 from both sides:
8 x + 22 − 22 = 6 − 22 8 x = −16.Notice how +22 has disappeared from the lefthandside and −22 has appeared on the righthandside.
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 + b − c). 
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 lefthandside by subtracting c x from both sides of the equation:
a x + b − c x = c x + d − c x a x − c x + b = d.Now transpose the b term to the righthandside by subtracting b from both sides:
a x − c x + b − b = d − b a x − c x = d − b.In the previous example we combined the x terms like this:
3 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 lefthandside. This puts the equation in the form:
x (a − c) = d − bAt this point the collecting of like terms is complete. All the terms containing x have been collected on the lefthandside of the equation and all the constant terms have been collected on the righthandside of the equation and x now appears just once. Divide both sides by a − c and we have the solution:
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. 
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 lefthandside. This gives:
38 − 14 x = 5 x.Now we can transpose the 14 x term to the righthandside 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 righthandside or the 5 x term to the lefthandside. 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.)
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.
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:

4 x − 3 = 6 x + 7.
−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:

7.2 x − 16 = 8.48Solution: First we will clear the decimals. Here are the required steps:
100 (7.2 x − 16) = 100 (8.48)
720 x − 1600 = 848.
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:

150 x = 100 x + 250.Solution: First we will clear the equation of large coefficients. Here are the steps:
50 (3 x) = 50 (2 x + 5).
3 x = 2 x + 5.
x = 5.Check the solution by substituting it back into the original equation. This gives 750 = 750, so the solution checks out.
Algebra Coach Exercises 