# Fundamentals in solving equations in one or more steps

Formulas are very common within physics and chemistry, for example, velocity equals distance divided by time. Thus we use the common symbols for velocity (v), distance (d) and time (t) and express it thus:

$\\ v=\frac{d}{t} \\$

We may simply describe a formula as being a variable and an expression separated by an equal sign between them. In other words a formula is the same as an equation.

Example:

A book club requires a membership fee of $10 in addition to the$2 levied for each book ordered. If we were to list the cost of ordering a number of books, it would look like:

 Number of books Cost 1 10 + 2 ∙ 1 = 12 2 10 + 2 ∙ 2 = 14 3 10 + 2 ∙ 3 = 16 4 10 + 2 ∙ 4 = 18 5 10 + 2 ∙ 5 = 20 x 10 + 2x

If we designate the total book club cost as C, we may derive the following formula for the expression:

$\\ C=10 + 2x \\$

If we then want to know how many books we may get from the book club for $30 we can either continue filling in the table above or use the properties of equations that we handled in the last section.  $\\ 30=10 + 2x \\$ C was the cost, i.e. it is now$30 $\\ 30\: {\color{green} -\, 10}=10+2x\, {\color{green} -\, 10} \\$ we subtract $10 from each side $\\ 20=2x \\$ simplify $\\ \frac{20}{{\color{green} 2}}=\frac{2x}{{\color{green} 2}} \\$ divide both sides by 2 to isolate x $\\ 10=x \\$ x equals 10 We may purchase 10 books for$30.

When we want to solve an equation including one unknown variable, as x in the example above, we always aim at isolating the unknown variable. You can say that we put everything else on the other side of the equal sign. It is always a good idea to first isolate the terms including the variable from the constants to begin with as we did above by subtracting or adding before dividing or multiplying away the coefficient in front of the variable. As long as you do the same thing on both sides of the equal sign you can do whatever you want and in which order you want.

Above we began by subtracting the constant on both sides. We could have begun by dividing by 2 instead. It would have looked like

$\\ \frac{30}{{\color{blue} 2}}=\frac{10+2x}{{\color{blue} 2}} \\$

$\\ \frac{30}{{\color{blue} 2}}=\frac{10}{{\color{blue} 2}}+\frac{2x}{{\color{blue} 2}} \\$

$\\ 15=5+x \\$

$\\ 15\, {\color{blue} -\, 5}=5+x\, {\color{blue} -\, 5} \\$

$\\ 10=x \\$

Again the same answer just proving the point.

If your equation contains like terms it is preferable to begin by combining the like terms before continuing solving the equation

Example

$\\ 5x+14+2x+2=30 \\$

Begin by combining the like terms (all terms including the same variable x and all constants)

$\\ \left (5x +2x \right )+\left (14+2 \right )=30 \\$

$\\ 7x+16=30 \\$

Now it's time to isolate the variable from the constant part. This is done by subtracting 16 from both sides

$\\ 7x+16\, {\color{green} -\, 16}=30\, {\color{green} -\, 16} \\$

$\\ 7x=14 \\$

Divide both sides by 7 to isolate the variable

$\\ \frac{7x}{{\color{green} 7}}=\frac{14}{{\color{green} 7}} \\$

$\\ x=2 \\$

If you have an equation where you have variables on both sides you do basically the same thing as before. You collect all like terms. Before you have worked by first collecting all constant terms on one side and keep the variable terms on the other side. The same applies here. You collect all constant terms on one side and the variable terms on the other side. It's usually a good idea to collect all variables on the side that has the variable with the highest coefficient i.e. in the example below there are more x:es on the left side (4x) compared to the right side (2x) and hence we collect all x:es on the left side.

Example

$\\ 4x+3 =2x+11 \\$

subtract 2x from both sides

$\\ 4x+3\, {\color{blue} -\, 2x}=2x+11\, {\color{blue} -\, 2x} \\$

Now it looks like any other equation

$\\ 2x+3=11 \\$

subtract 3 from both sides

$\\ 2x+3\, {\color{blue} -\, 3}=11\,{\color{blue} -\, 3} \\$

$\\ 2x=8 \\$

Divide by 2 on both sides

$\\ \frac{2x}{{\color{blue} 2}}=\frac{8}{{\color{blue} 2}} \\\\ x=4 \\$

In the beginning of this section we showed the formula for calculating the velocity where velocity (v) equals the distance (d) divided by time (t) or

$\\v=\frac{d}{t}$

If we by some chance want to know how far a truck drives in 3 hours at 60 miles per hour we can use the formula above and rewrite it to solve the distance, d.

$\\\frac{d}{t}\, {\color{green} \cdot \, t}=v\,{\color{green} \cdot \, t}$

$\\d=v\cdot t\\$

When that's done we can just put our numbers in the formula and calculate the answer

$d=60\cdot 3=180$

The truck travels 180 miles in 3 hours.

This holds true for all formulas and equations.

Videolesson: Solve the equation

$\\3\left (x + 2 \right )- 3 + x + 17 = 40\\$