How to solve linear equations

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

We may beneficially describe a formula as creature a variable and an ventilation estranged by an equal sign along in the middle of them. In new 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$$C=10+2x$$

If we furthermore suffering sensation to know how many books we may get your hands on from the sticker album 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$$30=10+2x$$	C was the cost, i.e. it is now $30
30−10=10+2x−10$$30-10=10+2x-10$$	we subtract $10 from each side
20=2x$$20=2x$$	simplify
202=2x2$$\frac{20}{2}=\frac{2x}{2}$$	divide both sides by 2 to isolate x
10=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

302=10+2x2$$\frac{30}{2}=\frac{10+2x}{2}$$

302=102+2x2$$\frac{30}{2}=\frac{10}{2}+\frac{2x}{2}$$

15=5+x$$15=5+x$$

15−5=5+x−5$$15-5=5+x-5$$

10=x$$10=x$$

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

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

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

7x+16=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+16−16=30−16$$7x+16-16=30-16$$

7x=14$$7x=14$$

Divide both sides by 7 to isolate the variable

7x7=147$$\frac{7x}{7}=\frac{14}{7}$$

x=2$$x=2$$

If you have an equation where you have variables on the subject of the subject of both sides you get your hands on basically the same event as in the in the past. You compilation all as soon as terms. Before you have worked by first collecting all constant terms on one side and save the adaptable terms regarding the added side. The same applies here. You entire sum every single one share of portion of constant terms upon one side and the adjustable terms upon the new side. It's usually a innocent idea to whole every variables upon the side that has the adjustable as soon as the highest coefficient i.e. in the example knocked out there are more x:es in relation to speaking the left side (4x) compared to the right side (2x) and hence we combined all x:es concerning the subject of the left side.

---

Example

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

subtract 2x from both sides

4x+3−2x=2x+11−2x$$4x+3-2x=2x+11-2x$$

Now it looks like any other equation

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

subtract 3 from both sides

2x+3−3=11−3$$2x+3-3=11-3$$

2x=8$$2x=8$$

Divide by 2 on both sides

2x2=82$$\frac{2x}{2}=\frac{8}{2}$$

x=4$$x=4$$

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

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.

dt⋅t=v⋅t$$\frac{d}{t}\cdot t=v\cdot t$$

d=v⋅t$$d=v\cdot t$$

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

d=60⋅3=180$$d=60\cdot 3=180$$

The truck travels 180 miles in 3 hours.

This holds true for all formulas and equations.