OpenMaths/Algebra/Rearranging Equations

Now that we have described the basic rules of negative and positive numbers and what to do when you add, subtract, multiply and divide them, we are ready to tackle some real mathematics problems! Earlier in this chapter, we wrote a general equation for calculating how much change (x) we can expect if we know how much an item costs (y) and how much we have given the cashier (z).

The equation is: $$x + y = z$$

So, if the price is $10 and you gave the cashier $15, then write $15 instead of z and $10 instead of y. $$x + 10 = 15$$

Now that we have written this equation down, how exactly do we go about finding what the change is? In mathematical terms, this is known as solving an equation for an unknown (x in this case). We want to re-arrange the terms in the equation, so that only x is on the left hand side of the = sign and everything else is on the right. The most important thing to remember is that an equation is like a set of weighing scales. In order to keep the scales balanced, whatever, is done to one side, must be done to the other.

Method: Rearranging Equations You can add, subtract, multiply or divide both sides of an equation by any number you want, as long as you always do it to both sides.

When you subtract a number from both sides of an equation, it looks just like you moved a positive number from one side and it became a negative on the other, which is exactly what happened. Likewise if you move a multiplied number from one side to the other, it looks like it changed to a divide. This is because you really just divided both sides by that number, and a number divided by itself is just 1

$$\begin{array}{rcl} a(5 + c) & = & 3a\\ a(5 + c) \div a & = & 3a \div a\\ \frac {a}{a} \times (5 + c) & = & 3 \times \frac{a}{a}\\ 1 \times (5 + c) & = & 3 \times 1\\ 5 + c & = & 3\,\!\\ c & = & 3 - 5\\ c & = & -2 \end{array}$$

Possible Errors
However you must be careful when doing this, as it is easy to make mistakes. The following is the WRONG thing to do

$$\begin{array}{rcl} 5a + c & = & 3a\\ 5 + c & = & 3 \end{array}$$

Can you see why it is wrong? It is wrong because we did not divide the c term by a as well. The correct thing to do is

$$\begin{array}{rcl} 5a + c & = & 3a\\ 5 + c \div a & = & 3\\ c \div a & = & 3 - 5\\ c \div a & = & -2 \end{array}$$