Solving other linear equations

What if we have an equation such as \({p}\) + 3 = 11? This time, in order to get \({p}\) on its own we would need to subtract 3 from both sides of the equation. This would leave us with \({p}\) = 8.

If we were told that \({L}\) + 9 = 14 we would subtract 9 from both sides leaving \({L}\) = 5.

If we had \({Q}\) + 3 = 1 we would subtract 3 from both sides leaving \({Q}\) = –2.

If we had \({S}\) + 12 = –4 we would subtract 12 from both sides, leaving \({S}\) = –16.

Similarly if we had an equation such as \({R}\) – 12 = 3, we now need to add 12 to both sides in order to get \({R}\) on its own on the left hand side of the equation. This would give \({R}\) = 15.

Very often solving equations requires more than one step, but the method is still what you do to one side of the equation, you must also do to the other side.

Take for example 2\({y}\) + 6 = 12. There are two steps that we have to take - we will have to divide to remove the 2 in front of the \({z}\) and then subtract to remove the 6.

Students often make the mistake of dividing both sides by 2 and writing:

\({y}\) + 6 = 6

This is not correct. The equation, when divided by 2, should read:

\({y}\) + 3 = 6

This is a very important difference. When you divide a side of the equation, you have to divide every term on that side of the equation. This is why it is often recommended that students perform the addition or subtraction first.

Let’s solve the equation:

2\({y}\) + 6 = 12

Subtracting 6 from both sides leaves:

2\({y}\) = 6

Dividing both sides of the equation by 2:

\({y}\) = 3

Example

Solve \(\frac{x}{3}~–~{2}={4}\)

Solution

Adding 2 to both sides gives:

\(\frac{x}{3}={6}\)

Multiplying both sides by 3:

\({x}={18}\)

Question

Solve \(\frac{3y}{4}={9}\)

Equations and formulae – WJECSolving other linear equations