Solving linear equations

#### Tier: #Foundation #Higher

Description

The variable (usually $x$) will either be on one side of the equation (one step equations) or both sides (two step equations)

1 step: $$3x+1=16$$ 2 step:$$3x+1 = x+7$$ Our task is to isolate $x$ by removing anything else that is on the same side of the equation. In the first example, $x$ has been multiplied by 3 and had 1 added to it in order to make 10. When solving an equation, we reverse the BIDMAS operations (SAMDIB) in the correct order and apply each change to both sides of the equation to keep it balanced.

So, if BIDMAS tells us to multiply by 3 first, then add 1, when we are solving the equation we reverse this and do the inverse operation (add becomes subtract, multiply becomes divide and vice-versa). We therefore subtract 1 from both sides first, then divide by 3 second. This is how it looks: $$3x=15$$ Here we subtracted 1 from both sides $$\frac{3x}{3}=\frac{15}{3}$$ Then we divided both sides by 3 $$x=5$$

For two step equations, we do a single setup move first and then it becomes a one step. We first identify which side of the equation has the 'biggest' (most positive) $x$. $$3x+1 = x+7$$ The left hand side (LHS) has $3x$ and the right hand side has $x$ or $1x$. As the LHS has the bigger $x$, we will collect all of the $x$ terms on that side. It is perfectly fine to do the right hand side but we will be dealing with more negative numbers.

Setup move: Remove the $x$ from the RHS by subtracting $x$ from both sides. $$3x+1-x=7$$ Collect the $x$ terms together: $$2x+1=7$$ We now have a one step equation to solve.

Links

Manipulating algebraic fractions Expanding single brackets Calculator use