Compared to what we’ve tackled so far in algebra this is a walk in the palk. Instead of rearranging or changing equations we are going to use just a fraction of the knowledge we’ve learnt in algebra so far in order to find the number value for letters in equations with no exponents.
A linear equation is where you have no algebraic exponent greater than 1. For example, quadratics have x squared terms. They are harder, this is simpler. We can solve linear equations through rearranging to make the letter be the subject on one side of the equation.
There’s nothing much to teach in this lesson other than to walk through some examples. Provided you avoid making silly mistakes these are easy marks to be won in the exam:
$$ 5x + 8 \color{red}{-8} = 12 \color{red}{-8} $$ $$ 5x \color{red}{\div 5} = 4 \color{red}{\div 5} $$ $$ x = \dfrac{4}{5} $$
This was so easy and took 2 steps of working! Just remember to balance the force…. Let’s try a slightly harder example:
$$ 7(x + 3) = 5x - 8 $$ $$ 7x + 21 \color{red}{-5x - 21} = 5x - 8 \color{red}{-5x - 21} $$ $$ \dfrac{2x}{\color{red}{2}} = \dfrac{-29}{\color{red}{2}} $$ $$ x = \dfrac{-29}{2} = -14.5 $$
Just like with the rest of algebra, we can make sure we have the correct answer by checking it with the original equation:
$$ \color{red}{-14.5} \times 7 + 21 = \color{red}{-14.5} \times 5 - 8 $$ $$ -80.5 = -80.5 $$
Let’s try one final linear equation:
$$ \dfrac{4x + 5}{2} = 3 $$
$$ \dfrac{4x + 5}{2} \color{red}{\times 2} = 3 \color{red}{\times 2} $$ $$ 4x + 5 \color{red}{-5} = 6 \color{red}{-5} $$ $$ \dfrac{4x}{\color{red}{4}} = \dfrac{1}{\color{red}{4}} $$ $$ x = 0.25 $$