⇒x = 7 + 2 (by transposing 2 to R.H.S.) ⇒y = 10 − 3 (by transposing 3 to R.H.S.) ⇒y= 7. 1.6 = y/1.5 ⇒1.6 x 1.5 = (y/1.5) x 1.5 [On multiplying both sides by 1.5] ...
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.
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