Formulas where Subject Appears Twice Flashcards

All 5 cards in this deck

What are the steps to make the subject of a formula where it appears twice?

e.g. make $m$ the subject of $f = \dfrac{3m+4}{m-1}$

1. Clear the fraction: $f(m-1) = 3m+4$
2. Expand and isolate the $m$ terms: $fm - 3m = f + 4$
3. Factorise and divide: $m = \dfrac{f+4}{f-3}$

True or false? To make $x$ the subject of $y = \dfrac{4(2x-7)}{5x+3}$, the correct first step is to multiply both sides by $5x+3$.

True. Clearing the denominator first gives $y(5x+3) = 4(2x-7)$.

True or false? $x = \dfrac{3y+28}{8-5y}$ and $x = \dfrac{-3y-28}{5y-8}$ are different answers.

False. They are equivalent — numerator and denominator have both been multiplied by $-1$.

The ratio $(y+x):(y-x)$ is equivalent to $k:1$. What equation does this give as a starting point for a "show that" rearrangement?

$\dfrac{y+x}{y-x} = k$, i.e. $y + x = k(y-x)$

What are the steps to show that an equation can be rearranged into a given form?

e.g. show that $x^3 + 2x - 6 = 0$ can be rearranged to give $x = \dfrac{6}{x^2+2}$

1. Move the term without the subject to the other side: $x^3 + 2x = 6$
2. Factorise out the subject: $x(x^2+2) = 6$
3. Divide by the bracket to reach the given result: $x = \dfrac{6}{x^2+2}$