Formulas where Subject Appears Twice Flashcards

All 6 cards in this deck

After clearing fractions, the new subject appears in two separate terms. What must you do?

Collect all terms in the subject on one side, factorise, then divide by the bracket.
e.g. $fm - 3m = f + 4 \Rightarrow m(f-3) = f+4$

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: $x = \dfrac{6}{x^2+2}$