Rearranging Formulas Flashcards

All 14 cards in this deck

What does it mean to "make $x$ the subject" of a formula?

Rearrange it so that $x$ stands alone on one side, appearing only once, in the form $x = \dots$

What are the steps to change the subject of a formula when the new subject appears once?

e.g. make $p$ the subject of $d = 3p + 4$

1. Undo the addition/subtraction: $d - 4 = 3p$
2. Undo the multiplication/division: $p = \dfrac{d-4}{3}$

What are the steps to make the subject of a formula when it is inside a square root?

e.g. make $k$ the subject of $y = \sqrt{2m-k}$

1. Square both sides: $y^2 = 2m - k$
2. Rearrange for the subject: $k = 2m - y^2$

When rearranging a formula you reach $r^2 = \dfrac{3V}{\pi h}$. What is the final step, and what must you remember?

Square root both sides: $r = \pm\sqrt{\dfrac{3V}{\pi h}}$ — take the positive root when $r$ is a length.

What are the steps to make the subject of a formula containing a fraction?

e.g. make $m$ the subject of $k = p + \dfrac{2m}{5}$

1. Isolate the fraction: $k - p = \dfrac{2m}{5}$
2. Multiply every term by the denominator: $5(k-p) = 2m$
3. Divide by the coefficient: $m = \dfrac{5(k-p)}{2}$

True or false? Multiplying both sides of $T = \dfrac{q}{2} + 5$ by $2$ gives $2T = q + 5$.

False. Every term must be multiplied by $2$, giving $2T = q + 10$.

What are the steps to find a missing quantity in a standard formula?

e.g. find $h$ from $V = \frac{1}{3}\pi r^2 h$ when $V$ and $r$ are known

1. Rearrange (or substitute first): multiply by $3$, then divide by $\pi r^2$
2. Gives $h = \dfrac{3V}{\pi r^2}$
3. Substitute the known values of $V$ and $r$ and evaluate

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

True or false? In a "show that" rearrangement question, checking that both forms give the same value for one number is enough to earn the marks.

False. You must show the algebraic steps leading to the given result exactly as printed.