Rearranging Formulas Flashcards

All 13 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

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? Checking that both forms give the same value for one number is enough to show that an equation rearranges to a given form.

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