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equals sign."},{"text":"The letter repeated most often on both sides of the equals sign.","correct":false,"explanation":"A letter appearing several times is not necessarily the subject. The subject is the single letter isolated on one side of the equals sign."},{"text":"The first letter written on the left side of the equals sign.","correct":false,"explanation":"The first letter written is not automatically the subject. The subject is identified by being isolated on one side of the equals sign."},{"text":"The number multiplying a letter on one side of the equals sign.","correct":false,"explanation":"A numerical factor multiplying a letter is its coefficient, not the subject. The subject is the single letter isolated on one side of the equals sign."}]},{"id":"46b2f10c-6eac-4fc9-b3d5-8aab0f338ff2","cardType":"basic","points":1,"front":"What is the balance rule for rearranging equations?","back":"Whatever is done to one side of the equation must be done to the other side.","mcqOptions":[{"text":"Whatever is done to one side of the equation must be done to the other side.","correct":true,"explanation":"Applying the same operation to both sides preserves equality, so the two expressions remain balanced while the required variable is isolated."},{"text":"Whatever is done to one side of the equation must be reversed on the other side.","correct":false,"explanation":"Using the inverse operation on the other side generally destroys equality. The same operation must be applied to both sides of the equation."},{"text":"Whatever is done to one side of the equation must not affect the other side.","correct":false,"explanation":"Changing only one side makes the two sides unequal. Whatever is done to one side must also be done to the other side."},{"text":"Whatever is added to one side of the equation must be multiplied on the other side.","correct":false,"explanation":"Swapping operations is not the balance rule and will not generally preserve equality. The same operation must be performed on both sides."}]},{"id":"d3f52e45-4d59-489d-8fa2-953d64e736f2","cardType":"cloze","points":null,"front":"In algebraic rearranging, [ ] is undone by subtracting and squaring is undone by square rooting.","back":"In algebraic rearranging, adding is undone by subtracting and squaring is undone by square rooting.","mcqOptions":[{"text":"adding","correct":true,"explanation":"Subtraction is the inverse operation of addition, so subtracting the same quantity reverses an addition while preserving the equation’s balance."},{"text":"multiplying","correct":false,"explanation":"Division reverses multiplication, not subtraction. The operation undone by subtracting is adding because addition and subtraction are inverse operations."},{"text":"squaring","correct":false,"explanation":"Square rooting reverses squaring, whereas subtraction does not. The operation undone by subtracting is adding, its inverse operation."},{"text":"subtracting","correct":false,"explanation":"Adding is used to undo subtraction, not the other way around. Subtracting undoes adding because these operations form an inverse pair."}]},{"id":"00712559-130d-4637-a615-da95f12a86fc","cardType":"cloze","points":null,"front":"In algebraic rearranging, adding is undone by subtracting and [ ] is undone by square rooting.","back":"In algebraic rearranging, adding is undone by subtracting and squaring is undone by square rooting.","mcqOptions":[{"text":"squaring","correct":true,"explanation":"Taking a square root reverses the operation of squaring, subject to considering both positive and negative roots in pure algebra."},{"text":"multiplying","correct":false,"explanation":"Multiplication is reversed by division, not by taking a square root. Square rooting specifically undoes squaring."},{"text":"cubing","correct":false,"explanation":"Cubing is reversed by taking a cube root rather than a square root. The operation undone by square rooting is squaring."},{"text":"square rooting","correct":false,"explanation":"Square rooting is reversed by squaring, so the stated direction is backwards. The operation that square rooting undoes is squaring."}]},{"id":"c26ee2fe-095c-4b94-89e5-d3ca6797e989","cardType":"basic","points":1,"front":"If the subject is affected by multiple operations, in what order should they be undone?","back":"From the outside in.","mcqOptions":[{"text":"From the outside in.","correct":true,"explanation":"The outermost operation was applied last, so it must be reversed first. Continuing towards the subject undoes the operations in reverse order."},{"text":"From the inside out.","correct":false,"explanation":"Starting with the operation closest to the subject reverses the construction order incorrectly. Operations should be undone from the outside in."},{"text":"From the products to sums.","correct":false,"explanation":"Multiplication is not always the first operation to reverse; the structure of the formula determines the order. Work from the outside in."},{"text":"From the left to right.","correct":false,"explanation":"The written left-to-right order does not reliably show which operation was applied last. Reverse operations structurally, working from the outside in."}]},{"id":"73fd5015-2c06-4908-b17f-cf0d54361c74","cardType":"cloze","points":null,"front":"When making $p$ the subject of $q^2 = p^2 + 5rs$, you must first isolate $p^2$ to get $\\text{[...]}$.","back":"When making $p$ the subject of $q^2 = p^2 + 5rs$, you must first isolate $p^2$ to get $\\boxed{q^2 - 5rs = p^2}$.","mcqOptions":[{"text":"$$q^2 - 5rs = p^2$","correct":true,"explanation":"Subtracting $5rs$ from both sides removes the added term beside $p^2$, leaving $p^2$ isolated while maintaining equality."},{"text":"$$q^2 + 5rs = p^2$","correct":false,"explanation":"Adding another $5rs$ does not cancel the existing positive term. Subtracting $5rs$ from both sides gives $q^2 - 5rs = p^2$."},{"text":"$$q^2 - rs = p^2$","correct":false,"explanation":"This forgets the coefficient $5$ when moving the whole term $5rs$. Subtracting the complete term gives $q^2 - 5rs = p^2$."},{"text":"$$5rs - q^2 = p^2$","correct":false,"explanation":"Reversing the subtraction changes the sign of the isolated expression. Subtracting $5rs$ from $q^2$ correctly gives $q^2 - 5rs = p^2$."}]},{"id":"8abe30c1-1106-4c9f-ace4-a1ee8ba31ef7","cardType":"basic","points":1,"front":"When is it necessary to use $\\pm$ when square rooting a formula to find the subject?","back":"In pure algebra (where the value could be negative).","mcqOptions":[{"text":"In pure algebra (where the value could be negative).","correct":true,"explanation":"In pure algebra, both positive and negative values may square to the same expression, so square rooting requires the two possibilities indicated by $b1$."},{"text":"In physical contexts (where the value must be positive).","correct":false,"explanation":"A physical context often restricts quantities such as length to non-negative values, so only the positive root may be valid. Pure algebra generally requires $b1$."},{"text":"In fractional formulae (where the value could be non-integer).","correct":false,"explanation":"The need for $b1$ comes from a squared variable having positive and negative roots, not from the presence of fractions. It is required in pure algebra."},{"text":"In linear algebra (where the value has power one).","correct":false,"explanation":"A linear subject has not been squared, so square rooting and $b1$ do not arise. In pure algebra, $b1$ is needed when reversing a square."}]},{"id":"f1553266-a030-4ee8-94d5-e0a85ba35463","cardType":"basic","points":1,"front":"Identify the coefficient of $k$ in the term $\\frac{1}{2}kt^2$.","back":"$$\\frac{1}{2}t^2$","mcqOptions":[{"text":"$$\\frac{1}{2}t^2$","correct":true,"explanation":"The coefficient of $k$ is everything multiplying it. In $rac{1}{2}kt^2$, removing $k$ leaves the factor $rac{1}{2}t^2$."},{"text":"$$\\frac{1}{2}$","correct":false,"explanation":"This includes only the numerical factor and overlooks the factor $t^2$. Everything multiplying $k$ forms its coefficient, namely $\\frac{1}{2}t^2$."},{"text":"$$t^2$","correct":false,"explanation":"This omits the numerical factor $\\frac{1}{2}$. The complete factor multiplying $k$, and therefore its coefficient, is $\\frac{1}{2}t^2$."},{"text":"$$2t^2$","correct":false,"explanation":"This incorrectly replaces the factor $\\frac{1}{2}$ with its reciprocal. The expression multiplying $k$ is $\\frac{1}{2}t^2$."}]},{"id":"1584007b-6273-4c13-b80b-5f2df02c524c","cardType":"basic","points":1,"front":"What is the first step to rearrange $D = nt + \\frac{1}{2}kt^2$ to make $k$ the subject?","back":"Subtract $nt$ from both sides to isolate the term with $k$.","mcqOptions":[{"text":"Subtract $nt$ from both sides to isolate the term with $k$.","correct":true,"explanation":"The $nt$ term does not contain $k$, so subtracting it from both sides leaves the term involving $k$ ready for further rearrangement."},{"text":"Divide both sides by $t^2$ to isolate the term with $k$.","correct":false,"explanation":"Dividing immediately by $t^2$ also affects the $nt$ term and does not isolate $k$. First subtract $nt$ from both sides to separate the term containing $k$."},{"text":"Subtract $\\frac{1}{2}kt^2$ from both sides to isolate the term with $k$.","correct":false,"explanation":"Subtracting the term containing $k$ moves it away rather than beginning to isolate it. The correct first step is to subtract $nt$ from both sides."},{"text":"Divide both sides by $n$ to isolate the term with $k$.","correct":false,"explanation":"Dividing by $n$ does not isolate the $k$ term because the two terms on the right have different coefficients. First subtract $nt$ from both sides."}]},{"id":"26ea065d-48a3-4337-a051-c43ea11e1325","cardType":"cloze","points":null,"front":"To factorise an expression like $4y + cy - dy$, you take out the common factor to get $\\text{[...]}$.","back":"To factorise an expression like $4y + cy - dy$, you take out the common factor to get $\\boxed{y(4 + c - d)}$.","mcqOptions":[{"text":"$$y(4 + c - d)$","correct":true,"explanation":"Every term contains the common factor $y$. Dividing $4y$, $cy$ and $-dy$ by $y$ gives $4$, $c$ and $-d$ inside the brackets."},{"text":"$$y(4 + c + d)$","correct":false,"explanation":"The negative sign before $dy$ has been lost during factorisation. Taking out the common factor $y$ correctly gives $y(4 + c - d)$."},{"text":"$$y(4c - d)$","correct":false,"explanation":"This incorrectly combines the separate terms $4y$ and $cy$ into a product involving $4c$. Factoring out $y$ gives $y(4 + c - d)$."},{"text":"$$y(4 + c) - d$","correct":false,"explanation":"The common factor has not been taken from the final term, because $-dy$ becomes $-d$ only inside the brackets. The correct result is $y(4 + c - d)$."}]},{"id":"e77caaa6-737b-4dc1-ad83-242b65cbf552","cardType":"basic","points":2,"front":"What is the necessary step if the subject appears more than once in a formula?","back":"

Collect all subject terms on one side
Factorise to isolate the subject

","mcqOptions":[{"text":"

Collect all subject terms on one side
Factorise to isolate the subject

","correct":true,"explanation":"When the subject occurs in multiple terms, those terms must first be brought together. Factoring out the subject then allows division by the remaining expression."},{"text":"

Cancel the repeated subject terms
Divide by the subject immediately

","correct":false,"explanation":"Repeated subject terms cannot simply be cancelled, and dividing by the subject may lose valid solutions. Collect the subject terms on one side, then factorise."},{"text":"

Substitute a value for the subject
Simplify only the numerical terms

","correct":false,"explanation":"Substitution does not rearrange a general formula, and simplifying constants cannot isolate repeated subject terms. They must be collected on one side and then factorised."},{"text":"

Move all constants to one side
Expand without collecting subject terms

","correct":false,"explanation":"Moving constants may help, but expanding alone leaves the subject in several separate terms. Collect all subject terms on one side and factorise them."}]},{"id":"3e3e2adf-9bff-4617-8d8a-6fe28fba78a1","cardType":"basic","points":1,"front":"What does \"clearing the fraction\" mean in algebraic rearranging?","back":"Multiplying both sides by the denominator.","mcqOptions":[{"text":"Multiplying both sides by the denominator.","correct":true,"explanation":"Multiplication by the denominator cancels the division represented by the fraction. Applying it to both sides preserves equality while removing that denominator."},{"text":"Dividing both sides by the denominator.","correct":false,"explanation":"Dividing by the denominator introduces an additional denominator rather than removing the fraction. Clearing it requires multiplying both sides by the denominator."},{"text":"Multiplying both sides by the numerator.","correct":false,"explanation":"Multiplying by the numerator does not cancel the denominator of a fraction. The fraction is cleared by multiplying both sides by its denominator."},{"text":"Subtracting the denominator from both sides.","correct":false,"explanation":"A denominator represents division, so subtraction cannot cancel it. Multiplying both sides by the denominator uses the inverse operation and clears the fraction."}]},{"id":"1446226e-3a73-4f36-a861-6b841dbc276a","cardType":"basic","points":3,"front":"Outline the steps to make $y$ the subject of $a = \\frac{5 + 3cy}{3y - 2}$.","back":"

Multiply by $(3y - 2)$
Expand brackets
Collect $y$ terms on one side
Factorise and divide

","mcqOptions":[{"text":"

Multiply by $(3y - 2)$
Expand brackets
Collect $y$ terms on one side
Factorise and divide

","correct":true,"explanation":"Multiplying by the denominator clears the fraction. Expansion then exposes every $y$ term, allowing them to be collected, factorised and divided to isolate $y$."},{"text":"

Divide by $(3y - 2)$
Expand brackets
Collect $y$ terms on one side
Factorise and divide

","correct":false,"explanation":"Dividing by the denominator makes the fractional structure more complicated instead of clearing it. Begin by multiplying by $(3y - 2)$, then expand, collect and factorise."},{"text":"

Multiply by $(5 + 3cy)$
Expand brackets
Collect $y$ terms on one side
Factorise and divide

","correct":false,"explanation":"The denominator, not the numerator, must be used to clear the fraction. Multiply by $(3y - 2)$ before expanding, collecting the $y$ terms and factorising."},{"text":"

Multiply by $(3y - 2)$
Cancel the $y$ terms
Collect constants on one side
Divide by $y$

","correct":false,"explanation":"The $y$ terms occur within different sums and cannot be cancelled directly. After clearing the fraction, expand, collect all $y$ terms, factorise and divide."}]},{"id":"c44d26fa-6951-49e5-9319-a36f7a90dd77","cardType":"cloze","points":null,"front":"The reciprocal of $q$ is $\\text{[...]}$.","back":"The reciprocal of $q$ is $\\boxed{\\frac{1}{q}}$.","mcqOptions":[{"text":"$$\\frac{1}{q}$","correct":true,"explanation":"The reciprocal of a non-zero quantity is found by dividing $1$ by that quantity. Therefore, multiplying $q$ by its reciprocal gives $1$."},{"text":"$$q$","correct":false,"explanation":"A quantity is not generally its own reciprocal; that happens only for special values such as $1$ and $-1$. The reciprocal of $q$ is $\\frac{1}{q}$."},{"text":"$$-\\frac{1}{q}$","correct":false,"explanation":"Taking a reciprocal does not change the sign of a quantity. The reciprocal must multiply by $q$ to give $1$, so it is $\\frac{1}{q}$."},{"text":"$$\\frac{1}{q^2}$","correct":false,"explanation":"Squaring the denominator confuses a reciprocal with a negative power of greater magnitude. The reciprocal is $q^{-1}$, which equals $\\frac{1}{q}$."}]},{"id":"70af4506-d4c2-483f-9feb-e4ca3b6221bd","cardType":"basic","points":1,"front":"How do you efficiently clear all fractions in a reciprocal formula like $\\frac{1}{p} = \\frac{1}{q} + \\frac{1}{r}$?","back":"Multiply every term by $pqr$.","mcqOptions":[{"text":"Multiply every term by $pqr$.","correct":true,"explanation":"The product $pqr$ is a common multiple of all three denominators. Multiplying every term by it cancels each reciprocal denominator and produces $qr = pr + pq$."},{"text":"Multiply every term by $pq$.","correct":false,"explanation":"The product $pq$ cancels the denominators $p$ and $q$ but leaves the term containing denominator $r$ as a fraction. Use $pqr$ to clear all three."},{"text":"Multiply every term by $q + r$.","correct":false,"explanation":"Adding denominators does not provide factors that cancel the reciprocal terms. Multiplying every term by the product $pqr$ clears the denominators $p$, $q$ and $r$."},{"text":"Multiply every term by $p + q + r$.","correct":false,"explanation":"The sum of the denominators cannot cancel any individual denominator by multiplication. Their product $pqr$ contains every required factor and clears all fractions."}]},{"id":"a37268d1-fc4e-4637-8502-b497cdfe467b","cardType":"basic","points":1,"front":"Why must brackets be expanded early when rearranging for a specific letter?","back":"To release the subject if it is trapped inside the brackets.","mcqOptions":[{"text":"To release the subject if it is trapped inside the brackets.","correct":true,"explanation":"Expanding distributes the outside factor across the bracket, turning the hidden subject term into a separate term that can be collected and isolated."},{"text":"To cancel the subject if it appears inside the brackets.","correct":false,"explanation":"A term inside a sum cannot be cancelled merely because brackets surround it. Expanding releases the subject into separate terms so it can be collected and isolated."},{"text":"To remove the subject from the formula completely.","correct":false,"explanation":"Rearrangement aims to isolate the subject, not eliminate it from the formula. Expanding brackets exposes its terms so the necessary inverse operations can be applied."},{"text":"To divide the subject by every term outside the brackets.","correct":false,"explanation":"Expansion requires multiplying each bracketed term by the outside factor, not dividing the subject. This releases the subject term so it can later be isolated."}]},{"id":"18fdd256-d283-4d81-8329-031d89ae3962","cardType":"cloze","points":null,"front":"In the formula $w = r + kt$, making $r$ the subject results in $\\text{[...]}$.","back":"In the formula $w = r + kt$, making $r$ the subject results in $\\boxed{r = w - kt}$.","mcqOptions":[{"text":"$$r = w - kt$","correct":true,"explanation":"Subtracting $kt$ from both sides reverses the addition of $kt$, leaving $r$ alone and giving $r = w - kt$."},{"text":"$$r = w + kt$","correct":false,"explanation":"Keeping $kt$ positive fails to undo its addition on the right-hand side. Subtracting $kt$ from both sides gives the correct result, $r = w - kt$."},{"text":"$$r = kt - w$","correct":false,"explanation":"This reverses the order of the subtraction, producing the negative of the required expression. Subtracting $kt$ from $w$ gives $r = w - kt$."},{"text":"$$r = w - k - t$","correct":false,"explanation":"The term $kt$ is a product, so it must be subtracted as a whole rather than split into separate terms. The correct rearrangement is $r = w - kt$."}]},{"id":"9feff379-358d-4b73-b54c-e0a9b064b9b6","cardType":"cloze","points":null,"front":"When rearranging $w = r + kt$ for $k$, after subtracting $r$, you must [ ] both sides by $t$.","back":"When rearranging $w = r + kt$ for $k$, after subtracting $r$, you must divide both sides by $t$.","mcqOptions":[{"text":"divide","correct":true,"explanation":"After subtracting $r$, the equation is $w-r=kt$. Since $k$ is multiplied by $t$, dividing both sides by $t$ isolates $k$."},{"text":"multiply","correct":false,"explanation":"Multiplying by $t$ would introduce another factor of $t$ rather than remove one. Since $k$ is multiplied by $t$, you must divide both sides by $t$."},{"text":"subtract","correct":false,"explanation":"Subtracting $t$ treats multiplication as though it were addition. In $w-r=kt$, the inverse of multiplying by $t$ is dividing both sides by $t$."},{"text":"add","correct":false,"explanation":"Adding $t$ does not cancel a factor that is multiplying $k$. To isolate $k$ in $w-r=kt$, divide both sides by $t$."}]},{"id":"9c20cabd-898d-4704-b915-8d2495344b46","cardType":"basic","points":1,"front":"What check should you perform at the end of a rearranging question?","back":"Ensure the subject appears exactly once and is completely alone on one side.","mcqOptions":[{"text":"Ensure the subject appears exactly once and is completely alone on one side.","correct":true,"explanation":"A correctly rearranged formula has the chosen subject isolated on one side, with no further occurrences of that subject elsewhere in the equation."},{"text":"Ensure the subject appears exactly twice and is completely alone on one side.","correct":false,"explanation":"If the subject appears twice, it has not usually been fully isolated. The final formula should contain the subject exactly once, completely alone on one side."},{"text":"Ensure the subject appears exactly once but remains inside brackets on one side.","correct":false,"explanation":"A subject remaining inside brackets is not completely isolated because other operations still act on it. It should appear exactly once and be alone on one side."},{"text":"Ensure the subject appears on both sides and is completely alone on one side.","correct":false,"explanation":"Leaving the subject on both sides means the rearrangement is incomplete. The subject should appear exactly once and be completely alone on one side."}]},{"id":"eb4ab949-aa2a-4830-be10-c269938f4b61","cardType":"cloze","points":null,"front":"If a formula involves a square root, the inverse operation to remove it is [ ] both sides.","back":"If a formula involves a square root, the inverse operation to remove it is squaring both sides.","mcqOptions":[{"text":"squaring","correct":true,"explanation":"Squaring reverses taking a square root because $(\\sqrt{x})^2=x$. Both sides must be squared to preserve the equality while removing the root."},{"text":"square rooting","correct":false,"explanation":"Taking another square root does not cancel the original root and instead introduces a fourth root. Squaring both sides is the inverse operation that removes a square root."},{"text":"doubling","correct":false,"explanation":"Doubling changes a value by multiplying it by $2$, which does not undo a square root. The root is removed by squaring both sides."},{"text":"halving","correct":false,"explanation":"Halving reverses multiplication by $2$, not taking a square root. To eliminate a square root while maintaining equality, square both sides of the formula."}]},{"id":"109f7379-4e19-486d-8d08-ef451899226d","cardType":"basic","points":2,"front":"Why is $x(a + b) = xa + b$ an incorrect expansion?","back":"The factor $x$ must multiply every term inside the bracket ($xa + xb$).","mcqOptions":[{"text":"The factor $x$ must multiply every term inside the bracket ($xa + xb$).","correct":true,"explanation":"The distributive law applies the outside factor to each term inside the bracket. Therefore, $x(a+b)$ expands to $xa+xb$, not $xa+b$."},{"text":"The factor $x$ must multiply only the first term inside the bracket ($xa + b$).","correct":false,"explanation":"This is the misconception shown in the proposed expansion: the second term has been missed. Distribution requires $x$ to multiply both terms, giving $xa+xb$."},{"text":"The factor $x$ must be added to every term inside the bracket ($(x + a) + (x + b)$).","correct":false,"explanation":"The factor outside the bracket indicates multiplication, not addition. Applying the distributive law correctly means multiplying both $a$ and $b$ by $x$, producing $xa+xb$."},{"text":"The factor $x$ must be squared with every term inside the bracket ($x^2a + x^2b$).","correct":false,"explanation":"Expanding a bracket does not create an extra factor of $x$. The single outside factor multiplies each term once, so the correct expansion is $xa+xb$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Rearranging Harder Formulae","title":"Rearranging Harder Formulae Flashcards","topbar":{"backHref":"/gcse/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"20bcccef-c52d-4528-b3b7-06ce35fba7e7","walled":"$L39"}],"$L3a","$L3b"]