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Here, $v$ is isolated on the left, while all other variables are on the right."},{"text":"It is the variable written first on one side of the equals sign.","correct":false,"explanation":"A variable’s position in the written formula does not determine the subject. The correct reason is that $v$ is isolated on one side of the equals sign."},{"text":"It is the variable with the greatest power on one side of the equals sign.","correct":false,"explanation":"The subject is not chosen by comparing powers of variables. Here, $v$ is the subject because it is isolated on one side of the equals sign."},{"text":"It is the variable calculated first from the values in the formula.","correct":false,"explanation":"The order in which values are calculated does not define the subject. A subject is the variable isolated on one side, which is $v$ in this formula."}]},{"id":"477e4771-266d-4b1a-a639-d5b45f2c65fe","cardType":"basic","points":1,"front":"What is the goal when asked to \"make $x$ the subject\" of a formula?","back":"To rearrange it so $x$ is alone on one side and the other side does not contain $x$.","mcqOptions":[{"text":"To rearrange it so $x$ is alone on one side and the other side does not contain $x$.","correct":true,"explanation":"Making $x$ the subject means completely isolating $x$. It must appear alone on one side, with an equivalent expression containing no $x$ on the other side."},{"text":"To rearrange it so $x$ is alone on one side and the other side also contains $x$.","correct":false,"explanation":"Leaving $x$ on both sides means it has not been fully isolated. The correct arrangement has $x$ alone on one side and no $x$ on the other."},{"text":"To rearrange it so $x$ is squared on one side and the other side does not contain $x$.","correct":false,"explanation":"Obtaining $x^2$ does not make $x$ itself the subject. Any remaining power must be undone so $x$ is alone and absent from the other side."},{"text":"To rearrange it so $x$ is in a fraction and the other side does not contain $x$.","correct":false,"explanation":"Placing $x$ inside a numerator or denominator does not necessarily isolate it. The goal is to have $x$ alone on one side and nowhere on the other."}]},{"id":"e74b7696-4b8f-4197-8923-03b77eb34a37","cardType":"basic","points":1,"front":"What is the inverse operation of squaring a variable?","back":"Taking the square root.","mcqOptions":[{"text":"Taking the square root.","correct":true,"explanation":"Taking a square root reverses squaring because $csqrt{x^2}$ returns the magnitude of $x$. When solving equations, both positive and negative roots may need consideration."},{"text":"Dividing by two.","correct":false,"explanation":"The exponent $2$ does not mean that the variable was multiplied by $2$, so division by $2$ will not undo it. Squaring is reversed by taking the square root."},{"text":"Taking the reciprocal.","correct":false,"explanation":"A reciprocal reverses multiplication by a quantity or a negative exponent, not squaring. The inverse operation needed to undo a square is taking the square root."},{"text":"Taking the cube root.","correct":false,"explanation":"A cube root reverses raising a variable to the power $3$, not the power $2$. The operation that reverses squaring is taking the square root."}]},{"id":"0890569b-4f09-4316-83f1-51778bb6c1bd","cardType":"basic","points":1,"front":"What is the inverse operation of square-rooting a variable?","back":"Squaring it.","mcqOptions":[{"text":"Squaring it.","correct":true,"explanation":"Squaring reverses the principal square-root operation: if $y=csqrt{x}$, then $y^2=x$. Therefore, squaring both sides removes a square root when rearranging a formula."},{"text":"Halving it.","correct":false,"explanation":"The root symbol does not mean division by $2$, so halving cannot undo it. A square root is removed by squaring the relevant expression or both sides."},{"text":"Reciprocating it.","correct":false,"explanation":"Taking a reciprocal changes $csqrt{x}$ into $1/csqrt{x}$ rather than recovering $x$. The inverse operation of square-rooting is squaring."},{"text":"Doubling it.","correct":false,"explanation":"A square root is not equivalent to dividing a value by $2$, so doubling does not reverse it. Squaring the square-rooted expression correctly returns the original value."}]},{"id":"05c04f86-2c1c-4867-b6c3-db73f322b5a4","cardType":"cloze","points":null,"front":"If $x^2 = 25$ in a pure algebra problem, then $\\text{[...]}$.","back":"If $x^2 = 25$ in a pure algebra problem, then $\\boxed{x = \\pm 5}$.","mcqOptions":[{"text":"$$x = \\pm 5$","correct":true,"explanation":"Squaring either $5$ or $-5$ gives $25$, so a pure algebraic equation requires both possible square roots: $x = \\pm 5$."},{"text":"$$x = 5$","correct":false,"explanation":"Taking only the positive square root omits $x=-5$, which also satisfies the equation because $(-5)^2=25$. Therefore, $x = \\pm 5$."},{"text":"$$x = -5$","correct":false,"explanation":"Taking only the negative square root omits $x=5$, which also satisfies the equation because $5^2=25$. Therefore, $x = \\pm 5$."},{"text":"$$x = \\pm 25$","correct":false,"explanation":"This fails to take the square root of $25$: squaring $25$ gives $625$, not $25$. The two solutions are $x = \\pm 5$."}]},{"id":"43a1075d-a961-4219-8014-5436200a0b18","cardType":"basic","points":1,"front":"When is it acceptable to ignore the negative root when rearranging a formula?","back":"When the variable represents a real-life physical quantity that cannot be negative (e.g., length, time).","mcqOptions":[{"text":"When the variable represents a real-life physical quantity that cannot be negative (e.g., length, time).","correct":true,"explanation":"Although algebra may produce two roots, context can exclude one. Quantities such as length and elapsed time cannot be negative, so only the physically meaningful positive root is retained."},{"text":"When the equation contains a squared variable that has two roots (e.g., $x^2=25$).","correct":false,"explanation":"Having two algebraic roots is precisely why both normally need to be stated. The negative root may be discarded only when the variable’s real-life meaning makes negative values impossible."},{"text":"When every value substituted into the formula is already positive (e.g., positive constants).","correct":false,"explanation":"Positive input values do not automatically rule out a negative solution after rearrangement. The negative root is ignored only if the target quantity itself cannot physically be negative."},{"text":"When the positive root is easier to use and simpler to write (e.g., a whole number).","correct":false,"explanation":"Convenience or simplicity is not a valid reason to reject a solution. A negative root can be excluded only when the context shows that the represented quantity cannot be negative."}]},{"id":"375b3a04-5562-4793-aad2-eb7597dae807","cardType":"basic","points":1,"front":"If the target variable is in the denominator of a fraction, what is the first step?","back":"Multiply both sides by that denominator to clear the fraction.","mcqOptions":[{"text":"Multiply both sides by that denominator to clear the fraction.","correct":true,"explanation":"Multiplying every term on both sides by the denominator cancels it from the fraction. This moves the target variable out of the denominator while preserving the equation’s equality."},{"text":"Divide both sides by that denominator to clear the fraction.","correct":false,"explanation":"Dividing by the denominator introduces another denominator factor rather than cancelling the original one. Multiplying both sides by that denominator clears the fraction correctly."},{"text":"Multiply only the numerator by that denominator to clear the fraction.","correct":false,"explanation":"Changing only the numerator alters one expression without applying an equivalent operation to the whole equation. The denominator is cleared by multiplying both complete sides by it."},{"text":"Square both sides using that denominator to clear the fraction.","correct":false,"explanation":"Squaring does not directly cancel a denominator and can unnecessarily introduce extra solutions. The appropriate first step is to multiply both sides by the denominator."}]},{"id":"5a1e6d81-2520-4a5d-9209-bfafa9323e7d","cardType":"basic","points":1,"front":"How do you remove the fraction from the term $\\frac{1}{2}at^2$ when rearranging?","back":"Multiply both sides of the equation by 2.","mcqOptions":[{"text":"Multiply both sides of the equation by 2.","correct":true,"explanation":"Multiplying by $2$ cancels the factor $cfrac{1}{2}$ because $2ctimescfrac{1}{2}=1$. Applying this to both sides preserves equality and leaves the term $at^2$."},{"text":"Divide both sides of the equation by 2.","correct":false,"explanation":"Dividing by $2$ introduces another factor of $cfrac{1}{2}$, making the term $cfrac{1}{4}at^2$. Multiplying both sides by $2$ cancels the original fraction."},{"text":"Multiply both sides of the equation by $cfrac{1}{2}$.","correct":false,"explanation":"Multiplying by another $cfrac{1}{2}$ makes the coefficient $cfrac{1}{4}$ rather than removing it. The reciprocal operation is multiplication by $2$ on both sides."},{"text":"Square both sides of the equation to remove the 2.","correct":false,"explanation":"Squaring affects every term and turns the coefficient into $cfrac{1}{4}$ rather than cancelling it. Multiplying both sides by $2$ directly removes the factor $cfrac{1}{2}$."}]},{"id":"bbf488f2-f4b2-4d16-a38a-dd39cee566ac","cardType":"cloze","points":null,"front":"To isolate a letter that appears more than once, you must [ ] the target terms, factorise the letter, and finally divide.","back":"To isolate a letter that appears more than once, you must collect the target terms, factorise the letter, and finally divide.","mcqOptions":[{"text":"collect","correct":true,"explanation":"Collecting places all terms containing the target letter on the same side, allowing that letter to be taken out as a common factor."},{"text":"expand","correct":false,"explanation":"Expanding removes brackets but does not bring all terms containing the target letter together. The correct first step is to collect those terms on the same side."},{"text":"cancel","correct":false,"explanation":"Cancelling is only valid for common factors in a fraction, not separate terms in an equation. The target terms must first be collected together."},{"text":"substitute","correct":false,"explanation":"Substitution replaces a variable with a known expression or value, so it does not isolate a repeated letter. The required first step is to collect its terms."}]},{"id":"bdf3a21a-072c-48c0-87cf-b8121a1d91c3","cardType":"cloze","points":null,"front":"To isolate a letter that appears more than once, you must collect the target terms, [ ] the letter, and finally divide.","back":"To isolate a letter that appears more than once, you must collect the target terms, factorise the letter, and finally divide.","mcqOptions":[{"text":"factorise","correct":true,"explanation":"Factorising takes the repeated letter outside a bracket, changing an expression such as $ax+bx$ into $x(a+b)$ so the letter appears once."},{"text":"cancel","correct":false,"explanation":"A letter cannot be cancelled when it occurs in separate additive terms. After collecting those terms, the correct process is to factorise the repeated letter."},{"text":"expand","correct":false,"explanation":"Expanding would undo the useful common-factor form and leave the letter appearing several times. The collected expression must instead be factorised."},{"text":"square","correct":false,"explanation":"Squaring the letter changes the equation and is not an inverse operation required here. The repeated letter should be taken out as a common factor by factorising."}]},{"id":"3c7b7e3c-0921-49c5-b048-45261d79ecd6","cardType":"cloze","points":null,"front":"To isolate a letter that appears more than once, you must collect the target terms, factorise the letter, and finally [ ].","back":"To isolate a letter that appears more than once, you must collect the target terms, factorise the letter, and finally divide.","mcqOptions":[{"text":"divide","correct":true,"explanation":"After factorising, the target letter is multiplied by a coefficient or bracket. Dividing both sides by that factor leaves the target letter isolated."},{"text":"multiply","correct":false,"explanation":"Multiplying would not remove the factor attached to the target letter. Once the letter has been factorised, divide by its remaining coefficient or bracket."},{"text":"subtract","correct":false,"explanation":"Subtraction removes an added term, not a factor multiplying the target letter. After factorising, the final step is division by that factor."},{"text":"add","correct":false,"explanation":"Addition reverses subtraction but cannot remove a factor multiplying the target letter. The factorised equation must finally be divided by the factor."}]},{"id":"9cce7571-cb47-4588-9406-de48f47b101c","cardType":"basic","points":1,"front":"Which algebraic process converts $ax + bx$ into $x(a + b)$?","back":"Factorising (taking out a common factor).","mcqOptions":[{"text":"Factorising (taking out a common factor).","correct":true,"explanation":"Both terms contain the common factor $x$. Taking $x$ outside the bracket leaves the remaining coefficients $a+b$, giving $x(a+b)$."},{"text":"Expanding (multiplying out the brackets).","correct":false,"explanation":"Expanding performs the reverse transformation, converting $x(a+b)$ into $ax+bx$. The given conversion takes out the common factor $x$, so it is factorising."},{"text":"Simplifying (combining like terms).","correct":false,"explanation":"The terms $ax$ and $bx$ cannot generally be combined because their coefficients are different algebraic expressions. Rewriting them with common factor $x$ is factorising."},{"text":"Cancelling (removing common factors).","correct":false,"explanation":"Cancelling applies to common factors in a quotient and would remove a factor entirely. Here, $x$ is retained outside a bracket through factorising."}]},{"id":"7afef3ef-5d30-4321-99b3-cd3f535341a8","cardType":"cloze","points":null,"front":"A formula rearranged correctly must have the subject appear [ ] and only on one side of the equation.","back":"A formula rearranged correctly must have the subject appear once and only on one side of the equation.","mcqOptions":[{"text":"once","correct":true,"explanation":"A correctly rearranged formula has a single occurrence of the subject on one side, with every other letter and operation placed on the other side."},{"text":"repeatedly","correct":false,"explanation":"A subject appearing repeatedly has not been fully isolated, even if every occurrence lies on the same side. It must appear once and alone as the subject."},{"text":"twice","correct":false,"explanation":"Two occurrences mean the subject has not yet been fully collected and factorised. A finished rearrangement requires the subject to appear once on only one side."},{"text":"three times","correct":false,"explanation":"Three occurrences show that terms containing the subject still need to be collected and factorised. The correctly isolated subject appears exactly once."}]},{"id":"f7f56ace-20e9-47b3-ab3f-3e85324c9332","cardType":"cloze","points":null,"front":"A formula rearranged correctly must have the subject appear once and only on [ ] of the equation.","back":"A formula rearranged correctly must have the subject appear once and only on one side of the equation.","mcqOptions":[{"text":"one side","correct":true,"explanation":"Isolation requires every occurrence of the subject to be confined to one side of the equation, with the subject ultimately appearing there exactly once."},{"text":"both sides","correct":false,"explanation":"If the subject remains on both sides, it has not been isolated. Its terms must be collected and rearranged until it appears once on one side only."},{"text":"neither side","correct":false,"explanation":"If the subject appears on neither side, it has been eliminated rather than isolated. A rearranged formula must display it once on one side."},{"text":"each side","correct":false,"explanation":"Placing the subject on each side leaves the rearrangement unfinished. The subject terms must be collected so that it occurs on one side only."}]},{"id":"4fad1d77-2975-49fe-8411-2dbf825c82be","cardType":"basic","points":2,"front":"What are the two final steps to find $y$ once you have isolated $\\frac{1}{y} = \\frac{1}{x} - \\frac{1}{z}$?","back":"

Combine the right-hand terms into a single fraction
Take the reciprocal (flip) of both sides

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

Combine the right-hand terms into a single fraction
Take the reciprocal (flip) of both sides

","correct":true,"explanation":"Combining gives $\\frac{1}{y}=\\frac{z-x}{xz}$. Taking the reciprocal of both sides then isolates the required subject as $y=\\frac{xz}{z-x}$."},{"text":"

Combine the right-hand terms by adding denominators
Take the reciprocal (flip) of both sides

","correct":false,"explanation":"Fractions are not combined by adding their denominators. Using the common denominator $xz$ gives $\\frac{z-x}{xz}$, after which reciprocating gives $y=\\frac{xz}{z-x}$."},{"text":"

Combine the right-hand terms into a single fraction
Take the negative of both sides

","correct":false,"explanation":"Taking the negative changes signs but leaves $y$ in a denominator. After combining the fractions, the reciprocal must be taken to obtain $y=\\frac{xz}{z-x}$."},{"text":"

Cancel the matching terms on the right-hand side
Multiply both sides by y

","correct":false,"explanation":"The terms $\\frac{1}{x}$ and $\\frac{1}{z}$ cannot be cancelled because they are subtracted, not common factors. Combine them, then reciprocate both sides."}]},{"id":"5505128f-46c5-457f-9d41-ab0d8b556985","cardType":"basic","points":2,"front":"Make $k$ the subject of $m = n + kp$.","back":"$$k = \\frac{m - n}{p}$","mcqOptions":[{"text":"$$k = \\frac{m - n}{p}$","correct":true,"explanation":"Subtracting $n$ gives $m-n=kp$. Dividing both sides by $p$ then leaves $k$ alone, producing $k=\\frac{m-n}{p}$."},{"text":"$$k = \\frac{m + n}{p}$","correct":false,"explanation":"Adding $n$ repeats the original addition instead of reversing it. Subtract $n$ from both sides first, then divide by $p$ to obtain $k=\\frac{m-n}{p}$."},{"text":"$$k = (m - n)p$","correct":false,"explanation":"Multiplying by $p$ does not undo the multiplication in $kp$ and introduces an extra factor. After subtracting $n$, divide by $p$ instead."},{"text":"$$k = \\frac{m}{p} - n$","correct":false,"explanation":"This divides only $m$ by $p$, although $p$ multiplies $k$ after the whole difference $m-n$ is formed. The entire numerator must be divided by $p$."}]},{"id":"1c61035f-5e16-4c49-a73b-04cedf7d0e95","cardType":"basic","points":2,"front":"Make $P$ the subject of $R^2 = P^2 + 5qt$.","back":"$$P = \\pm\\sqrt{R^2 - 5qt}$","mcqOptions":[{"text":"$$P = \\pm\\sqrt{R^2 - 5qt}$","correct":true,"explanation":"Subtracting $5qt$ gives $P^2=R^2-5qt$. Taking the square root requires both positive and negative roots, so $P=\\pm\\sqrt{R^2-5qt}$."},{"text":"$$P = \\sqrt{R^2 - 5qt}$","correct":false,"explanation":"Taking only the positive square root loses one possible value of $P$. Since $P^2=R^2-5qt$, the complete answer is $P=\\pm\\sqrt{R^2-5qt}$."},{"text":"$$P = \\pm\\sqrt{R^2 + 5qt}$","correct":false,"explanation":"Adding $5qt$ is a sign error: it must be subtracted from both sides. This gives $P^2=R^2-5qt$, then $P=\\pm\\sqrt{R^2-5qt}$."},{"text":"$$P = \\pm(R^2 - 5qt)$","correct":false,"explanation":"This omits the square-root operation when undoing the square. After obtaining $P^2=R^2-5qt$, take both square roots to get $P=\\pm\\sqrt{R^2-5qt}$."}]},{"id":"b7707a7a-e3e6-4b04-9a8a-a96aaed94ff5","cardType":"basic","points":2,"front":"Make $n$ the subject of $d = wt + \\frac{1}{2}nt^2$.","back":"$$n = \\frac{2(d - wt)}{t^2}$","mcqOptions":[{"text":"$$n = \\frac{2(d - wt)}{t^2}$","correct":true,"explanation":"Subtracting $wt$ gives $d-wt=\\frac{1}{2}nt^2$. Multiplying by 2 and dividing by $t^2$ produces $n=\\frac{2(d-wt)}{t^2}$."},{"text":"$$n = \\frac{2(d - wt)}{t}$","correct":false,"explanation":"This divides by $t$ rather than the full factor $t^2$. After multiplying by 2, dividing by $t^2$ gives $n=\\frac{2(d-wt)}{t^2}$."},{"text":"$$n = \\frac{d - wt}{2t^2}$","correct":false,"explanation":"Dividing by 2 does not undo the coefficient $\\frac{1}{2}$; the equation must instead be multiplied by 2. Therefore $n=\\frac{2(d-wt)}{t^2}$."},{"text":"$$n = \\frac{2d - wt}{t^2}$","correct":false,"explanation":"The factor 2 must multiply the whole bracket $d-wt$, not only $d$. Correct distribution gives the numerator $2d-2wt$, so $n=\\frac{2(d-wt)}{t^2}$."}]},{"id":"d4895b61-e59a-421e-b419-beb1dee72c43","cardType":"basic","points":1,"front":"What is the result of factorising $2px - 2qx$ to make $x$ the subject?","back":"$$x(2p - 2q)$ (or $2x(p - q)$)","mcqOptions":[{"text":"$$x(2p - 2q)$ (or $2x(p - q)$)","correct":true,"explanation":"Both terms contain the common factor $2x$. Taking it outside the bracket leaves $p-q$, so the expression is $2x(p-q)$, equivalently $x(2p-2q)$."},{"text":"$$x(2p + 2q)$ (or $2x(p + q)$)","correct":false,"explanation":"The subtraction sign must remain when the common factor is extracted. Expanding this option gives $2px+2qx$, whereas the correct factorisation is $x(2p-2q)$."},{"text":"$$x(2p - 4q)$ (or $2x(p - 2q)$)","correct":false,"explanation":"Only one factor of 2 is common to both original terms. Expanding this option gives $2px-4qx$, so the correct factorisation is $x(2p-2q)$."},{"text":"$$x(2pq)$ (or $2x(pq)$)","correct":false,"explanation":"Factorising extracts common factors; it does not multiply $p$ and $q$ together. The shared factor is $2x$, leaving $p-q$, so the result is $2x(p-q)$."}]},{"id":"01ebd1de-dbf7-4292-b5fd-c3f938dbc9f4","cardType":"basic","points":3,"front":"Make $x$ the subject of $\\frac{a}{b} = \\frac{3x}{x + 4}$.","back":"$$x = \\frac{4a}{3b - a}$","mcqOptions":[{"text":"$$x = \\frac{4a}{3b - a}$","correct":true,"explanation":"Cross-multiplication gives $a(x+4)=3bx$. Expanding and collecting the terms containing $x$ gives $4a=x(3b-a)$, hence $x=\\frac{4a}{3b-a}$."},{"text":"$$x = \\frac{4a}{3b + a}$","correct":false,"explanation":"The $ax$ term changes sign when it is subtracted from both sides, producing $3bx-ax=x(3b-a)$. Thus the denominator is $3b-a$, not $3b+a$."},{"text":"$$x = \\frac{a}{3b - 4a}$","correct":false,"explanation":"The 4 comes from expanding $a(x+4)$, so it multiplies $a$ in the numerator rather than appearing in the denominator. The correct result is $x=\\frac{4a}{3b-a}$."},{"text":"$$x = \\frac{4a}{a - 3b}$","correct":false,"explanation":"This reverses the subtraction in the collected coefficient of $x$. Since $3bx-ax=x(3b-a)$, the correct denominator is $3b-a$, giving $x=\\frac{4a}{3b-a}$."}]},{"id":"30f1f597-44a0-4868-a06c-3455429532d9","cardType":"cloze","points":null,"front":"Cross-multiplication is a shortcut used when one [ ] is equal to another fraction.","back":"Cross-multiplication is a shortcut used when one fraction is equal to another fraction.","mcqOptions":[{"text":"fraction","correct":true,"explanation":"Cross-multiplication applies when an equality has one fraction on each side. It multiplies each numerator by the denominator of the fraction on the opposite side."},{"text":"numerator","correct":false,"explanation":"A numerator is only the top part of a fraction, not the entire quantity on one side. The missing term is fraction because cross-multiplication links two equal fractions."},{"text":"denominator","correct":false,"explanation":"A denominator is only the bottom part of a fraction. The shortcut is used when one fraction equals another fraction, allowing opposite numerators and denominators to be multiplied."},{"text":"decimal","correct":false,"explanation":"A decimal would first need to be expressed appropriately as a fraction before this shortcut could be applied. The stated condition is that one fraction equals another fraction."}]},{"id":"6fb93991-4d43-41a6-835f-fe39dc0b55b8","cardType":"cloze","points":null,"front":"Cross-multiplication is a shortcut used when one fraction is equal to another [ ].","back":"Cross-multiplication is a shortcut used when one fraction is equal to another fraction.","mcqOptions":[{"text":"fraction","correct":true,"explanation":"The shortcut is designed for an equation consisting of two equal fractions. Multiplying across removes their denominators while preserving the equality between the two sides."},{"text":"numerator","correct":false,"explanation":"The second quantity must be a complete fraction, not merely its numerator. Cross-multiplication works by multiplying each fraction’s numerator by the other fraction’s denominator."},{"text":"denominator","correct":false,"explanation":"A denominator alone is not the second fraction in the equality. The blank is fraction because the method compares two fractions and multiplies their components across."},{"text":"decimal","correct":false,"explanation":"A decimal is not the required form for direct cross-multiplication. The blank is fraction, giving an equality between two fractions whose denominators can be cleared."}]},{"id":"476b8f64-5a5e-4ebd-9ad2-8d299735401c","cardType":"basic","points":1,"front":"In the term $7x$, what is the name for the number 7?","back":"The coefficient.","mcqOptions":[{"text":"The coefficient.","correct":true,"explanation":"A coefficient is the numerical factor multiplying a variable. In $7x$, the variable $x$ is multiplied by 7, making 7 its coefficient."},{"text":"The variable.","correct":false,"explanation":"The variable is the symbol whose value can change, which is $x$ in this term. The number 7 multiplying it is called the coefficient."},{"text":"The constant.","correct":false,"explanation":"A constant is a term with no variable attached. Because 7 multiplies $x$ in $7x$, it is the coefficient rather than a constant term."},{"text":"The exponent.","correct":false,"explanation":"An exponent indicates a power and would be written above the variable, such as the 2 in $x^2$. The number 7 in $7x$ is the coefficient."}]},{"id":"fdb710e8-257f-4325-95a0-8e6b010ac88a","cardType":"basic","points":1,"front":"Why is the \"balance method\" safer than just \"moving\" terms across the equals sign?","back":"It helps prevent sign errors by ensuring the same operation is applied to both sides.","mcqOptions":[{"text":"It helps prevent sign errors by ensuring the same operation is applied to both sides.","correct":true,"explanation":"Performing the same inverse operation on both sides preserves the equality. Writing that operation explicitly reduces the risk of accidentally changing a term’s sign incorrectly."},{"text":"It helps prevent fraction errors by ensuring every denominator is cancelled immediately.","correct":false,"explanation":"Cancelling every denominator immediately is neither always possible nor the defining safety feature. The balance method prevents sign errors by applying the same valid operation to both sides."},{"text":"It makes rearranging faster by allowing terms to change sides without an inverse operation.","correct":false,"explanation":"Terms cannot simply change sides without an inverse operation; that shorthand often causes sign mistakes. The balance method is safer because the same operation is explicitly applied to both sides."},{"text":"It ensures the unknown stays positive by keeping it on the left-hand side.","correct":false,"explanation":"Keeping an unknown on the left does not determine whether its value is positive. The method’s safety comes from applying identical operations to both sides, which prevents sign errors."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Rearranging Harder Formulae","title":"Rearranging Harder Formulae Flashcards","topbar":{"backHref":"/igcse/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"630e8c65-f90a-4515-b224-9b5d1ce4f533","walled":"$L39"}],"$L3a","$L3b"]