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For example, $x$ is the variable in $3x + 2$."},{"text":"A number that multiplies a letter.","correct":false,"explanation":"This describes a coefficient, which is the numerical multiplier of a variable. A variable is a letter that stands for a number."},{"text":"A term that has no letter.","correct":false,"explanation":"This describes a constant term, such as 5 or -3. A variable is a letter used to stand for a number."},{"text":"A symbol that means an operation.","correct":false,"explanation":"Symbols such as + and − indicate operations, rather than variables. A variable is a letter that represents a number."}]},{"id":"edafd0d1-c49c-4cc0-bc05-f5ef955d8b57","cardType":"basic","points":1,"front":"In the term $6m$, what is the coefficient?","back":"6","mcqOptions":[{"text":"6","correct":true,"explanation":"The coefficient is the numerical factor multiplying the variable. Since $6m$ means $6 \\times m$, its coefficient is 6."},{"text":"0","correct":false,"explanation":"Choosing 0 confuses the coefficient with an absent constant term. In $6m$, the number multiplying $m$ is 6, so the coefficient is 6."},{"text":"1","correct":false,"explanation":"A coefficient of 1 is implied only when a variable appears alone, as in $m$. In $6m$, the written multiplier and coefficient is 6."},{"text":"12","correct":false,"explanation":"This unnecessarily doubles the numerical multiplier. The term already means $6 \\times m$, so its coefficient is 6, not 12."}]},{"id":"a341fd42-e72e-4bce-8b29-c87b77b6812b","cardType":"cloze","points":null,"front":"A [ ] is a term with no variable, such as 5 or -3.","back":"A constant is a term with no variable, such as 5 or -3.","mcqOptions":[{"text":"constant","correct":true,"explanation":"A constant is a term with a fixed numerical value and no variable. Examples include 5, -3 and 12."},{"text":"coefficient","correct":false,"explanation":"A coefficient is a number that multiplies a variable, such as 6 in $6m$. A term containing no variable is called a constant."},{"text":"variable","correct":false,"explanation":"A variable is a letter representing a number, so it is not a term with no variable. Such a term is called a constant."},{"text":"expression","correct":false,"explanation":"An expression can contain variables, constants and operations, so it is not specifically a term without a variable. The required term is constant."}]},{"id":"1ca80c7e-9cad-40ac-aedd-c14b33a9d972","cardType":"basic","points":1,"front":"Simplify $x + x + x$.","back":"$$3x$","mcqOptions":[{"text":"$$3x$","correct":true,"explanation":"The three terms are like terms, each with an implied coefficient of 1. Adding their coefficients gives $1 + 1 + 1 = 3$, producing $3x$."},{"text":"$$x^3$","correct":false,"explanation":"This confuses repeated addition with repeated multiplication: $x^3$ means $x \\times x \\times x$. Adding three identical $x$ terms gives $3x$."},{"text":"$$3 + x$","correct":false,"explanation":"Counting the terms as 3 but retaining only one $x$ does not combine their coefficients. Adding three coefficients of 1 gives $3x$."},{"text":"$$2x$","correct":false,"explanation":"This misses one of the three $x$ terms. Their coefficients add as $1 + 1 + 1 = 3$, so the simplified result is $3x$."}]},{"id":"533af2a3-9232-4d99-8144-a20644848d8e","cardType":"cloze","points":null,"front":"In algebra, a single $x$ means $\\text{[...]}$, not zero $x$.","back":"In algebra, a single $x$ means $\\boxed{1x}$, not zero $x$.","mcqOptions":[{"text":"$$1x$","correct":true,"explanation":"When no coefficient is written before a variable, the coefficient is understood to be 1. Therefore, $x$ means exactly the same as $1x$."},{"text":"$$0x$","correct":false,"explanation":"A coefficient of 0 would make the whole term equal to zero. A variable written alone has an implied coefficient of 1, so $x$ means $1x$."},{"text":"$$x^2$","correct":false,"explanation":"This confuses an omitted coefficient with an omitted index. A single $x$ has coefficient 1 and means $1x$, whereas $x^2$ means $x \\times x$."},{"text":"$$1 + x$","correct":false,"explanation":"An implied coefficient represents multiplication, not addition. A variable written alone means $1 \\times x$, written as $1x$, rather than $1 + x$."}]},{"id":"70f6817b-3e76-4a71-9a2a-1d4c331ad18e","cardType":"cloze","points":null,"front":"Like terms must have exactly the same [ ].","back":"Like terms must have exactly the same variable part.","mcqOptions":[{"text":"variable part","correct":true,"explanation":"Like terms have identical variables raised to identical powers. Their coefficients may differ, but the complete variable part must be exactly the same."},{"text":"coefficient","correct":false,"explanation":"Coefficients do not need to match: for example, $2x$ and $5x$ are like terms. It is their variable part that must be identical."},{"text":"numerical value","correct":false,"explanation":"Like terms can have different values because their coefficients may differ. They are classified by having exactly the same variable part, including any powers."},{"text":"number of terms","correct":false,"explanation":"The number of terms in an expression does not determine whether individual terms are alike. Like terms must instead have exactly the same variable part."}]},{"id":"47aba6b9-2ef9-4f0a-b152-a9882c4d8e06","cardType":"basic","points":2,"front":"Simplify $k + k + 9$.","back":"$$2k + 9$","mcqOptions":[{"text":"$$2k + 9$","correct":true,"explanation":"The two $k$ terms are like terms, so their implied coefficients add to 2. The constant 9 is unlike them and remains unchanged."},{"text":"$$k + 9$","correct":false,"explanation":"This loses one of the two $k$ terms by failing to add their coefficients. Since $k + k = 2k$, the result is $2k + 9$."},{"text":"$$2k + 18$","correct":false,"explanation":"Only the two like $k$ terms should be combined. The 9 appears once and must not be doubled, so the result is $2k + 9$."},{"text":"$$11k$","correct":false,"explanation":"This incorrectly combines the constant 9 with the variable terms as though all three were alike. Only $k + k$ combines, giving $2k + 9$."}]},{"id":"a875cdae-d480-497c-9791-83b67223449a","cardType":"basic","points":2,"front":"Why can $3a$ and $3b$ not be collected together?","back":"They are unlike terms (they have different variables).","mcqOptions":[{"text":"They are unlike terms (they have different variables).","correct":true,"explanation":"Like terms must have exactly the same variable part. Although both coefficients are 3, one term contains $a$ and the other contains $b$, so they cannot be collected."},{"text":"They have different coefficients (one coefficient is 3 and the other is 1).","correct":false,"explanation":"Both terms have coefficient 3, so this misidentifies their numerical multipliers. They cannot be collected because their variables, $a$ and $b$, are different."},{"text":"They are constant terms (their values do not change).","correct":false,"explanation":"Both terms contain variables, so neither is a constant term. They cannot be collected because $a$ and $b$ give them different variable parts."},{"text":"They are like terms (both contain one variable).","correct":false,"explanation":"Having one variable each is not enough to make terms alike; the variables must be identical. Since one has $a$ and the other has $b$, they are unlike terms."}]},{"id":"6dc8e070-ed5b-4c94-a298-f6f783a19a3a","cardType":"basic","points":2,"front":"Are $x$ and $x^2$ like terms? Why?","back":"No; they have different powers.","mcqOptions":[{"text":"No; they have different powers.","correct":true,"explanation":"Like terms must contain exactly the same variables raised to the same powers. Here, $x$ has power $1$, while $x^2$ has power $2$."},{"text":"Yes; they have the same variable.","correct":false,"explanation":"Sharing the variable $x$ is not sufficient for terms to be alike. Their powers must also match, so $x$ and $x^2$ are not like terms."},{"text":"Yes; they have different powers.","correct":false,"explanation":"Different powers make terms unlike, rather than like. Since $x=x^1$ and $x^2$ have different exponents, they cannot be collected together."},{"text":"No; they have different variables.","correct":false,"explanation":"Both terms contain the same variable, $x$, so the variables are not different. They are unlike because the powers of $x$ are different."}]},{"id":"f6c97b1b-bfb7-46c4-84cf-cc7a8a8b0a8d","cardType":"basic","points":2,"front":"Identify the four terms in $8c - 5d + 2c - d$.","back":"

$8c$
$-5d$
$2c$
$-d$

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

$8c$
$-5d$
$2c$
$-d$

","correct":true,"explanation":"Terms are separated by addition or subtraction signs, and each preceding sign belongs to its term. Therefore the four terms include the negative signs on both $d$-terms."},{"text":"

$8c$
$5d$
$2c$
$d$

","correct":false,"explanation":"This drops both subtraction signs, incorrectly making the $d$-terms positive. A sign belongs to the term after it, so the terms are $8c$, $-5d$, $2c$, and $-d$."},{"text":"

$8c$
$-5d$
$-2c$
$-d$

","correct":false,"explanation":"The plus sign before $2c$ makes that term positive, not negative. The four terms are $8c$, $-5d$, $2c$, and $-d$."},{"text":"

$-8c$
$-5d$
$2c$
$-d$

","correct":false,"explanation":"The first term has no written sign, so it is positive rather than negative. The terms are $8c$, $-5d$, $2c$, and $-d$."}]},{"id":"8da1aff5-4615-4575-8d5f-dda3d731468c","cardType":"cloze","points":null,"front":"In an expression, the sign before a term [ ].","back":"In an expression, the sign before a term belongs to that term.","mcqOptions":[{"text":"belongs to that term","correct":true,"explanation":"A plus or minus sign immediately before a term determines whether that entire term is positive or negative, so it must remain attached when identifying or rearranging terms."},{"text":"belongs to the previous term","correct":false,"explanation":"This reverses how signs are attached in an expression. A sign belongs to the term immediately after it, which is why $-5d$ is one complete term."},{"text":"can always be ignored","correct":false,"explanation":"Ignoring signs changes the values of negative terms and often produces an incorrect simplification. The sign before a term belongs to that term and must move with it."},{"text":"applies only to the variable","correct":false,"explanation":"The sign affects the entire term, including its coefficient and variable part. For example, the minus sign in $-5d$ applies to the whole term."}]},{"id":"1c6b8c63-7b92-482e-ba1b-ec0bc1faf963","cardType":"basic","points":2,"front":"Simplify $10a - 3b + 4a - 5b$.","back":"$$14a - 8b$","mcqOptions":[{"text":"$$14a - 8b$","correct":true,"explanation":"Collecting like terms gives $10a+4a=14a$ and $-3b-5b=-8b$. Terms containing different variables cannot be combined with each other."},{"text":"$$14a + 2b$","correct":false,"explanation":"This treats $-3b-5b$ as though the second negative were positive. Adding two negative coefficients gives $-8$, so the simplified expression is $14a-8b$."},{"text":"$$6a - 8b$","correct":false,"explanation":"This subtracts the coefficient $4$ from $10$ despite the plus sign before $4a$. The $a$-terms total $14a$, giving $14a-8b$."},{"text":"$$14a - 2b$","correct":false,"explanation":"This finds the difference between $3$ and $5$ instead of adding two negative coefficients. Since $-3-5=-8$, the result is $14a-8b$."}]},{"id":"f5965113-37dc-415b-a7ee-a8347b9b456d","cardType":"cloze","points":null,"front":"The simplified way to write $a \\times b$ is $\\text{[...]}$.","back":"The simplified way to write $a \\times b$ is $\\boxed{ab}$.","mcqOptions":[{"text":"$$ab$","correct":true,"explanation":"In algebra, adjacent letters represent multiplication, so the multiplication sign between $a$ and $b$ can be omitted and the product written as $ab$."},{"text":"$$a+b$","correct":false,"explanation":"Replacing multiplication with addition changes the operation and therefore the expression’s value. The product of $a$ and $b$ is written in simplified form as $ab$."},{"text":"$$a^b$","correct":false,"explanation":"Writing $a^b$ makes $b$ an exponent, meaning a power rather than a product. Multiplication of the two variables is written simply as $ab$."},{"text":"$$2ab$","correct":false,"explanation":"The coefficient $2$ would double the original product, but no factor of $2$ appears in $a\\times b$. The simplified product is therefore $ab$."}]},{"id":"2879ff06-318c-4bac-901a-85d64c146969","cardType":"basic","points":1,"front":"Simplify $5a \\times 2b$.","back":"$$10ab$","mcqOptions":[{"text":"$$10ab$","correct":true,"explanation":"Multiply the numerical coefficients to get $5\\times2=10$, then include both variable factors. This gives the product $10ab$."},{"text":"$$7ab$","correct":false,"explanation":"This adds the coefficients instead of multiplying them. Since the operation is multiplication, $5\\times2=10$, and including both variables gives $10ab$."},{"text":"$$10a+b$","correct":false,"explanation":"This multiplies the numbers but incorrectly changes multiplication between the variable factors into addition. Both variables remain factors, so the product is $10ab$."},{"text":"$$10a^2b$","correct":false,"explanation":"The factor $a$ appears only once, so it should not be squared. Multiplying the coefficients and retaining one of each variable gives $10ab$."}]},{"id":"07b705dc-0ba5-4180-a381-4311faf00aa2","cardType":"basic","points":2,"front":"Why is $3a \\times 4b$ not $7ab$?","back":"Coefficients must be multiplied ($3 \\times 4$), not added.","mcqOptions":[{"text":"Coefficients must be multiplied ($3 \\times 4$), not added.","correct":true,"explanation":"Because the terms are being multiplied, their numerical coefficients must also be multiplied. Thus $3\\times4=12$, making the product $12ab$ rather than $7ab$."},{"text":"Coefficients must be added ($3 + 4$), not multiplied.","correct":false,"explanation":"Adding the coefficients is the misconception that produces $7ab$. The multiplication sign requires $3\\times4=12$, so the correct product is $12ab$."},{"text":"Coefficients must be subtracted ($4 - 3$), not multiplied.","correct":false,"explanation":"Subtraction is not justified when the original operation is multiplication. The coefficients must be multiplied, giving $3\\times4=12$ and therefore $12ab$."},{"text":"Coefficients must be concatenated ($3$ and $4$), not multiplied.","correct":false,"explanation":"Placing coefficient digits together would incorrectly produce $34$ and is not an algebraic multiplication rule. The coefficients multiply to $12$, so the result is $12ab$."}]},{"id":"799306e0-9933-4aef-a7e2-d6cdc0275e8d","cardType":"cloze","points":null,"front":"The expression $y^3$ means $\\text{[...]}$.","back":"The expression $y^3$ means $\\boxed{y \\times y \\times y}$.","mcqOptions":[{"text":"$$y \\times y \\times y$","correct":true,"explanation":"An exponent shows how many times the base is used as a factor. The power $y^3$ therefore contains three factors of $y$ multiplied together."},{"text":"$$y \\times 3$","correct":false,"explanation":"This treats the exponent as a multiplier, confusing $y^3$ with $3y$. An exponent of $3$ means multiplying three factors: $y\\times y\\times y$."},{"text":"$$y + y + y$","correct":false,"explanation":"Repeated addition of $y$ gives $3y$, not $y^3$. A power represents repeated multiplication, so $y^3$ means $y\\times y\\times y$."},{"text":"$$y \\times y$","correct":false,"explanation":"This uses only two factors of $y$, which represents $y^2$. The exponent $3$ requires three factors, giving $y\\times y\\times y$."}]},{"id":"5adc5a0b-0ce4-473b-9c5f-e926503502ad","cardType":"basic","points":1,"front":"Simplify $r \\times r \\times r$.","back":"$$r^3$","mcqOptions":[{"text":"$$r^3$","correct":true,"explanation":"Multiplying three identical factors of $r$ means that $r$ is used as a factor three times, which is written in index notation as $r^3$."},{"text":"$$r^2$","correct":false,"explanation":"An exponent of 2 represents only two factors of $r$. The expression contains three identical factors multiplied together, so index notation gives $r^3$."},{"text":"$$3r$","correct":false,"explanation":"Writing $3r$ confuses repeated multiplication with repeated addition. Three factors of $r$ are multiplied, so the simplified expression is $r^3$, not a multiple of $r$."},{"text":"$$r^4$","correct":false,"explanation":"Adding the powers incorrectly gives an exponent one too large. There are exactly three factors of $r$, so the correct index is 3 and the answer is $r^3$."}]},{"id":"207b5341-8c0f-43e2-99a0-05d57d655d08","cardType":"basic","points":1,"front":"Simplify $4x^2 + 2x^2 - x^2$.","back":"$$5x^2$","mcqOptions":[{"text":"$$5x^2$","correct":true,"explanation":"All three terms contain $x^2$, so their coefficients can be combined. Since $4+2-1=5$, the expression simplifies to $5x^2$."},{"text":"$$6x^2$","correct":false,"explanation":"The coefficients have been added without subtracting the final coefficient: $4+2-1=5$, not 6. Since the terms are like terms, the answer is $5x^2$."},{"text":"$$5x^4$","correct":false,"explanation":"Combining like terms changes only their coefficients; it does not add the powers of $x$. Calculating $4+2-1$ gives the correct result $5x^2$."},{"text":"$$5x$","correct":false,"explanation":"The power of $x$ remains 2 when like terms are added or subtracted. Combining the coefficients gives $4+2-1=5$, so the answer is $5x^2$."}]},{"id":"53487f49-2d82-462f-8d91-c46fab1ff656","cardType":"basic","points":1,"front":"Simplify $\\frac{12h}{3}$.","back":"$$4h$","mcqOptions":[{"text":"$$4h$","correct":true,"explanation":"Dividing the coefficient 12 by 3 gives 4, while the factor $h$ remains unchanged. Therefore, $rac{12h}{3}=4h$."},{"text":"$$4$","correct":false,"explanation":"Cancelling $h$ is invalid because there is no factor of $h$ in the denominator. Only $12$ is divided by 3, giving the correct answer $4h$."},{"text":"$$9h$","correct":false,"explanation":"Subtracting 3 from 12 gives 9, but the fraction requires division rather than subtraction. Since $12\\div3=4$, the correct simplification is $4h$."},{"text":"$$36h$","correct":false,"explanation":"Multiplying 12 by 3 gives 36, but the fraction bar indicates division. Dividing the coefficient by 3 gives $12\\div3=4$, so the answer is $4h$."}]},{"id":"412be5ac-a9f3-4e74-b63c-25ea7b2fa56d","cardType":"basic","points":1,"front":"Simplify $\\frac{15d}{d}$.","back":"15","mcqOptions":[{"text":"15","correct":true,"explanation":"The common factor $d$ cancels because $d/d=1$ for $d\\ne0$. This leaves the numerical coefficient 15, so the simplified result is 15."},{"text":"$$15d$","correct":false,"explanation":"Cancelling the common factor $d$ removes it from both numerator and denominator. Since $d/d=1$ for $d\\ne0$, the expression simplifies to 15, not $15d$."},{"text":"1","correct":false,"explanation":"Although $d/d=1$, the coefficient 15 must remain. Cancelling only the common factor $d$ therefore leaves 15 rather than reducing the entire numerator to 1."},{"text":"0","correct":false,"explanation":"Division by the same non-zero variable gives 1, not 0. The remaining coefficient is 15, so $rac{15d}{d}$ simplifies to 15."}]},{"id":"6bfef6cf-cdbf-424a-a6a0-07d239994b8e","cardType":"basic","points":2,"front":"Simplify $\\frac{8y + 4y}{2}$.","back":"

Collect: $12y / 2$
Answer: $6y$

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

Collect: $12y / 2$
Answer: $6y$

","correct":true,"explanation":"The like terms in the numerator combine to give $8y+4y=12y$. Dividing this result by 2 gives $12y\\div2=6y$."},{"text":"

Collect: $12y / 2$
Answer: $12y$

","correct":false,"explanation":"The numerator is collected correctly, but it must still be divided by 2. Since $12y\\div2=6y$, the final answer is $6y$, not $12y$."},{"text":"

Collect: $8y / 2$
Answer: $4y$

","correct":false,"explanation":"The like terms in the numerator should be added, so $8y+4y=12y$, not $8y$. Dividing the collected numerator by 2 then gives $6y$."},{"text":"

Collect: $12y / 4$
Answer: $3y$

","correct":false,"explanation":"The denominator remains 2; the coefficient 4 in the numerator is part of the term $4y$, not an additional divisor. Thus $12y\\div2=6y$."}]},{"id":"0a088d3d-023e-4608-ae1f-4a528e20d169","cardType":"basic","points":2,"front":"Simplify $7g + 4h - 2g + 3h$.","back":"$$5g + 7h$","mcqOptions":[{"text":"$$5g + 7h$","correct":true,"explanation":"Collecting like terms separately gives $7g-2g=5g$ and $4h+3h=7h$. Therefore, the whole expression simplifies to $5g+7h$."},{"text":"$$9g + 7h$","correct":false,"explanation":"The subtraction of $2g$ has been overlooked. Combining the $g$-terms gives $7g-2g=5g$, while the $h$-terms give $4h+3h=7h$."},{"text":"$$5g + h$","correct":false,"explanation":"The $h$-coefficients should be added, not subtracted: $4+3=7$. The $g$-terms give $7g-2g=5g$, so the result is $5g+7h$."},{"text":"$$12g + 2h$","correct":false,"explanation":"Unlike terms cannot be combined by adding all their coefficients. The $g$-terms and $h$-terms must be collected separately, producing $5g+7h$."}]}]}],"examCountdown":null,"heading":"Simplifying Algebra","title":"Simplifying Algebra Flashcards","topbar":{"backHref":"/igcse/maths/oxford-aqa/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"f5b9b22e-91ea-4cc6-a442-b698f8897a25","walled":"$L39"}],"$L3a","$L3b"]