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they are written in different forms."},{"text":"same number of terms","correct":false,"explanation":"The two sides do not need the same number of terms. They must have the same value, such as $2+3$ and $5$."},{"text":"different values","correct":false,"explanation":"Different values would make the equation false. The equals sign states that the expression on each side has the same value."},{"text":"answer written next","correct":false,"explanation":"An equals sign does not simply introduce an answer, as it may connect two expressions. It means that both sides have the same value."}]},{"id":"33e738fc-e468-4eb2-9e01-bb9b2fb7e968","cardType":"cloze","points":null,"front":"The balance rule: doing the [ ] to [ ] keeps an equation true.","back":"The balance rule: doing the same operation to both sides keeps an equation true.","mcqOptions":[{"text":"The balance rule: doing the same operation to both sides keeps an equation true.","correct":true,"explanation":"Applying the same operation to both sides preserves equality because both sides are changed equally, maintaining the balance represented by the equation."},{"text":"The balance rule: doing the same operation to one side keeps an equation true.","correct":false,"explanation":"Changing only one side usually destroys the equality. The balance rule requires the same operation to be applied to both sides."},{"text":"The balance rule: doing the different operations to both sides keeps an equation true.","correct":false,"explanation":"Applying different operations changes the sides by unequal amounts or factors. An equation remains balanced when the same operation is performed on both sides."},{"text":"The balance rule: doing the inverse operations to opposite sides keeps an equation true.","correct":false,"explanation":"Using inverse operations on opposite sides does not preserve equality because the sides change differently. Instead, perform the same operation on both sides."}]},{"id":"00b05ff2-b5ec-4062-b58e-95e469657f16","cardType":"cloze","points":null,"front":"Adding is undone by [ ]; multiplying is undone by [ ].","back":"Adding is undone by subtracting; multiplying is undone by dividing.","mcqOptions":[{"text":"Adding is undone by subtracting; multiplying is undone by dividing.","correct":true,"explanation":"Subtraction reverses addition, while division reverses multiplication. These inverse operations allow terms and coefficients to be removed when solving equations."},{"text":"Adding is undone by subtracting; multiplying is undone by subtracting.","correct":false,"explanation":"Subtraction correctly reverses addition, but it does not reverse multiplication. Multiplication must be undone by dividing by the same non-zero number."},{"text":"Adding is undone by dividing; multiplying is undone by dividing.","correct":false,"explanation":"Division correctly reverses multiplication, but it does not reverse addition. Addition must be undone by subtracting the same amount."},{"text":"Adding is undone by multiplying; multiplying is undone by adding.","correct":false,"explanation":"Multiplying does not reverse addition, and adding does not reverse multiplication. The required inverse operations are subtracting and dividing respectively."}]},{"id":"76df22c7-8e0d-49bf-b5ac-2aa41fc40b5a","cardType":"basic","points":1,"front":"Find the missing number: $9+\\Box=16$.","back":"$7$.","mcqOptions":[{"text":"$7$.","correct":true,"explanation":"Subtracting $9$ from both sides gives the missing number: $16-9=7$. Checking confirms that $9+7=16$."},{"text":"$25$.","correct":false,"explanation":"Adding $9$ and $16$ instead of finding their difference produces $25$. The missing number is $16-9=7$."},{"text":"$-7$.","correct":false,"explanation":"Reversing the subtraction as $9-16$ creates an unnecessary negative answer. The required calculation is $16-9=7$."},{"text":"$6$.","correct":false,"explanation":"This is an off-by-one subtraction error. Taking $9$ from $16$ gives $7$, and $9+7$ checks correctly as $16$."}]},{"id":"18193f23-3cd9-46d6-aad9-cbdbdb06ad57","cardType":"basic","points":1,"front":"Solve $x+5=18$.","back":"$x=13$.","mcqOptions":[{"text":"$x=13$.","correct":true,"explanation":"Subtracting $5$ from both sides isolates $x$, giving $x=18-5=13$. Substitution confirms that $13+5=18$."},{"text":"$x=23$.","correct":false,"explanation":"Adding $5$ to $18$ applies the inverse operation in the wrong direction. Subtracting $5$ from both sides gives $x=13$."},{"text":"$x=90$.","correct":false,"explanation":"Multiplying $18$ by $5$ confuses an additive equation with a multiplicative one. Since $5$ is added, subtract it to obtain $x=13$."},{"text":"$x=3.6$.","correct":false,"explanation":"Dividing $18$ by $5$ does not undo adding $5$. The correct inverse operation is subtraction, so $x=18-5=13$."}]},{"id":"7b3eb533-8054-460f-8cfe-b1920b4863d5","cardType":"basic","points":2,"front":"Solve $6-n=14$.","back":"$n=-8$.","mcqOptions":[{"text":"$n=-8$.","correct":true,"explanation":"Subtracting $6$ from both sides gives $-n=8$. Multiplying both sides by $-1$ then gives $n=-8$."},{"text":"$n=8$.","correct":false,"explanation":"Stopping at $-n=8$ and dropping the negative sign gives this tempting result. Multiplying by $-1$ shows that $n=-8$."},{"text":"$n=20$.","correct":false,"explanation":"Adding $6$ and $14$ ignores that the equation contains $-n$. Rearranging gives $-n=8$, and therefore $n=-8$."},{"text":"$n=-20$.","correct":false,"explanation":"Combining $6$ and $14$ with the wrong sign leads to this value. Correctly subtracting $6$ gives $-n=8$, so $n=-8$."}]},{"id":"d5d2a333-ce53-4497-bc23-1c884b11a614","cardType":"cloze","points":null,"front":"A letter next to a number means [ ]; a fraction bar means [ ].","back":"A letter next to a number means multiplication; a fraction bar means division.","mcqOptions":[{"text":"A letter next to a number means multiplication; a fraction bar means division.","correct":true,"explanation":"In algebra, adjacency indicates multiplication, so $3x$ means $3\\times x$. A fraction bar indicates that the numerator is divided by the denominator."},{"text":"A letter next to a number means addition; a fraction bar means division.","correct":false,"explanation":"A number written beside a letter is multiplied by it, not added to it. The fraction bar does correctly represent division."},{"text":"A letter next to a number means multiplication; a fraction bar means multiplication.","correct":false,"explanation":"Adjacency correctly represents multiplication, but a fraction bar represents division rather than multiplication. For example, $\\frac{x}{2}$ means $x\\div2$."},{"text":"A letter next to a number means division; a fraction bar means multiplication.","correct":false,"explanation":"These operations have been reversed. A number beside a letter indicates multiplication, while a fraction bar indicates division."}]},{"id":"7aa95689-df9f-4f0c-9c9a-b4445e812e23","cardType":"basic","points":1,"front":"Solve $8p=56$.","back":"$p=7$.","mcqOptions":[{"text":"$p=7$.","correct":true,"explanation":"Dividing both sides of $8p=56$ by $8$ isolates the variable, giving $p=56\\div8=7$."},{"text":"$p=48$.","correct":false,"explanation":"Subtracting $8$ from $56$ treats $8p$ as though it meant $p+8$. Since $8p$ is multiplication, divide to get $p=7$."},{"text":"$p=448$.","correct":false,"explanation":"Multiplying $56$ by $8$ repeats the operation instead of undoing it. Division by $8$ is required, giving $p=7$."},{"text":"$p=64$.","correct":false,"explanation":"Adding $8$ to $56$ does not undo the multiplication in $8p$. Dividing both sides by $8$ gives $p=7$."}]},{"id":"6a59707d-d307-470f-ace6-db05bb02e633","cardType":"basic","points":1,"front":"Solve $\\frac{q}{3}=7$.","back":"$q=21$.","mcqOptions":[{"text":"$q=21$.","correct":true,"explanation":"Multiplying both sides of $\\frac{q}{3}=7$ by $3$ gives $q=21$. Substitution confirms that $\\frac{21}{3}=7$."},{"text":"$q=\\frac{7}{3}$.","correct":false,"explanation":"Dividing $7$ by $3$ applies the inverse operation in the wrong direction. Multiply both sides by $3$ to obtain $q=21$."},{"text":"$q=14$.","correct":false,"explanation":"Multiplying by $2$ instead of the denominator $3$ leaves the equation unbalanced. The correct calculation is $7\\times3$, so $q=21$."},{"text":"$q=24$.","correct":false,"explanation":"Adding $3$ treats the denominator as though it were a subtracted term. Multiplication by $3$ is required, giving $q=21$."}]},{"id":"9c4870c3-a332-4415-8684-509fa4cd5d6e","cardType":"basic","points":2,"front":"How can you check a proposed solution to an equation?","back":"Substitute it into the original equation; both sides must match.","mcqOptions":[{"text":"Substitute it into the original equation; both sides must match.","correct":true,"explanation":"Replacing the variable with the proposed value tests the complete original equation. The value is a solution only when both sides evaluate to the same result."},{"text":"Round it before putting it into the original equation; both sides must be close.","correct":false,"explanation":"Rounding can conceal an incorrect value, and being close is not sufficient for an exact equation. Substitute the proposed value unchanged; both sides must match exactly."},{"text":"Substitute it into one side of the original equation; that side must simplify.","correct":false,"explanation":"Checking only one side gives no comparison and cannot establish equality. Substitute the value throughout the original equation and verify that both sides have equal values."},{"text":"Rearrange the original equation until the proposed solution appears on one side.","correct":false,"explanation":"Rearranging may reproduce an algebraic error and does not directly verify the proposed value. Substitute it into the original equation and check that both sides match."}]},{"id":"50a4d5a3-28ff-425a-938d-e85302ac384a","cardType":"cloze","points":null,"front":"In a two-step equation, usually undo [ ] before [ ].","back":"In a two-step equation, usually undo addition or subtraction before multiplication or division.","mcqOptions":[{"text":"In a two-step equation, usually undo addition or subtraction before multiplication or division.","correct":true,"explanation":"Inverse operations are normally applied in reverse order. Since multiplication or division acts on the variable before addition or subtraction, the added or subtracted term is undone first."},{"text":"In a two-step equation, usually undo multiplication or division before addition or subtraction.","correct":false,"explanation":"This follows the original order rather than reversing it. Usually remove addition or subtraction first, then undo multiplication or division to isolate the variable."},{"text":"In a two-step equation, usually undo the operation nearest the variable before the operation furthest from the variable.","correct":false,"explanation":"Starting nearest the variable reverses the required solving order. Undo the outer operation first - usually addition or subtraction - before undoing multiplication or division."},{"text":"In a two-step equation, usually undo operations on the right-hand side before operations on the left-hand side.","correct":false,"explanation":"Equations are solved by applying inverse operations to both sides, not by clearing one side first. Usually undo addition or subtraction before multiplication or division."}]},{"id":"1e6772f3-c17a-43a9-8baa-361e62ea3cb9","cardType":"basic","points":2,"front":"Solve $5a-3=27$.","back":"$a=6$.","mcqOptions":[{"text":"$a=6$.","correct":true,"explanation":"Adding $3$ to both sides gives $5a=30$. Dividing both sides by $5$ then gives $a=6$, which satisfies the original equation."},{"text":"$a=\\frac{24}{5}$.","correct":false,"explanation":"Subtracting $3$ from $27$ repeats the subtraction instead of reversing it. Add $3$ to obtain $5a=30$, then divide by $5$ to get $a=6$."},{"text":"$a=5$.","correct":false,"explanation":"This effectively divides before correctly removing the constant term. Add $3$ first to make $5a=30$, then divide by $5$, giving $a=6$."},{"text":"$a=30$.","correct":false,"explanation":"Adding $3$ correctly produces $5a=30$, but the coefficient $5$ still needs to be removed. Dividing by $5$ gives the correct solution $a=6$."}]},{"id":"e17bfb3c-6062-4aea-9e70-b00821642318","cardType":"basic","points":2,"front":"Solve $2x+9=3$.","back":"$x=-3$.","mcqOptions":[{"text":"$x=-3$.","correct":true,"explanation":"Subtracting $9$ from both sides gives $2x=-6$. Dividing by $2$ gives $x=-3$, and substitution produces $-6+9=3$."},{"text":"$x=3$.","correct":false,"explanation":"The negative sign is lost when subtracting $9$ from both sides. Since $3-9=-6$, the equation becomes $2x=-6$, so $x=-3$."},{"text":"$x=-6$.","correct":false,"explanation":"This stops after finding $2x=-6$ and does not remove the coefficient of $x$. Dividing both sides by $2$ gives the correct answer $x=-3$."},{"text":"$x=6$.","correct":false,"explanation":"This both loses the negative sign and omits the final division by $2$. Subtracting $9$ gives $2x=-6$, so the solution is $x=-3$."}]},{"id":"5c2f955d-56ae-42f4-b8d2-927553609ec1","cardType":"basic","points":1,"front":"Expand $4(x-3)$ correctly.","back":"$4x-12$.","mcqOptions":[{"text":"$4x-12$.","correct":true,"explanation":"The multiplier $4$ applies to both terms inside the brackets: $4\\times x=4x$ and $4\\times(-3)=-12$, giving $4x-12$."},{"text":"$4x-3$.","correct":false,"explanation":"Only the $x$ term has been multiplied by $4$. The multiplier must distribute to both terms, so $4(x-3)=4x-12$."},{"text":"$4x-7$.","correct":false,"explanation":"Subtracting $4$ from $3$ does not distribute the multiplier. Multiply both $x$ and $-3$ by $4$ to obtain $4x-12$."},{"text":"$x-12$.","correct":false,"explanation":"The constant has been multiplied by $4$, but the $x$ term has not. Distribution applies to every term, so the correct expansion is $4x-12$."}]},{"id":"a1b0da8a-fc2f-4897-82bf-09c4d5869c3b","cardType":"basic","points":2,"front":"Solve $3(g-4)=21$.","back":"$g=11$.","mcqOptions":[{"text":"$g=11$.","correct":true,"explanation":"Dividing both sides by $3$ gives $g-4=7$. Adding $4$ to both sides then gives $g=11$, which makes $3(11-4)=21$."},{"text":"$g=3$.","correct":false,"explanation":"Subtracting $4$ after finding $g-4=7$ uses the wrong inverse operation. Add $4$ instead, giving the correct solution $g=11$."},{"text":"$g=9$.","correct":false,"explanation":"Adding $4$ directly to $21$ before removing the outer multiplier mixes the operation order. Divide by $3$ first, then add $4$ to obtain $g=11$."},{"text":"$g=25$.","correct":false,"explanation":"Adding $4$ to $21$ ignores the multiplier outside the brackets. First divide by $3$ to get $g-4=7$, then add $4$, giving $g=11$."}]},{"id":"367ed859-f6cb-417c-800e-c93d019f651f","cardType":"basic","points":2,"front":"Solve $\\frac{r}{5}-2=4$.","back":"$r=30$.","mcqOptions":[{"text":"$r=30$.","correct":true,"explanation":"Adding $2$ to both sides gives $\\frac{r}{5}=6$. Multiplying both sides by $5$ then gives $r=30$."},{"text":"$r=10$.","correct":false,"explanation":"Multiplying the added $2$ by $5$ ignores the right-hand side value $4$. First obtain $\\frac{r}{5}=6$, then multiply by $5$ to get $r=30$."},{"text":"$r=22$.","correct":false,"explanation":"Multiplying $4$ by $5$ before adding $2$ fails to reverse the operations in the correct order. Add $2$ first, then multiply by $5$, giving $r=30$."},{"text":"$r=2$.","correct":false,"explanation":"Subtracting $2$ from $4$ repeats the subtraction instead of undoing it. Add $2$ to get $\\frac{r}{5}=6$, then multiply by $5$, so $r=30$."}]},{"id":"8df6bbc1-c06b-418b-a971-4072c7fbbdf7","cardType":"basic","points":2,"front":"Solve $\\frac{h+2}{3}=8$.","back":"$h=22$.","mcqOptions":[{"text":"$h=22$.","correct":true,"explanation":"Multiplying both sides by $3$ gives $h+2=24$. Subtracting $2$ from both sides then gives $h=22$."},{"text":"$h=26$.","correct":false,"explanation":"Adding $2$ after obtaining $h+2=24$ reverses the required operation. Subtracting $2$ isolates the variable, giving $h=22$."},{"text":"$h=6$.","correct":false,"explanation":"Dividing $8$ by $3$ does not undo division of the entire numerator by $3$. Multiplying by $3$ gives $h+2=24$, so $h=22$."},{"text":"$h=24$.","correct":false,"explanation":"Multiplying both sides by $3$ gives the value of $h+2$, not of $h$. A further subtraction of $2$ is needed, giving $h=22$."}]},{"id":"88f83a44-4d22-4343-870a-f8b023d01551","cardType":"basic","points":2,"front":"Solve $\\frac{3x}{4}=9$.","back":"$x=12$.","mcqOptions":[{"text":"$x=12$.","correct":true,"explanation":"Multiplying both sides by $4$ gives $3x=36$. Dividing both sides by $3$ then gives $x=12$."},{"text":"$x=\\frac{27}{4}$.","correct":false,"explanation":"Multiplying $9$ by $\\frac{3}{4}$ applies the coefficient again instead of undoing it. Multiplication by the reciprocal $\\frac{4}{3}$ gives $x=12$."},{"text":"$x=9$.","correct":false,"explanation":"Treating the fraction coefficient as if it were $1$ ignores both multiplication by $3$ and division by $4$. Using the reciprocal $\\frac{4}{3}$ gives $x=12$."},{"text":"$x=16$.","correct":false,"explanation":"Multiplying $9$ by $4$ but failing to divide by $3$ leaves the coefficient only partly undone. Since $3x=36$, the correct result is $x=12$."}]},{"id":"2c45e9e6-0973-4510-9423-b82bb4f7745e","cardType":"cloze","points":null,"front":"Like terms have the [ ]; variable terms should be collected on [ ].","back":"Like terms have the same variable part; variable terms should be collected on one side.","mcqOptions":[{"text":"Like terms have the same variable part; variable terms should be collected on one side.","correct":true,"explanation":"Like terms contain identical variables raised to identical powers, so they share the same variable part. Collecting variable terms on one side allows the equation to be simplified and solved."},{"text":"Like terms have the same coefficient; variable terms should be collected on one side.","correct":false,"explanation":"Coefficients do not need to match for terms to be like terms; for example, $3x$ and $7x$ are like terms. They must have the same variable part."},{"text":"Like terms have the same variable part; variable terms should be collected on both sides.","correct":false,"explanation":"Leaving variable terms on both sides does not complete the collection step. They should be moved to one side, while their identical variable parts determine whether they are like terms."},{"text":"Like terms have the same coefficient; variable terms should be collected on both sides.","correct":false,"explanation":"This confuses coefficients with variable parts and leaves variables uncollected. Like terms share the same variable part, and variable terms should be gathered on one side of the equation."}]},{"id":"f480b9c6-d038-4692-8cff-312812c4c346","cardType":"basic","points":3,"front":"Solve $7x+5=3x+25$.","back":"$x=5$.","mcqOptions":[{"text":"$x=5$.","correct":true,"explanation":"Subtracting $3x$ and $5$ from both sides gives $4x=20$. Dividing both sides by $4$ gives $x=5$."},{"text":"$x=10$.","correct":false,"explanation":"This typically comes from mishandling the difference between the variable coefficients. Subtracting $3x$ gives $4x+5=25$, then subtracting $5$ and dividing by $4$ gives $x=5$."},{"text":"$x=\\frac{15}{2}$.","correct":false,"explanation":"This results from adding the constants instead of subtracting $5$ from $25$. Correctly collecting terms gives $4x=20$, so dividing by $4$ gives $x=5$."},{"text":"$x=-5$.","correct":false,"explanation":"A sign reversal while moving terms can produce a negative result. Keeping equivalent operations on both sides gives $4x=20$, so the solution is positive: $x=5$."}]},{"id":"1993f65e-2c7a-4ea2-bd9a-c06dd83096d9","cardType":"basic","points":3,"front":"Solve $12-3s=s-8$.","back":"$s=5$.","mcqOptions":[{"text":"$s=5$.","correct":true,"explanation":"Adding $3s$ and $8$ to both sides gives $20=4s$. Dividing both sides by $4$ gives $s=5$."},{"text":"$s=-5$.","correct":false,"explanation":"A sign error when moving $-3s$ or $-8$ can make the result negative. Correct collection gives $20=4s$, so $s=5$."},{"text":"$s=20$.","correct":false,"explanation":"The equation $20=4s$ is an intermediate step, not the final solution. Dividing both sides by the coefficient $4$ gives $s=5$."},{"text":"$s=4$.","correct":false,"explanation":"Dividing $20$ by the sum of the coefficient magnitudes, $5$, is a coefficient error. The collected equation is $20=4s$, so $s=5$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Solving Equations","title":"Solving Equations Flashcards","topbar":{"backHref":"/igcse/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"4e58b58c-4b38-4218-9cd3-3b7094bcb76d","walled":"$L39"}],"$L3a","$L3b"]