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They therefore satisfy both linear equations simultaneously, making that point the graphical solution."},{"text":"The point where either straight line crosses the $y$-axis.","correct":false,"explanation":"A $y$-intercept satisfies only the equation represented by its own line. The simultaneous solution is where the two straight lines cross each other."},{"text":"The gradient shared by the two straight lines when they are parallel.","correct":false,"explanation":"Parallel lines may share a gradient, but distinct parallel lines never intersect and therefore have no simultaneous solution. A graphical solution is the lines’ intersection point."},{"text":"The region lying between the two straight lines on the graph.","correct":false,"explanation":"A region between two lines does not give a coordinate satisfying both equations. The solution is the single point where the two straight lines intersect."}]},{"id":"60d83b44-fae1-4c39-ba16-598c2ccfb0fe","cardType":"basic","points":1,"front":"What does a solution pair $(x, y)$ represent in simultaneous equations?","back":"The single pair of values that makes both equations true at the same time.","mcqOptions":[{"text":"The single pair of values that makes both equations true at the same time.","correct":true,"explanation":"A simultaneous solution must satisfy every equation in the system. Substituting its $x$-value and $y$-value makes both equations true at once."},{"text":"The separate pair of values that makes either equation true at a particular time.","correct":false,"explanation":"Satisfying either equation alone is insufficient because many points lie on each individual line. The solution pair must make both equations true simultaneously."},{"text":"The single value of $x$ that makes one equation true without using $y$.","correct":false,"explanation":"A solution pair contains both an $x$-value and a $y$-value, not just one unknown. Together, the values must satisfy both equations."},{"text":"The pair of intercepts where each equation crosses an axis on the graph.","correct":false,"explanation":"Axis intercepts describe where individual lines meet the axes, not necessarily each other. The solution pair gives the coordinates satisfying both equations simultaneously."}]},{"id":"ed63922e-53fe-4eb8-bacd-37580d00a6bb","cardType":"basic","points":2,"front":"How do you verify if a coordinate like $(4, 1)$ is the solution to a system of equations?","back":"Substitute the values into both equations to see if they hold true.","mcqOptions":[{"text":"Substitute the values into both equations to see if they hold true.","correct":true,"explanation":"Replacing $x$ with $4$ and $y$ with $1$ must produce a true statement in each equation. Only then is $(4, 1)$ the simultaneous solution."},{"text":"Substitute the values into one equation to see if it holds true.","correct":false,"explanation":"A coordinate can satisfy one equation without satisfying the whole system. Substitute $x=4$ and $y=1$ into both equations and check both statements."},{"text":"Add the two values in both equations to see if their total is 5.","correct":false,"explanation":"The fact that $4+1=5$ does not verify a system unless an equation specifically requires it. Substitute each value for its corresponding variable in both equations."},{"text":"Plot the values on one line to see if the coordinate lies on it.","correct":false,"explanation":"Lying on one line proves that only one equation is satisfied. The coordinate must lie on both lines, checked by substitution into both equations."}]},{"id":"09001819-a458-40bd-b7f2-56e53be2da83","cardType":"cloze","points":null,"front":"To eliminate a variable with [ ] signs (e.g. $+3y$ and $-3y$), you add the equations together.","back":"To eliminate a variable with opposite signs (e.g. $+3y$ and $-3y$), you add the equations together.","mcqOptions":[{"text":"opposite","correct":true,"explanation":"Equal coefficients with opposite signs cancel when added: $+3y+(-3y)=0$. This eliminates $y$ while preserving the equality of the combined equation."},{"text":"the same","correct":false,"explanation":"Adding coefficients with the same sign would combine rather than eliminate them, such as $3y+3y=6y$. Opposite signs are required when adding equations."},{"text":"positive","correct":false,"explanation":"Two positive coefficients add together instead of cancelling. To eliminate by addition, the coefficients must have equal magnitude and opposite signs, such as $+3y$ and $-3y$."},{"text":"negative","correct":false,"explanation":"Two negative coefficients also combine rather than cancel, such as $-3y+(-3y)=-6y$. Addition eliminates a variable when its coefficients have opposite signs."}]},{"id":"5009b2f6-ddc0-48ad-8aa9-e9205828c42c","cardType":"cloze","points":null,"front":"To eliminate a variable with [ ] sign (e.g. $2x$ and $2x$), you subtract one equation from the other.","back":"To eliminate a variable with the same sign (e.g. $2x$ and $2x$), you subtract one equation from the other.","mcqOptions":[{"text":"the same","correct":true,"explanation":"Equal coefficients with the same sign cancel under subtraction: $2x-2x=0$. This removes the variable and leaves an equation in the other unknown."},{"text":"opposite","correct":false,"explanation":"Subtracting a negative coefficient changes it to positive, so opposite signs combine rather than cancel. Opposite coefficients should be added; same-sign coefficients should be subtracted."},{"text":"a different","correct":false,"explanation":"Different coefficients do not cancel directly when one equation is subtracted from the other. First make their magnitudes equal, then use subtraction if their signs are the same."},{"text":"a negative","correct":false,"explanation":"Being negative alone does not determine the operation; two equal negative coefficients have the same sign and can cancel by subtraction. The key condition is the same sign."}]},{"id":"c6e10f3f-9d84-492b-afb1-c4f70743bca5","cardType":"basic","points":1,"front":"What is a coefficient in an algebraic term like $4x$?","back":"The number multiplying the variable (the 4).","mcqOptions":[{"text":"The number multiplying the variable (the 4).","correct":true,"explanation":"In the term $4x$, the numerical factor attached to $x$ is $4$. This number is called the coefficient of the variable."},{"text":"The variable being multiplied by the number (the x).","correct":false,"explanation":"The $x$ is the variable because its value can change. The coefficient is the numerical factor multiplying it, which is $4$ in $4x$."},{"text":"The power applied to the variable (the 1).","correct":false,"explanation":"The implicit power of $x$ in $4x$ is $1$, but that is its exponent. The coefficient is the multiplying number, which is $4$."},{"text":"The term formed by multiplying the number and variable (the 4x).","correct":false,"explanation":"The whole expression $4x$ is an algebraic term, not its coefficient. The coefficient is only the numerical factor multiplying $x$, namely $4$."}]},{"id":"1af192ce-822f-43a3-9439-4357896baf4d","cardType":"basic","points":1,"front":"When subtracting equations, what is the result of $5y - (-y)$?","back":"$$6y$","mcqOptions":[{"text":"$$6y$","correct":true,"explanation":"Subtracting a negative changes the operation to addition, so $5y-(-y)=5y+y=6y$. The two like terms then combine normally."},{"text":"$$4y$","correct":false,"explanation":"This results from treating $-(-y)$ as $-y$ and calculating $5y-y$. The two negatives make a positive, so the correct result is $6y$."},{"text":"$$5y$","correct":false,"explanation":"This incorrectly assumes that the two negative signs cancel the entire second term. They change subtraction to addition, giving $5y+y=6y$."},{"text":"$$-6y$","correct":false,"explanation":"This combines the coefficients as $5+1$ but incorrectly gives the result a negative sign. Since $-(-y)=+y$, the correct result is positive $6y$."}]},{"id":"78257a11-b3e4-4519-97c2-ee333dee2807","cardType":"basic","points":2,"front":"If neither variable has a matching coefficient, how can you prepare the equations for elimination?","back":"Multiply one (or both) whole equations by a constant to make one coefficient match.","mcqOptions":[{"text":"Multiply one (or both) whole equations by a constant to make one coefficient match.","correct":true,"explanation":"Multiplying every term on both sides preserves an equation’s solutions. Suitable multipliers can create equal or opposite coefficients, allowing one variable to be eliminated."},{"text":"Multiply only one variable term by a constant to make one coefficient match.","correct":false,"explanation":"Changing only one term alters the equation and therefore its solutions. Every term on both sides of one or both equations must be multiplied by the constant."},{"text":"Change one coefficient directly without altering any other terms in the equation.","correct":false,"explanation":"Changing a coefficient alone creates a different equation with different solutions. Instead, multiply the entire equation by a constant so that equality is preserved."},{"text":"Add a constant to one variable term until its coefficient matches the other.","correct":false,"explanation":"Adding to a variable term does not scale its coefficient correctly and changes the equation. Multiply one or both whole equations by a suitable constant instead."}]},{"id":"e995f73c-b4fc-4a3e-b6f7-1e7412e745b4","cardType":"basic","points":1,"front":"What is the common mistake when multiplying an equation to match coefficients?","back":"Forgetting to multiply every term on both sides of the equals sign.","mcqOptions":[{"text":"Forgetting to multiply every term on both sides of the equals sign.","correct":true,"explanation":"Multiplying an equation by a factor means applying it to every term on both sides, preserving equality. Missing even one term changes the equation and produces an incorrect solution."},{"text":"Multiplying only the terms containing variables on both sides of the equals sign.","correct":false,"explanation":"This leaves constants unchanged even though the factor applies to the entire equation. The correct method is multiplying every term on both sides of the equals sign."},{"text":"Changing the equals sign to a plus sign after multiplying both sides.","correct":false,"explanation":"Multiplication does not change the operation joining the two sides; the equals sign must remain. The common mistake is failing to multiply every term on both sides."},{"text":"Dividing the equation by the coefficient instead of multiplying every term.","correct":false,"explanation":"Dividing may be useful in another rearrangement, but it does not match coefficients when multiplication is required. The correct precaution is multiplying every term on both sides."}]},{"id":"29aa0d31-30e9-4ddc-95bf-f6ef31aa316b","cardType":"basic","points":1,"front":"Define standard form for linear simultaneous equations.","back":"$$ax + by = c$, where $x$ and $y$ terms are on the left and the constant is on the right.","mcqOptions":[{"text":"$$ax + by = c$, where $x$ and $y$ terms are on the left and the constant is on the right.","correct":true,"explanation":"This arrangement places both variable terms on the left and the constant term on the right, giving the standard linear form $ax + by = c$."},{"text":"$$ax - by = c$, where $x$ and $y$ terms are on the right and the constant is on the left.","correct":false,"explanation":"The variables and constant have been placed on opposite sides from the required arrangement, and the minus sign is not part of the general definition. The correct form is $ax + by = c$ with variables left and constant right."},{"text":"$$a + bx + cy = 0$, where all terms are on the left and there is no constant on the right.","correct":false,"explanation":"Although equations can sometimes be rearranged this way, it is not the stated standard form for two-variable linear equations. The correct form is $ax + by = c$, with variables left and constant right."},{"text":"$$x + y = ab + c$, where the variables are on the left and all coefficients are on the right.","correct":false,"explanation":"Coefficients belong with their variable terms rather than being moved to the constant side. The required standard form is $ax + by = c$, with both variable terms left and the constant right."}]},{"id":"485441ad-d320-4017-9f0e-59b20f3063f2","cardType":"cloze","points":null,"front":"The standard form of $3x = 10 + y$ is $\\text{[...]}$.","back":"The standard form of $3x = 10 + y$ is $\\boxed{3x - y = 10}$.","mcqOptions":[{"text":"$$3x - y = 10$","correct":true,"explanation":"Subtracting $y$ from both sides of $3x = 10 + y$ places the $y$ term on the left, leaving the constant $10$ on the right."},{"text":"$$3x + y = 10$","correct":false,"explanation":"The $y$ term changes sign when it moves from the right to the left, so it must become negative. The correct standard form is $3x - y = 10$."},{"text":"$$3x - y = -10$","correct":false,"explanation":"Moving $y$ across changes its sign, but the positive constant $10$ stays positive on the right. The correct rearrangement is $3x - y = 10$."},{"text":"$$3x + y = -10$","correct":false,"explanation":"This changes the signs of both terms incorrectly. Only $y$ changes sign when subtracted from both sides, giving the correct form $3x - y = 10$."}]},{"id":"6f609683-2281-49fc-944b-a652ac8163d3","cardType":"basic","points":1,"front":"How can you simplify an equation containing decimals (e.g., 4.5) before solving?","back":"Multiply the whole equation by 10 to clear the decimal point.","mcqOptions":[{"text":"Multiply the whole equation by 10 to clear the decimal point.","correct":true,"explanation":"Multiplying every term in the equation by 10 changes tenths into whole numbers, such as $4.5$ into $45$, while preserving equality and making the equations easier to solve."},{"text":"Multiply only the decimal term by 10 and leave the other terms unchanged.","correct":false,"explanation":"Applying the factor to only one term changes the equation unevenly. The correct method is to multiply the whole equation by 10 so both sides remain equal."},{"text":"Divide the whole equation by 10 to move the decimal point left.","correct":false,"explanation":"Dividing by 10 would make the decimal smaller rather than clear it, changing $4.5$ to $0.45$. Multiplying the whole equation by 10 clears the decimal point."},{"text":"Round each decimal term to the nearest whole number before solving.","correct":false,"explanation":"Rounding changes the exact equation and can alter the solutions. Instead, multiply the whole equation by 10 to remove the decimals without losing accuracy."}]},{"id":"8051a9b3-b954-49af-aca3-325b3315b665","cardType":"cloze","points":null,"front":"If an equation contains 4.5, multiplying the whole equation by 10 changes it to [ ].","back":"If an equation contains 4.5, multiplying the whole equation by 10 changes it to 45.","mcqOptions":[{"text":"45","correct":true,"explanation":"Multiplying $4.5$ by 10 moves the decimal point one place to the right, giving $45$. Applying the same multiplication to the whole equation preserves its equality."},{"text":"0.45","correct":false,"explanation":"Moving the decimal point one place left corresponds to dividing by 10, not multiplying by 10. Multiplying $4.5$ by 10 gives the correct result, $45$."},{"text":"450","correct":false,"explanation":"This moves the decimal point two places right, which would mean multiplying by 100. Multiplying $4.5$ by 10 moves it one place right to give $45$."},{"text":"4.05","correct":false,"explanation":"This incorrectly inserts a zero rather than moving the decimal point one place right. The calculation $4.5 \\times 10$ gives $45$."}]},{"id":"190ec7e1-f47f-4fc5-a8e8-0ddd11fef511","cardType":"basic","points":1,"front":"What is the very first step when solving a word problem using simultaneous equations?","back":"Define variables (assign letters) for the unknown values.","mcqOptions":[{"text":"Define variables (assign letters) for the unknown values.","correct":true,"explanation":"Assigning variables to the unknown quantities gives the problem a clear algebraic structure. These variables can then be used to form the two simultaneous equations."},{"text":"Solve one equation before identifying the unknown values.","correct":false,"explanation":"An equation cannot be formed or solved reliably until the unknown quantities have been identified and represented. The first step is to define variables by assigning letters."},{"text":"Substitute numerical values for the unknowns immediately.","correct":false,"explanation":"The unknown values are what the problem asks you to find, so inserting numbers before forming equations is unjustified. First define variables for those unknown values."},{"text":"Eliminate one variable before translating the information into equations.","correct":false,"explanation":"Elimination is a solving method used after the equations have been formed. The first step is to define variables, allowing the worded relationships to be expressed algebraically."}]},{"id":"538aa107-51f4-4057-ab07-04ce8fd0fa0c","cardType":"basic","points":1,"front":"If you solve a word problem using pence, but the question asks for prices, what must you remember?","back":"Convert the final answers back into pounds (£).","mcqOptions":[{"text":"Convert the final answers back into pounds (£).","correct":true,"explanation":"Since 100 pence make one pound, divide each final pence answer by 100 and report the prices in pounds, matching the units requested by the question."},{"text":"Convert the final answers into pence (p).","correct":false,"explanation":"The calculations were already performed in pence, so leaving the answers in pence does not match a question asking for prices in pounds. The final answers must be converted back into pounds (£)."},{"text":"Multiply the final answers by 100 and write them in pounds (£).","correct":false,"explanation":"Multiplying by 100 makes the numerical values larger, whereas converting pence to pounds requires dividing by 100. The final answers should be converted into pounds (£)."},{"text":"Round the final answers to the nearest whole pound (£).","correct":false,"explanation":"Rounding can lose the exact price and does not perform the required unit conversion. Divide the pence values by 100 to convert the final answers into pounds (£)."}]},{"id":"d6a8a4c5-04ab-4279-9392-857424a2ce6d","cardType":"basic","points":2,"front":"After using elimination to find the value of $x$, how do you find the value of $y$?","back":"Substitute the value of $x$ back into one of the original equations.","mcqOptions":[{"text":"Substitute the value of $x$ back into one of the original equations.","correct":true,"explanation":"Replacing $x$ with its known value in either original equation leaves an equation containing only $y$. Solving that equation gives the corresponding value of $y$."},{"text":"Substitute the value of $y$ back into one of the original equations.","correct":false,"explanation":"At this stage the known value is $x$, so substituting $y$ reverses the order of the method and does not provide the needed value. Substitute the value of $x$ to find $y$."},{"text":"Eliminate the value of $x$ from one of the original equations.","correct":false,"explanation":"Elimination is used with both equations to remove a variable, but after finding $x$, the efficient next step is substitution. Put the value of $x$ into an original equation to find $y$."},{"text":"Divide the value of $x$ by one of the original equations.","correct":false,"explanation":"An equation is not divided by a value in this step, and doing so does not isolate $y$. Substitute the known value of $x$ into an original equation, then solve for $y$."}]},{"id":"4b13e8cf-3870-409a-bf10-0bcc198f3f4c","cardType":"cloze","points":null,"front":"In the elimination method, you [ ] or subtract equations to remove one variable.","back":"In the elimination method, you add or subtract equations to remove one variable.","mcqOptions":[{"text":"add","correct":true,"explanation":"Adding the two equations can make opposite coefficients cancel, removing one variable and leaving an equation that contains only the other variable."},{"text":"multiply","correct":false,"explanation":"Multiplication may first be used to create matching coefficients, but the equations are then added or subtracted. The missing word is add."},{"text":"divide","correct":false,"explanation":"Dividing an equation can simplify its coefficients, but it does not describe the combining step in elimination. The missing word is add."},{"text":"factorise","correct":false,"explanation":"Factorising rewrites an expression as a product and is not the operation used to combine equations. In elimination, you add or subtract equations."}]},{"id":"5476cb46-7364-412b-82ed-e87425e91faf","cardType":"cloze","points":null,"front":"In the elimination method, you add or [ ] equations to remove one variable.","back":"In the elimination method, you add or subtract equations to remove one variable.","mcqOptions":[{"text":"subtract","correct":true,"explanation":"Subtracting equations with equal coefficients can make one pair of variable terms cancel, leaving an equation in just one variable."},{"text":"multiply","correct":false,"explanation":"Multiplying may be used beforehand to make coefficients equal or opposite, but the equations are combined by adding or subtracting. The missing word is subtract."},{"text":"divide","correct":false,"explanation":"Dividing can simplify an individual equation, but it is not the alternative to adding equations in the elimination method. The correct term is subtract."},{"text":"expand","correct":false,"explanation":"Expanding removes brackets within an expression rather than combining two equations to cancel a variable. In elimination, equations are added or subtracted."}]},{"id":"cab239b7-aafe-4465-945c-34851a588529","cardType":"basic","points":2,"front":"In the equations $4x + 3y = 26$ and $x - 3y = -1$, what is the result of adding them?","back":"$$5x = 25$","mcqOptions":[{"text":"$$5x = 25$","correct":true,"explanation":"Adding gives $4x+x=5x$, while $3y+(-3y)=0$ and $26+(-1)=25$. Therefore the resulting equation is $5x=25$."},{"text":"$$3x = 27$","correct":false,"explanation":"This incorrectly subtracts the $x$ terms and treats $-1$ as positive. Adding correctly gives $4x+x=5x$ and $26+(-1)=25$, so $5x=25$."},{"text":"$$5x = 27$","correct":false,"explanation":"The variable terms have been combined correctly, but the constants have not: $26+(-1)$ is $25$, not $27$. The correct result is $5x=25$."},{"text":"$$5x = 26$","correct":false,"explanation":"This cancels the $y$ terms but overlooks the $-1$ on the right-hand side. Since $26+(-1)=25$, the correct result is $5x=25$."}]},{"id":"d3c7ae9a-6a9c-4e05-9155-87da1745fedb","cardType":"basic","points":1,"front":"What does it mean to eliminate a variable?","back":"To remove it from the system so you are left with an equation in only one variable.","mcqOptions":[{"text":"To remove it from the system so you are left with an equation in only one variable.","correct":true,"explanation":"Elimination combines the equations so one variable’s terms cancel. This produces a simpler equation involving only the remaining variable, which can then be solved."},{"text":"To isolate it in the system so you are left with that variable on one side.","correct":false,"explanation":"Isolating a variable describes rearrangement or substitution, not elimination. To eliminate a variable means to remove it, leaving an equation in only one variable."},{"text":"To replace it in the system with a different variable from the other equation.","correct":false,"explanation":"Variables are not replaced by different variables during elimination. Instead, the equations are combined so one variable is removed, leaving an equation in the other variable."},{"text":"To divide it from the system so both equations are written in their simplest form.","correct":false,"explanation":"Elimination does not mean dividing a variable out of the equations. It means combining the equations to remove one variable and obtain a one-variable equation."}]},{"id":"1a2a588f-e469-4f2d-aa98-ef0988cac33a","cardType":"cloze","points":null,"front":"Rearranging $x + 4y = 12$ to match $3x - y = 10$ involves multiplying the first equation by [ ] to make the $x$ coefficients match.","back":"Rearranging $x + 4y = 12$ to match $3x - y = 10$ involves multiplying the first equation by 3 to make the $x$ coefficients match.","mcqOptions":[{"text":"3","correct":true,"explanation":"Multiplying every term in $x+4y=12$ by $3$ gives $3x+12y=36$, so its $x$ coefficient matches the $3x$ coefficient in the other equation."},{"text":"$$\\frac{1}{3}$","correct":false,"explanation":"Multiplying by $\\frac{1}{3}$ changes the first $x$ coefficient to $\\frac{1}{3}$ rather than $3$. The required multiplier is $3$."},{"text":"$$-3$","correct":false,"explanation":"Multiplying by $-3$ produces an $x$ coefficient of $-3$, which is opposite rather than matching. To make both $x$ coefficients $3$, multiply by $3$."},{"text":"4","correct":false,"explanation":"Multiplying by $4$ makes the first $x$ coefficient $4$, so it does not match the second equation’s coefficient. The correct multiplier is $3$."}]}]}],"examCountdown":null,"heading":"Simultaneous Equations","title":"Simultaneous Equations Flashcards","topbar":{"backHref":"/igcse/maths/cie/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"cc3e5f10-278f-482d-8131-37f903fc7c36","walled":"$L39"}],"$L3a","$L3b"]