You are turning a worded problem into a pair of simultaneous equations. What must you do before writing either equation down?

Choose a letter for each of the two unknowns and write down exactly what each one stands for, together with its units. Both equations and the final answer depend on those two definitions being fixed from the start.

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Simultaneous Equations (Cambridge (CIE) O Level Maths): Flashcards

Exam code: 4024

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FrontSolving Simultaneous Equations
Define linear simultaneous equations.

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Define linear simultaneous equations.
Linear simultaneous equations are two equations in the same two unknowns that are satisfied by the same pair of values at the same time.
Linear means the highest power of each unknown is 1, so terms such as $x^{2}$ or $y^{2}$ never appear.
Why do two unknowns need two different equations?
A single equation such as $3 x + 2 y = 11$ is satisfied by infinitely many pairs of values.
A second equation narrows those down to the one pair that works in both equations at once.
Before a variable can be eliminated from a pair of simultaneous equations, what must be true of its two coefficients?
The two coefficients must be equal in size, although their signs may be different.
Multiply every term of one or both equations by a suitable number to make that happen: multiplying $2 x - y = 5$ by 3 turns its $x$ term into $6 x$ .
Complete the rule for eliminating a variable from a pair of simultaneous equations:
When the signs in front of the matching terms are the same you $______$ one equation from the other, and when the signs are different you $______$ the equations together.
The completed rule is:
When the signs in front of the matching terms are the same you subtract one equation from the other, and when the signs are different you add the equations together.
When you subtract $6 x - 3 y = 15$ from $6 x + 4 y = 22$ , why is the $y$ term $7 y$ and not $y$ ?
Subtracting takes away the whole of the second equation, so the $y$ terms give $4 y - (- 3 y)$ .
Subtracting a negative is the same as adding, so this comes to $4 y + 3 y = 7 y$ .
You have eliminated $y$ from a pair of simultaneous equations and found $x = 3$ . What do you do next?
Substitute $x = 3$ into one of the original equations and solve that for $y$ .
Substituting the pair into the other original equation then checks them, because a correct solution must satisfy both equations.
How do you solve a pair of simultaneous equations by substitution?
Rearrange one equation to make an unknown the subject, for example $y = 2 x - 5$ .
Replace every $y$ in the other equation by $(2 x - 5)$ in brackets, solve the resulting equation for $x$ , then put that value back into $y = 2 x - 5$ .
True or False?
A pair of linear simultaneous equations can be solved by plotting both equations and finding where the lines cross.
True.
The $x$ and $y$ coordinates of the point of intersection are exactly the values that satisfy both equations.
For example $2 x - y = 3$ and $3 x + y = 7$ cross at $(2, 1)$ , so $x = 2$ and $y = 1$ .
You are turning a worded problem into a pair of simultaneous equations. What must you do before writing either equation down?
Choose a letter for each of the two unknowns and write down exactly what each one stands for, together with its units.
Both equations and the final answer depend on those two definitions being fixed from the start.
True or False?
Three apples and two bananas cost £1.80 in total. If $x$ is the price of an apple in pence and $y$ is the price of a banana in pence, then $3 x + 2 y = 1.80$ is the correct equation.
False.
Because $x$ and $y$ are measured in pence, the total on the right must be in pence as well, so the equation is $3 x + 2 y = 180$ instead.
Working in pounds all the way through would also be correct, giving $3 x + 2 y = 1.80$ with $x$ and $y$ both measured in pounds.
Taking $b$ as the price of one bagel and $s$ as the price of one sausage roll, both in pounds, complete the equation for six bagels and twelve sausage rolls costing £9 in total:
$6______+ 12______= 9$
The completed equation is:
$6 b + 12 s = 9$
Each number of items multiplies the price of one of that item, and the two costs add to give the total.
Why must the two equations you form from a worded problem be genuinely different?
Two equations that are multiples or rearrangements of each other carry exactly the same information, so they cannot pin down two unknowns.
For example $x + y = 19$ and $2 x + 2 y = 38$ are really one equation, and every pair of numbers adding to 19 still satisfies both of them.
Two numbers have a sum of 19 and a difference of 5. How do you find their product?
Form and solve the pair $x + y = 19$ and $x - y = 5$ simultaneously, which gives $x = 12$ and $y = 7$ .
The product is then $x y = 12 \times 7 = 84$ , so a question can require simultaneous equations without ever using the word.
In a worded problem $x$ is the price of an apple in pence and $y$ is the price of a banana in pence. You have found $x = 40$ and $y = 30$ , so why is that not yet a complete answer?
The values only mean something once they are put back into the context: an apple costs 40 pence and a banana costs 30 pence.
Until that is said, $x$ and $y$ are only labels and the two numbers carry no meaning.

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