Define simultaneous equations .

Simultaneous equations are two or more equations in more than one unknown, solved together. Solving them means finding the pairs of values that make every equation true at the same time.

What does it mean for an equation in two unknowns to be linear ?

None of the unknowns is raised to any power other than one. Its graph is a straight line , which is where the name comes from.

What does the solution of a pair of linear simultaneous equations look like on a graph?

It is the point where the two lines cross . The coordinates of that point are the pair of values you are solving for.

True or False? A pair of linear simultaneous equations always has a solution.

False. Two parallel lines never meet, so a pair like that has no solution at all.

Once you have found both unknowns, how should you check the answer to a pair of simultaneous equations?

Substitute both values into the equation you did not use to find the second unknown. Using the same equation again would only confirm your arithmetic rather than the solution.

True or False? Elimination will solve any pair of linear simultaneous equations.

True. Elimination always works on a linear pair. Substitution is often quicker, particularly when one equation already has an unknown as its subject, so which you use is a matter of choice.

Which method must you use for quadratic simultaneous equations?

Substitution. Elimination will not work on them. Rearrange the linear equation, then substitute it into the quadratic one.

How many solution pairs does one linear and one quadratic equation usually have?

Usually two. There can be one, or none at all, so the number is not guaranteed.

A LevelMathsEdexcelPureFlashcardsAlgebra & FunctionsSimultaneous Equations

Simultaneous Equations (Edexcel A Level Maths: Pure): Flashcards

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FrontLinear Simultaneous Equations using Elimination
Define simultaneous equations.

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Cards in this collection (19)

Define simultaneous equations.
Simultaneous equations are two or more equations in more than one unknown, solved together.
Solving them means finding the pairs of values that make every equation true at the same time.
What does it mean for an equation in two unknowns to be linear?
None of the unknowns is raised to any power other than one.
Its graph is a straight line, which is where the name comes from.
What does the solution of a pair of linear simultaneous equations look like on a graph?
It is the point where the two lines cross.
The coordinates of that point are the pair of values you are solving for.
True or False?
A pair of linear simultaneous equations always has a solution.
False.
Two parallel lines never meet, so a pair like that has no solution at all.
What must you do before you can eliminate an unknown?
Make the coefficients of that unknown match, by multiplying one or both equations by a constant.
For $- 2 x + 4 y = 5$ and $4 x - 5 y = - 7$ , doubling the first gives $- 4 x + 8 y = 10$ , which matches the $4 x$ .
In elimination, if the matching coefficients have the same sign you $______$ the equations, and if they have different signs you $______$ them.
In elimination, if the matching coefficients have the same sign you subtract the equations, and if they have different signs you add them.
Getting this the wrong way round leaves both unknowns still in the equation.
Once you have found both unknowns, how should you check the answer to a pair of simultaneous equations?
Substitute both values into the equation you did not use to find the second unknown.
Using the same equation again would only confirm your arithmetic rather than the solution.
What is the first move when solving simultaneous equations by substitution?
Rearrange one equation to make one of the unknowns the subject.
If an equation is already in that form, such as $y = 1 - 9 x$ , you can use it as it stands.
Complete the substitution of $y = 1 - 9 x$ into $3 x + 5 y = - 9$ :
$3 x + 5 (______) = - 9$
$3 x + 5 (1 - 9 x) = - 9$
The whole expression replaces $y$ , and it needs brackets so that the $5$ multiplies all of it.
Why must the rearranged expression be substituted into the other equation, rather than the one it came from?
Because putting it back into its own equation gives a statement that is always true, such as $0 = 0$ , which tells you nothing.
Only the second equation brings in new information about the unknowns.
True or False?
Elimination will solve any pair of linear simultaneous equations.
True.
Elimination always works on a linear pair.
Substitution is often quicker, particularly when one equation already has an unknown as its subject, so which you use is a matter of choice.
Substitution has given you the value of one unknown. How do you find the other?
Substitute it into the rearranged equation from the first step, which already has an unknown as its subject.
With $y = 1 - 9 x$ and $x = \frac{1}{3}$ , this gives $y = 1 - 3 = - 2$ .
What makes an equation quadratic?
It contains terms of degree two and no higher, with no unknowns raised to negative or fractional powers.
So $y = 5 x^{2} - 2 x + 3$ is quadratic, but $y = \sqrt{x} - 5$ is not, because $\sqrt{x}$ is a fractional power.
True or False?
$y^{2} + 4 x y - x^{2} = - 7$ is a quadratic equation.
True.
The term $4 x y$ has degree two, because the powers of $x$ and $y$ add to two.
Nothing in the equation goes above degree two, so it is quadratic.
Which method must you use for quadratic simultaneous equations?
Substitution. Elimination will not work on them.
Rearrange the linear equation, then substitute it into the quadratic one.
How many solution pairs does one linear and one quadratic equation usually have?
Usually two.
There can be one, or none at all, so the number is not guaranteed.
For $2 x - 3 y = 23$ and $3 y^{2} = 4 x^{2} + 11$ , why is it easier to rearrange to $2 x = 3 y + 23$ than to $x = \frac{3 y + 23}{2}$ ?
Because the quadratic contains $4 x^{2}$ , which is ${(2 x)}^{2}$ .
Stopping at $2 x$ lets you substitute straight into that square, with no fraction to square out.
Complete the substitution of $2 x = 3 y + 23$ into $3 y^{2} = 4 x^{2} + 11$ :
$3 y^{2} = {(______)}^{2} + 11$
$3 y^{2} = {(3 y + 23)}^{2} + 11$
Expanding gives $3 y^{2} = 9 y^{2} + 138 y + 540$ , which tidies to $y^{2} + 23 y + 90 = 0$ .
You have found two values of $y$ . How do you complete the solution?
Substitute each one into the rearranged linear equation to get its matching value of $x$ .
Then state the answers as pairs, making clear which $x$ goes with which $y$ .

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