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.

True or False? A region on a graph can only be bounded by straight lines.

False. The inequalities can be quadratic as well as linear, so a curve can form one of the boundaries.

A shaded region is given on a graph. What do you need before you can write down its inequalities?

The equation of each boundary , worked out from where each line or curve crosses the axes. Only once you have those equations can you decide which way each inequality points.

A LevelMathsCambridge (CIE)Pure 1FlashcardsAlgebra & FunctionsInequalities & Simultaneous Equations

Inequalities & Simultaneous Equations (Cambridge (CIE) A Level Maths: Pure 1): Flashcards

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FrontLinear Simultaneous Equations using Elimination
Define simultaneous equations.

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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 $y$ .
Define linear inequality.
A linear inequality is an inequality in which every term has degree $1$ or $0$ , so there are no squared or higher powers, and no fractional or negative powers.
Unlike an equation, it is satisfied by a range of values rather than a single one.
On a number line, what does a filled circle mean, and an open one?
A filled circle means the endpoint is included, so it goes with $\leq$ and $\geq$ .
An open circle means the endpoint is excluded, so it goes with $<$ and $>$ .
Complete the inequality that the interval $(4, 8]$ describes:
$4______x______8$
The completed inequality is:
$4 < x \leq 8$
The round bracket at $4$ gives a strict inequality, and the square bracket at $8$ allows equality.
Why is $\infty$ always written with a round bracket in interval notation?
Because a square bracket would mean the endpoint is included, and $\infty$ is not a value that $x$ can take.
So $x \leq 2$ is written $(- \infty, 2]$ , open at the infinite end and closed at $2$ .
True or False?
Multiplying both sides of an inequality by $- 1$ reverses the inequality sign.
True.
Multiplying or dividing by any negative number reverses the sign, so $- x > 3$ becomes $x < - 3$ .
Adding or subtracting a negative number does not reverse it, which is why rearranging is usually the safer route.
When solving $8 - 3 x \geq 5 x - 4$ , why collect the $x$ terms on the right rather than the left?
It keeps the coefficient of $x$ positive, so the inequality sign never has to be reversed.
Collecting on the left would give $- 8 x \geq - 12$ , and dividing by $- 8$ would then flip $\geq$ to $\leq$ .
How do you solve two inequalities such as $19 - 2 x \geq 5$ and $9 < x + 5$ together?
Solve each one separately, then find the values that satisfy both.
Here $x \leq 7$ and $x > 4$ , and a number line shows the overlap is $4 < x \leq 7$ .
Why is sketching the graph the key step in solving a quadratic inequality?
The roots alone do not say which values of $x$ satisfy the inequality, only where the curve meets the axis.
The sketch shows where the curve is above the axis and where it is below, which is what the inequality is asking about.
True or False?
You can multiply both sides of an inequality by $x$ , provided you reverse the inequality sign.
False.
You do not know whether $x$ is positive or negative, so you cannot know whether the sign should be reversed at all.
Multiplying by $x^{2}$ avoids that, since a square is never negative, but it can introduce extra values that do not satisfy the original inequality.
When solving $5 - 5 x^{2} \leq 7 + 4 x - 8 x^{2}$ , why rearrange it to $3 x^{2} - 4 x - 2 \leq 0$ ?
So that the $x^{2}$ term is positive, which makes the sketch a $\cup$ shape every time.
Working from one standard shape means the region above or below the axis can be read off without re-deriving it.
Complete the solution of the inequality $(x - 2) (x + 1) > 0$ :
$x <______$ or $x >______$
The completed solution is:
$x < - 1$ or $x > 2$
The brackets give roots of $- 1$ and $2$ , and the curve rises above the axis on the two outer sections.
A quadratic with a positive $x^{2}$ term has roots $x_{1} < x_{2}$ . Which values of $x$ make it negative, and which make it positive?
It is negative between the roots, $x_{1} < x < x_{2}$ , where the curve dips below the axis.
It is positive outside them, $x < x_{1}$ or $x > x_{2}$ , where the two arms rise above the axis.
Why is it worth dividing $2 x^{2} - 2 x - 4 > 0$ through by $2$ before solving it?
The numbers become smaller and factorise more easily, giving $(x - 2) (x + 1) > 0$ .
Dividing by a positive number leaves the inequality sign unchanged, so nothing is lost.
How can you recognise an inequality whose solution is a region on a graph?
It has two variables, $x$ and $y$ , rather than one.
The solution is then an area of the graph rather than a range on a number line.
When drawing inequalities on a graph, use a $______$ line for $<$ or $>$ , and a $______$ line for $\leq$ or $\geq$ .
When drawing inequalities on a graph, use a dotted line for $<$ or $>$ , and a solid line for $\leq$ or $\geq$ .
The solid line shows that the boundary itself is part of the region.
How do you decide which side of a line satisfies an inequality?
Test a point on each side, by substituting its coordinates into the inequality.
The origin is usually easiest: $(0, 0)$ satisfies $y < x^{2} + 1$ , so the side containing the origin is the one you want.
True or False?
A region on a graph can only be bounded by straight lines.
False.
The inequalities can be quadratic as well as linear, so a curve can form one of the boundaries.
When shading a region on a graph, which area do you shade, and how is the answer shown?
Shade the unwanted areas, then label the region left unshaded, usually $R$ .
Some questions ask instead for the wanted region to be shaded, so check the wording before you start.
A shaded region is given on a graph. What do you need before you can write down its inequalities?
The equation of each boundary, worked out from where each line or curve crosses the axes.
Only once you have those equations can you decide which way each inequality points.

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