How do you find the region satisfying three inequalities at once?

Draw all three boundary lines, mark the correct side of each, then take the area that is on the correct side of every one. Shading the unwanted sides and leaving the region clear works just as well, and keeps a region bounded by several lines readable.

When integer solutions are wanted, what if the optimal corner is not a whole-number point?

Take the point with whole-number coordinates nearest to that corner, and check it still lies inside the region. The best whole-number point need not sit at a corner at all, so the corner only tells you where to start looking.

IGCSEFurther MathsEdexcelFlashcardsEquations, Identities & InequalitiesInequalities

Inequalities (Edexcel IGCSE Further Pure Maths): Flashcards

Exam code: 4PM1

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FrontSolving Inequalities
Define linear inequality.

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Cards in this collection (16)

Define linear inequality.
A linear inequality contains only constant terms and terms in $x$ , and possibly $y$ , with no $x^{2}$ or any higher power.
So $3 x + 1 > 12$ is linear, while $3 x^{2} > 12$ is a quadratic inequality instead.
What must you do when you multiply or divide an inequality by a negative number?
Flip the inequality sign, so $<$ becomes $>$ and $\leq$ becomes $\geq$ .
Starting from $1 < 2$ and multiplying both sides by $- 1$ gives $- 1 > - 2$ , which shows why the reversal is needed.
Why must you never multiply or divide an inequality by $x$ ?
Because $x$ could be positive or negative, so you would have no way of knowing whether to flip the sign.
Rearrange by adding and subtracting instead, moving every term to one side, which is always safe.
How do you solve $- 7 \leq 3 x - 1 < 2$ ?
Do the same thing to all three parts at once: adding $1$ throughout gives $- 6 \leq 3 x < 3$ .
Dividing all three parts by $3$ , which is positive so nothing flips, gives $- 2 \leq x < 1$ .
How is $- 2 \leq x < 1$ written in set notation as two sets?
As an intersection, since both conditions have to hold at once: $\{x : x \geq - 2\} \cap \{x : x < 1\}$ .
The colon is read as 'such that', so the first set is ' $x$ such that $x$ is at least $- 2$ '.
Complete the set notation for $x$ less than $3$ or greater than or equal to $5$ :
$\{x : x < 3\}______\{x : x \geq 5\}$
The completed statement is:
$\{x : x < 3\} \cup \{x : x \geq 5\}$
The union symbol is the right one because the two pieces are separate and $x$ lies in one or the other, never in both.
What is the first thing to do when solving a quadratic inequality?
Rearrange it so that everything is on one side and the $x^{2}$ term is positive.
Then find the roots of the matching equation and sketch the U-shaped curve, which is what shows you which region to take.
For a U-shaped quadratic with roots $x_{1} < x_{2}$ , when is it positive and when negative?
It is positive outside the roots and negative between them.
So $> 0$ gives $x < x_{1}$ or $x > x_{2}$ , in two separate pieces, while $< 0$ gives the single interval $x_{1} < x < x_{2}$ .
True or False?
Multiplying an inequality through by $x^{2}$ is allowed.
True.
A square is never negative, so multiplying by $x^{2}$ cannot reverse the inequality the way multiplying by $x$ might.
The catch is that $x^{2}$ can be zero, which is how the step can introduce extra solutions that do not satisfy the original, so check whatever you get.
True or False?
A point lying on a dashed boundary line satisfies the inequality that line represents.
False.
A dashed line marks a strict inequality, $<$ or $>$ , so the points on the line itself are excluded from the region.
A solid line, used for $\leq$ or $\geq$ , does include its own points.
Complete the rule for which side of a line each inequality describes:
The region for $y > m x + c$ lies $______$ the line, and the region for $x < a$ lies to the $______$ of it.
The completed rule is:
The region for $y > m x + c$ lies above the line, and the region for $x < a$ lies to the left of it.
In each case the sign points the way: greater means above or to the right, less means below or to the left.
How do you decide which side of a line satisfies an inequality?
Take any point not on the line and substitute its coordinates into the inequality.
If it holds, that side is the one you want: testing $(0, 0)$ in $3 x + 2 y \geq 12$ gives the false statement $0 \geq 12$ , so the region lies the other way.
How do you find the region satisfying three inequalities at once?
Draw all three boundary lines, mark the correct side of each, then take the area that is on the correct side of every one.
Shading the unwanted sides and leaving the region clear works just as well, and keeps a region bounded by several lines readable.
How do you write down the inequalities that define a shaded region?
Find the equation of each boundary line first, then replace each $=$ with an inequality sign.
The direction of each sign depends on whether the shaded region lies above or below that particular line, and the line style tells you whether it is strict.
Where in a region does a function such as $P = 20 x - 3 y$ take its greatest value?
At one of the corners, where two boundary lines intersect.
Substitute each corner's coordinates into $P$ and pick the largest: for corners $(0, 0)$ , $(1, 6)$ and $(3.5, 3.5)$ the values are $0$ , $2$ and $59.5$ .
When integer solutions are wanted, what if the optimal corner is not a whole-number point?
Take the point with whole-number coordinates nearest to that corner, and check it still lies inside the region.
The best whole-number point need not sit at a corner at all, so the corner only tells you where to start looking.

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