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Solving Inequalities (Edexcel GCSE Maths: Foundation): Flashcards

Exam code: 1MA1

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FrontInterpreting Inequalities
What is a strict inequality?

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What is a strict inequality?
A strict inequality is one that does not allow the two sides to be equal, so the symbols are $<$ and $>$ .
For example $x > 5$ does not include $5$ itself, whereas $x \geq 5$ does.
Complete the list of integers that satisfy $3 < x \leq 6$ :
$x =______, 5,______$
The completed list is:
$x = 4, 5, 6$
$3$ is left out because the inequality is strict at that end, while $6$ is included because it is not.
True or False?
$x > 3$ means that the smallest possible value of $x$ is $4$ .
False.
Unless the question says that $x$ is an integer, $x$ can take any value greater than $3$ , such as $3.1$ or $π$ .
There is no smallest such value at all.
What is the smallest integer that satisfies $x > 6.5$ ?
The smallest integer is $7$ .
$6.5$ is not itself an integer, and the next whole number above it is $7$ .
List all the integers that satisfy $- 4 \leq x < 2$ .
$x = - 4, - 3, - 2, - 1, 0, 1$ .
Remember that zero and negative whole numbers are integers too, so the list does not begin at $1$ .
True or False?
There are infinitely many integers that satisfy $x > 2$ .
True.
Only one end is fixed, so the integers $3, 4, 5, 6, \dots$ carry on without limit.
An inequality needs two end points before the list of integers is finite.
How do you find the integers that satisfy both $0 < x < 5$ and $x \geq 3$ ?
List the integers for each inequality separately, then pick out the values that appear in both lists.
Here that gives $1, 2, 3, 4$ and $3, 4, 5, 6, \dots$ , so the answer is $x = 3$ and $x = 4$ .
Complete the rule for drawing an inequality on a number line:
For $<$ or $>$ , use an $______$ circle.
For $\leq$ or $\geq$ , use a $______$ circle.
The completed rule is:
For $<$ or $>$ , use an open circle.
For $\leq$ or $\geq$ , use a closed circle.
An open circle shows that the end point is left out, and a closed circle shows that it is included.
How do you show $- 2 \leq x < 1$ on a number line?
Draw a closed circle at $- 2$ and an open circle at $1$ , then join them with a horizontal line.
The line between the circles shows that every value in between satisfies the inequality.
How do you show $x > 5$ on a number line?
Draw an open circle at $5$ and a horizontal arrow pointing to the right.
There is no second circle, because the inequality has only one end point and the values carry on without limit.
For $t < 3$ , which way does the arrow on the number line point, and why?
The arrow points to the left.
Every value less than $3$ satisfies the inequality, and those values lie to the left of $3$ on the line.
True or False?
On a number line, the left-hand end point of an inequality always has a closed circle.
False.
Each end point is decided only by the symbol at that end, not by which side it is on.
So $- 3 < x \leq 4$ is open on the left and closed on the right, while $- 2 \leq x < 1$ is the other way round.
True or False?
Solving $2 x - 5 \leq 21$ uses the same steps as solving $2 x - 5 = 21$ .
True.
The same steps work, but the inequality sign is kept in place of the equals sign all the way through.
Changing it to an equals sign would change the meaning of the problem.
When must you flip the direction of an inequality sign?
Whenever you multiply or divide both sides by a negative number.
For example $1 < 2$ is true, but multiplying both sides by $- 1$ gives $- 1 > - 2$ , with the sign reversed.
Complete the working for $2 x - 5 \leq 21$ :
$2 x \leq______$ and so $x \leq______$
The completed working is:
$2 x \leq 26$ and so $x \leq 13$
Both steps divide or add a positive number, so the sign stays as it is throughout.
Solve $5 - 2 x \leq 21$ .
$x \geq - 8$ .
Subtracting $5$ gives $- 2 x \leq 16$ , and dividing by $- 2$ reverses the sign to give $x \geq - 8$ .
How do you solve a double inequality such as $- 7 \leq 3 x - 1 < 2$ ?
Do the same thing to all three parts at once.
Adding $1$ throughout gives $- 6 \leq 3 x < 3$ , and dividing throughout by $3$ gives $- 2 \leq x < 1$ .
True or False?
When solving an inequality, you can divide both sides by $x$ just as you would divide by a number.
False.
A letter can stand for a positive or a negative value, and you would not know whether the sign had to be reversed.
Rearrange by adding and subtracting instead, so that every term ends up on one side.

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