How is solving a linear inequality like solving a linear equation?

You use exactly the same inverse operations, but keep the inequality sign all the way through. Replacing the sign with an equals sign changes the meaning of the problem, so never do it.

GCSEFurther MathsAQAFlashcards2. Algebra & FunctionsSolving Inequalities

Solving Inequalities (AQA GCSE Further Maths): Flashcards

Exam code: 8365

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How is solving a linear inequality like solving a linear equation?

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How is solving a linear inequality like solving a linear equation?
You use exactly the same inverse operations, but keep the inequality sign all the way through.
Replacing the sign with an equals sign changes the meaning of the problem, so never do it.
True or False?
Dividing both sides of an inequality by $- 2$ leaves the sign unchanged.
False.
Multiplying or dividing by a negative number flips the inequality sign round.
For instance $1 < 2$ becomes $- 1 > - 2$ once both sides are multiplied by $- 1$ .
Why must you never divide an inequality by $x$ ?
Because $x$ could be positive or negative, so you would not know whether to flip the sign.
The safest route is to add and subtract until all the terms are on one side.
Solve $5 - 2 x \leq 21$ .
Subtracting 5 from both sides gives $- 2 x \leq 16$ .
Dividing by $- 2$ flips the sign, giving $x \geq - 8$ .
Complete the number-line convention:
Use an $______$ circle for $<$ or $>$ , and a $______$ circle for $\leq$ or $\geq$ .
The completed convention is:
Use an open circle for $<$ or $>$ , and a solid circle for $\leq$ or $\geq$ .
The solid circle shows the end value is included; the open circle shows it is not.
How do you solve $- 7 \leq 3 x - 1 < 2$ ?
Do the same thing to all three parts each time, so adding 1 gives $- 6 \leq 3 x < 3$ .
Dividing all three parts by 3 then gives $- 2 \leq x < 1$ .
How do you write $- 2 \leq x < 1$ in set notation?
Write it as two sets joined by the intersection symbol: $\{x : x \geq - 2\} \cap \{x : x < 1\}$ .
Each end value needs its own set, with $\cap$ used for "and" and $\cup$ used for "or".
What form does the solution to a quadratic inequality take?
A range of $x$ values, such as $1 \leq x \leq 4$ , rather than single values.
Sometimes it is two separate ranges, joined by the word "or".
How do you solve a quadratic inequality?
Factorise, sketch the curve showing its $x$ -intercepts, then shade the parts of it on the required side of the $x$ -axis.
The solution is the range of $x$ lying underneath the shaded parts.
Complete the shading rule for a positive $x^{2}$ term:
Shade the curve above the $x$ -axis when the inequality is $______0$ , and below when it is $______0$ .
The completed rule is:
Shade above the $x$ -axis when the inequality is $> 0$ , and below when it is $< 0$ .
The same applies with $\geq$ and $\leq$ ; only the strictness of the final answer changes.
Solve $2 x^{2} - 2 x + 5 > 9$ .
Rearranging and dividing by 2 gives $x^{2} - x - 2 > 0$ , which factorises to $(x + 1) (x - 2) > 0$ .
The curve is above the axis outside its intercepts, so the answer is $x < - 1$ or $x > 2$ .
True or False?
The solution to $(x - 2) (x - 5) \leq 0$ is $x \leq 2$ or $x \geq 5$ .
False.
Those two separate ranges are where the curve sits above the axis, which answers the $\geq 0$ version instead.
For $\leq 0$ the answer is the single range between the intercepts, $2 \leq x \leq 5$ .
Solve $4 - x^{2} > 0$ .
The intercepts come from $4 - x^{2} = 0$ , giving $x = \pm 2$ , and the curve is an upside-down U shape.
That shape sits above the axis between its intercepts, so the answer is $- 2 < x < 2$ .
Why rearrange $x^{2} - x < 6$ before solving it?
Because the shading rule needs the inequality compared with zero, so it becomes $x^{2} - x - 6 < 0$ .
Choose the side that leaves a positive $x^{2}$ term, which keeps the curve a u-shape.

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