What do expanding brackets and multiplying by a common denominator both achieve?

Both turn the problem into an ordinary linear equation with no brackets and no fractions left in it. From that point it is solved in the usual way, by undoing the operations attached to the unknown.

True or False? You can add or subtract terms to both sides of a linear inequality in exactly the same way as you do to a linear equation .

True. You can add or subtract terms to both sides of a linear inequality in exactly the same way as you do to a linear equation .

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Linear Equations & Inequalities (Cambridge (CIE) O Level Maths): Flashcards

Exam code: 4024

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FrontSolving Linear Equations
Define a linear equation in one unknown.

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Cards in this collection (26)

Define a linear equation in one unknown.
A linear equation in one unknown is one that can be written in the form $a x + b = c$ , where $a$ , $b$ and $c$ are numbers.
The highest power of $x$ is 1, so a linear equation never contains a term such as $x^{2}$ .
Complete the sentence about undoing operations when solving an equation:
Addition is undone by $______$ and multiplication is undone by $______$ since these pairs of operations are inverses of one another.
The completed sentence is:
Addition is undone by subtraction and multiplication is undone by division since these pairs of operations are inverses of one another.
A student solves $4 x + 8 = 12$ by dividing by 4 and writes $x + 8 = 3$ . What has gone wrong?
Dividing a side by 4 means dividing every term on that side, so the 8 has to be divided as well.
The line should read $x + 2 = 3$ .
True or False?
There is only one correct order in which to undo the operations when solving a linear equation.
False.
Solving $2 x + 1 = 9$ by subtracting 1 first and by dividing by 2 first both lead to $x = 4$ .
Subtracting first is usually easier because it keeps the numbers whole, but neither order is wrong.
How do you begin solving $2 - 3 x = 10$ ?
Add $3 x$ to both sides so that the term in $x$ becomes positive, giving $2 = 10 + 3 x$ .
The same move can be seen by reordering the left-hand side as $- 3 x + 2$ , since $2 - 3 x$ and $- 3 x + 2$ are the same expression.
True or False?
The solution of a linear equation is always a whole number.
False.
Solving $3 x + 10 = 2$ gives $3 x = - 8$ and so $x = - \frac{8}{3}$ .
A solution can be negative, fractional or both, and an answer that is not a whole number is no reason to suspect a mistake.
What does it mean to say that $x = 6$ is the solution of $5 x - 8 = 22$ ?
It means that replacing $x$ by 6 makes the two sides equal, since $5 \times 6 - 8 = 22$ .
That is why putting an answer back into the original equation tests whether it is right.
What is the first thing to do when solving $2 (x - 3) = 10$ ?
Expand the bracket to get $2 x - 6 = 10$ , which is then an ordinary linear equation.
Dividing both sides by 2 first also works and gives $x - 3 = 5$ , but expanding is the route that always works whatever the numbers are.
Complete the rule for solving an equation that contains fractions:
Multiply $______$ term on both sides by the lowest common $______$ of the fractions, so that the denominators cancel.
The completed rule is:
Multiply every term on both sides by the lowest common denominator of the fractions, so that the denominators cancel.
You solve $\frac{x}{5} + 4 = \frac{9}{2}$ by multiplying by 10. Which term is most often forgotten, and what does the equation become?
The 4 is the term most often forgotten, because it is not written as a fraction and looks as though the multiplication does not apply to it.
With the 4 included the equation becomes $2 x + 40 = 45$ .
How do you start solving $\frac{4}{x - 2} = 3$ ?
Multiply both sides by the denominator $(x - 2)$ , which cancels the fraction and leaves $4 = 3 (x - 2)$ .
Expanding the bracket then gives the ordinary linear equation $4 = 3 x - 6$ .
True or False?
An answer of $x = \frac{5}{2}$ may be left exactly as it is, rather than being turned into a decimal or a mixed number.
True.
An improper fraction is an acceptable final form unless the question asks for something else, and $2.5$ or $2 \frac{1}{2}$ would be accepted just as readily.
What do expanding brackets and multiplying by a common denominator both achieve?
Both turn the problem into an ordinary linear equation with no brackets and no fractions left in it.
From that point it is solved in the usual way, by undoing the operations attached to the unknown.
What is the first step in solving $4 x - 7 = 11 + x$ ?
Collect the terms in $x$ onto one side by subtracting $x$ from both sides, which gives $3 x - 7 = 11$ .
The equation then has the unknown in one place only and is solved in the usual way.
In an equation with terms in $x$ on both sides, which of the two should you remove first, and why?
Remove the smaller of the two, so that the term left behind is positive.
In $4 - 5 x = 6 x - 29$ that means adding $5 x$ to both sides, which leaves $11 x$ on the right and no negative term in $x$ to deal with.
Complete the line of working that collects the terms in $x$ in $7 x + 2 = 3 x + 18$ :
$______x + 2 = 18$
The completed line is:
$4 x + 2 = 18$
On the left $7 x - 3 x$ leaves $4 x$ , and on the right the $3 x$ has gone.
True or False?
An equation with the unknown on both sides can be solved by collecting the unknowns on either side, left or right.
True.
Both choices lead to the same solution, so $4 x - 7 = 11 + x$ gives $x = 6$ whichever side the terms in $x$ are collected on.
You are solving $4 - 5 x = 6 x - 29$ and the unknown ends up on the right. Should you swap the two sides over?
Swapping the sides part-way through is unnecessary and easy to get wrong, so leave the terms where they are and carry on to $33 = 11 x$ and then $3 = x$ .
Swap the sides only at the very end, so that the answer is presented as $x = 3$ .
Define the word inequality in algebra.
An inequality compares a left-hand side to a right-hand side and states which one is bigger, using the symbols $<, >, \leq, \geq$ .
Explain the meaning of the word linear in linear inequality.
The word linear in linear inequality means that the terms in the inequality are either constant numbers or terms in $x$ , but not terms in $x^{2}$ or $x^{3}$ etc.
These are examples of linear inequalities:
- $x + 2 > 5$
- $2 x < x - 1$
True or False?
You can add or subtract terms to both sides of a linear inequality in exactly the same way as you do to a linear equation.
True.
You can add or subtract terms to both sides of a linear inequality in exactly the same way as you do to a linear equation.
True or False?
You can multiply or divide both sides of a linear inequality in exactly the same way as you do to a linear equation.
False.
You can multiply or divide both sides of a linear inequality in exactly the same way as you do to a linear equation as long as you multiply or divide by positive numbers.
If, however, you multiply or divide both sides by negative numbers, you have to flip the direction of the inequality sign.
E.g. You can divide $\begin{array}{rcl} 2 x & < & 4 \end{array}$ by 2 to get $\begin{array}{rcl} x & < & 2 \end{array}$ .
You can divide $\begin{array}{rcl} - 2 x & < & 4 \end{array}$ by -2, but you must flip the inequality to get $\begin{array}{rcl} x & > & - 2 \end{array}$ .
How do number lines highlight the difference between strict inequalities (such as $x < 3$ ) and non-strict inequalities (such as $x \leq 3$ )?
Number lines show an open circle for strict inequalities, e.g. $x < 3$ , and a closed circle for non-strict inequalities, e.g. $x \leq 3$ .
True or False?
The number line representing " $x < 1$ or $x > 3$ " consists of two separate arrows pointing outwards in opposite directions.
True.
The number line representing " $x < 1$ or $x > 3$ " consists of two separate arrows pointing outwards in opposite directions.
True or False?
The diagram below shows the inequality $- 4 < x \leq 2$ .
False.
The number line in the diagram does not show the inequality $- 4 < x \leq 2$ .
It has two open circles which indicate the inequality $- 4 < x < 2$ .
Explain how you would solve an inequality in the form $- 10 < a x + b < 10$ .
To solve an inequality in the form $- 10 < a x + b < 10$ ,
1. Subtract $b$ from all three parts.
2. Then divide all three parts by $a$ .
E.g. $- 10 < 2 x + 6 < 10 - 16 < 2 x < 16 - 8 < x < 8$
An alternative method is to split into two different inequalities, $- 10 < 2 x + 6$ and $2 x + 6 < 10$ , then solve these individually.

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