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Mathematics 8.

Table of contents
Algebra
Algebraic expressions (revision)
How to solve equations and inequalities? (revision)
Multiplication of algebraic expressions formed by multiple terms
The square of a sum or a difference (supplementary material)
The product of a sum and a difference (supplementary material)
Factorisation
Algebraic fractions (supplementary material)
Solving equations by factorisation
Mixed exercises
Word problems
Sets, combinatorics
Geometry I.
Three-dimensional geometry
Statistics, probability
Geometry II.
Functions, sequences
How to solve equations and inequalities (revision)
An equation expresses the equality between two algebraic expressions, together with a defined domain.
The solution of the equation is the element of the domain for which the equality holds true.
If no domain is given, we consider the largest possible set in which the expressions are defined as the domain.
Example 1
Solve the following equations over the set of rational numbers.
a) \latex{ 8-4a+10 = 3a+2+a }                  b) \latex{ \frac{y}{2}+\frac{4}{5}=1.2 }
Solution
a)  \latex{ 8-4a+10 = 3a+2+a }   combine the like terms 
\latex{ 18-4a = 4a+2 \;\;\;\;\;\;\;\;\;\;\;\;\;/ -4a }
\latex{ 18-8a = 2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;/ -18 }
\latex{ -8a = -16 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;/ \div (-8) }
\latex{a = 2}                        rational number
Check: 
left side: \latex{ 8 - 4 \times 2 + 10 = 8 - 8 + 10 = 10 };
right side: \latex{ 3 \times 2 + 2 + 2 = 6 + 2 + 2 = 10 \\[10pt]}.
Answer: The solution is \latex{\underline{\underline{ a = 2}} }.

b)  \latex{ \frac{y}{2}+\frac{4}{5} =1.2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; / -\frac{4}{5} }

  \latex{ \frac{y}{2} =1.2-0.8 }

  \latex{ \frac{y}{2} =0.4 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; / \times 2 }

  \latex{ \frac{y}{2} =0.8 }                        rational number

 

Check:  \latex{ \frac{0.8}{2}+\frac{4}{5}=0.4+0.8=1.2\\[10pt] }.
Answer: The solution is \latex{\underline{\underline{ y=0.8}} }.

 

 

If you add the same number or the same algebraic term to both sides of an equation, the sides remain equal.

If you subtract the same number or the same algebraic term from both sides of an equation, the sides remain equal.

 

 

If you multiply both sides of an equation by the same non-zero number or expression, the sides remain equal.

If you divide both sides of an equation by the same non-zero number or expression, the sides remain equal.

 

\latex{\frac{4}{5}=0.8}

The equations were solved using the balance method, which means performing the same operations on both sides of the equation.
Solving and checking an equation:
no

Determine the value

of the variable

Is the solution an

element of the domain?

Check
Answer
The equation has no solution
yes
Example 2
Write in each coloured square the sum of the two numbers connected to it diagonally in the row above. What number does \latex{ x } represent?
\latex{ 59 }
\latex{ 4 }
\latex{ 5 }
\latex{ x }
\latex{ 7 }
\latex{ 9 }
Solution
\latex{\fcolorbox{black}{#fabfba}{\textcolor{#fabfba}{O}}} \latex{ = 5+x }
\latex{\fcolorbox{black}{#c7ddf3}{\textcolor{#c7ddf3}{O}}} \latex{ = x+7 }
\latex{\fcolorbox{black}{#fdd1b0}{\textcolor{#fdd1b0}{O}}} \latex{ = 9+ \fcolorbox{black}{#fabfba}{\textcolor{#fabfba}{O}} = 9 + (5+x) = 9+5+x = 14+x}
\latex{\fcolorbox{black}{#c0e2ca}{\textcolor{#c0e2ca}{O}}} \latex{ = \fcolorbox{black}{#fabfba}{\textcolor{#fabfba}{O}} + \fcolorbox{black}{#c7ddf3}{\textcolor{#c7ddf3}{O}} = (5+x) + (x+7) = 5+x + x+7 = 2x + 12 }
\latex{\fcolorbox{black}{#dacae3}{\textcolor{#dacae3}{O}}} \latex{ = \fcolorbox{black}{#fdd1b0}{\textcolor{#fdd1b0}{O}} + \fcolorbox{black}{#c0e2ca}{\textcolor{#c0e2ca}{O}} = (14 + x) + (2x+12) = 14+x+2x+12 = 3x + 26. }
The value of the purple square is \latex{ 59 }, thus                                            Check:
\latex{ 3x + 26 = 59\;\;\;\;\;\;\;\;\;\;\;\;\; / -26 }
          \latex{ 3x = 33 \;\;\;\;\;\;\;\;\;\;\;\;/ \div 3 }
            \latex{\underline{\underline{ x=11}} }
Answer: The value of \latex{ x } is \latex{ 11 }.
\latex{ 59 }
\latex{ 4 }
\latex{ 5 }
\latex{ x }
\latex{ 7 }
\latex{ 25 }
\latex{ 9 }
\latex{ 16 }
\latex{ 18 }
\latex{ 34 }
Example 3
Three numbers were written on a piece of paper. Aaron subtracted the second number from the first, then subtracted the third number from that result. Ben subtracted the third number from the second, then subtracted that result from the first number. The number Ben got was \latex{ 12 } more than the number Aaron got. What was the third number on the paper?
Ben's result is \latex{ 12 } more than Aaron's.
\latex{ \Updownarrow }
If you add \latex{ 12 } to Aaron's number, you get Ben's result.
\latex{ \Updownarrow }
Solution
Let the numbers be denoted by letters: the first number by \latex{\large a}, the second by \latex{\large b}, and the third by \latex{\large c}. Aaron performed the operation \latex{ (a-b) -c }, while Ben performed the operation \latex{ a - (b-c) }.  
Ben's result is \latex{ 12 } more than Aaron's, so if you add \latex{ 12 } to Aaron's result, the two expressions will be equal.

The difference between Ben's result and Aaron's result is \latex{ 12 }.

\latex{ (a - b) - c + 12 = a - (b - c) }
\latex{ a - b - c + 12 = a - b + c }
\latex{ - b - c + 12 = - b + c }
  \latex{ - c + 12 =c }
        \latex{ 12 = 2c }
        \latex{\underline{\underline{ 6=c }}}

Expand the brackets.

\latex{ / - a }

\latex{ / + b }

\latex{ / + c }

\latex{ / \div 2 }

\latex{(a-b)=a-b}

\latex{-(b-c)=-b+c}

 
Check:       Aaron: \latex{ (a - b) - 6 = a - b - 6 }
  Ben: \latex{ a - (b - 6) = a - b + 6 }
You can check your answer to a word problem by substituting the solution back into the text or the problem.
Ben's number is indeed \latex{ 12 } more than Aaron's, since
\latex{ a - b + 6 - (a - b - 6) = a - b + 6 - a + b + 6 = 12. }
 
Answer: The third number is \latex{ 6 }.
Try the exercise with different numbers.
Before using the balance method, it is recommended to expand the brackets and combine the like terms on both sides of the equation.
Example 4
The sum of one-half, one-third, and one-quarter of a number is one-tenth. What is the number?
Solution
Let the number be denoted by \latex{ x }. According to the text, the equation is the following: 
 
\latex{ \frac{x}{2} + \frac{x}{3} + \frac{x}{4} = \frac{1}{10} }
 
Combine the like terms by expressing the fractions with a common denominator.
 
\latex{ \frac{6x+4x+3x}{12} = \frac{1}{10} }
 
\latex{ \frac{13x}{12} = \frac{1}{10} }
 
\latex{ 13x = \frac{\overset{6}{\cancel{12}}}{\underset{5}{\cancel{10}} } }
 
\latex{ x = \frac{6}{5 \times 13} }
 
\latex{\underline{\underline{x = \frac{6}{65}}}}
 
Check:
One-half of \latex{\frac{6}{65}}  is   \latex{ \frac{1}{\underset{1}{\cancel{2}} } \times \frac{\overset{3}{\cancel{6}}}{65} = \frac{3}{65} }
its one-third is   \latex{ \frac{1}{\underset{1}{\cancel{3}} } \times \frac{\overset{2}{\cancel{6}}}{65} = \frac{2}{65} }
its one-fourth is   \latex{ \frac{1}{\underset{2}{\cancel{4}} } \times \frac{\overset{3}{\cancel{6}}}{65} = \frac{3}{130} }
 
\latex{ \frac{3}{65} + \frac{2}{65} + \frac{3}{130} = \frac{6+4+3}{130} = \frac{13}{130} = \frac{1}{10}}

Answer: The number is \latex{\frac{6}{65}}.

One-half of \latex{ x } is:

\latex{ \frac{1} {2}\times x=\frac{x}{2} }

 

one-third of \latex{ x } is:

\latex{ \frac{1} {3}\times x=\frac{x}{3} }

 

one-quarter of \latex{ x } is:

\latex{ \frac{1} {4}\times x=\frac{x}{4} }

 

Example 5
Solve the following equations over the set of integers.
 
a) \latex{ \frac{x}{3} - \frac{3}{4} = \frac{1}{4} + \frac{x}{2}}                 b) \latex{ \frac{x}{2} - \frac{x-3}{2} = 3}
Solution
a)
    \latex{ \frac{x}{3} - \frac{3}{4} = \frac{1}{4} + \frac{x}{2} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; /- \frac{x}{3}}

         \latex{ - \frac{3}{4} = \frac{1}{4}+ \frac{x}{2} - \frac{x}{3}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; /- \frac{1}{4}} 

\latex{ - \frac{3}{4} - \frac{1}{4} = \frac{x}{2} - \frac{x}{3} }

       \latex{ - \frac{4}{4} = \frac{3x-2x}{6} }

          \latex{ - 1 = \frac{x}{6}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; /\times 6 }

          \latex{\underline{\underline{- 6 = x}}}                                           integer
 
Check:           left side: \latex{ \frac{-6}{3}-\frac{3}{4}=-2-\frac{3}{4}=-\frac{11}{4} }
 
                        rigth side:  \latex{ \frac{1}{4}+\frac{-6}{2}=\frac{1}{4}-3=\frac{1-12}{4}=-\frac{11}{4} \\[10pt]}
Answer:                                \latex{\underline{\underline{ x=-6}} }

b) 
             \latex{ \frac{x}{2}-\frac{x-3}{2} = 3 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; /\times 2 }
 
\latex{ 2 \times \bigg( \frac{x}{2}-\frac{x-3}{2} \bigg) = 2 \times 3 }
 
\latex{ \cancel{2}\times \frac{x}{\cancel{2}} - \cancel{2} \times \frac{x-3}{\cancel{2}} = 6 }
 
             \latex{ x-(x-3)=6 }
                \latex{ x-x+3=6 }
                               \latex{ 3 = 6}
 
\latex{ 3 \neq 6 }. Therefore, this equation has no solution.

 

 

 

 

 

 

 

 

 

 

 

You can check the solution of an equation by replacing the variables with the value of the solution on both sides. If the sides are equal after substitution, the solution is correct.

 

In the numerator, the algebraic expression formed by a group of terms should be put into parentheses after expansion. 

The balance method can also be used to rearrange formulas in physics.  
For example, in uniform linear motion, \latex{ speed } is the quotient of \latex{ distance } and \latex{ time }: \latex{ v=\frac{s}{t} }.
If you multiply both sides of the equation by \latex{ t }, you get \latex{ t \times v = s }. This means that \latex{ distance } is the product of \latex{ speed } and \latex{ time. }
Then, if you divide both sides of the equation by \latex{ v }, you get \latex{ t = \frac{s}{v} }. This means that \latex{ time } is the quotient of \latex{ distance } and \latex{ speed. }
Inequalities describe the relationship between expressions that are not equal. The following symbols are used to express inequalities: \latex{\lt;\, \leq;\, \gt;\, \geq;\, \neq}. 
The domain should also be defined when solving inequalities.

\latex{ v=\frac{s}{t} }          \latex{ / \times t}

\latex{ v \times t = s }    \latex{ / \div v }

\latex{ t=\frac{s}{v} }

Example 6
Solve the following inequalities.
a) \latex{ 3x - 2 \lt 16 }, over the set of positive integers. 
b) \latex{7 - 2x\leq 11}, over the set of rational numbers.
Solution
a)                            \latex{ 3x- 2 \lt 16 }                        \latex{ /+2 }
                                       \latex{ 3x \lt 18}                      \latex{ /\div3 }
                                         \latex{\underline{\underline{x \lt 6}}}
 
Solving the inequality: \latex{ x } is an element of the set of positive integers, and \latex{ x \lt 6 }. Therefore, the solution set is  \latex{ \left\{1, 2, 3, 4, 5\right\} }. 
The solution of the inequality is represented on a number line.
\latex{7}
\latex{-1}
\latex{0}
\latex{1}
\latex{2}
\latex{3}
\latex{4}
\latex{5}
\latex{6}

b)                                \latex{ 7- 2x \leq 11 }                     \latex{ /-7 }

                                  \latex{ - 2x \leq 4}                     \latex{ /\div(-2) }

                                        \latex{\underline{\underline{x \geq -2}}}

 

Solving the inequality: \latex{ x } is a rational number, and \latex{ x\geq -2 }. The solutions of the inequality are the rational numbers shown in red on the number line.

\latex{4}
\latex{-2}
\latex{-1}
\latex{0}
\latex{1}
\latex{2}
\latex{3}

\latex{ -2 \lt 3 }     \latex{ / \times (-1) }

    \latex{ 2 \gt -3 }

 

Multiplying both sides of an inequality by a negative number reverses the direction of the inequality sign.

 

\latex{ -x \leq 2 }     \latex{ / \times (-1) }

    \latex{ x \geq -2 }

 

The inequality has an infinite number of solutions. 

The inequality can be checked by substitution only if it has a finite number of solutions.
The balance method can also be used to solve inequalities. Remember that multiplying or dividing both sides by a negative number reverses the direction of the inequality.
Exercises
{{exercise_number}}. Solve the following equations over the set of integers.
  1. \latex{ 5x - 7 = 18 }
  1. \latex{ 3 = 7 - 2x }
  1. \latex{4 - 3x = x + 16}
  1. \latex{ 11 - 4x = 11 + 3x }
  1. \latex{ 8x + 41 = 11x + 16 }
  1. \latex{ x + 1 + 2x+ 2 + 3x+ 3 = 4x + 4 }
  1. \latex{ 3x + 7 - 4x + 9 = 5x + 13 - 8x - 1 }
  1. \latex{ \frac{x}{5} +2 = 12-x }
  1. \latex{ 2x + \frac{5}{3} = \frac{x}{6} }
  1. \latex{ \frac{x}{4} + \frac{3}{5} = \frac{5x}{20} +1}
{{exercise_number}}. Solve the following equations over the set of rational numbers.
  1. \latex{ \frac{x}{4} - 1 = \frac{37}{5}-5x}
  1. \latex{ \frac {3}{2}x + 3 = 5x+1-x}
  1. \latex{ 2(x + 2) - 1 = 4}
  1. \latex{ 4x + 2(x + 3) = 6 }
  1. \latex{ 3(x - 3) + 4(x + 1) = 2(x + 9) }
  1. \latex{ 5 - 2(x + 4) = x + 6 }
  1. \latex{ 5(x + 4) - 3(x - 2) = 20 }
  1. \latex{ 2- \frac{x+2}{6} =1 }
  1. \latex{ \frac{x+1}{3} + \frac{x-1}{2} =x-1 }
  1. \latex{ \frac{x}{3} + \frac{x-2}{4} - \frac{x+2}{5} = \frac{x}{6} +3 }
{{exercise_number}}. Match each equation with its solution.
\latex{ \frac{x}{2} - \frac{x-2}{12} = \frac{x+7}{3} }
\latex{ x + 7 = 5x - 3 }
\latex{ -2 }
\latex{ (x - 1)x -1 = x^2 + 2x + 5 }
\latex{ 3(x + 2) - 5(x + 3) = 9( x\, –\, 1) }
\latex{ 26 }
\latex{ 2.5 }
\latex{ 0 }
{{exercise_number}}. Solve the following equations over the set of natural numbers.
  1. \latex{ 3x - 1 \lt 8 }
  1. \latex{ 2x - 5 \leq x + 4 }
  1. \latex{ 4 - 2x \gt x - 11 }
  1. \latex{ 5 - 3x \geq 5 }
  1. \latex{ 6x + 6 - 4x \lt 8x - 18 }
  1. \latex{ 3(x - 1) + 4x \gt 3(2x - 5) }
{{exercise_number}}. Solve the following equations over the set of rational numbers. Represent each solution on the number line.
  1. \latex{ 6x-7 \geq 5}
  1. \latex{ 3 + 3x \lt x + 17 }
  1. \latex{ 18 - 6x \geq 5x + 7 }
  1. \latex{ 2(3 - x) + 5(x - 1) \leq x + 8 }
  1. \latex{ \frac{x}{3} -4 \lt \frac{x}{6} }
  1. \latex{ \frac{2-3x}{5} + \frac{x-1}{3} +1 \gt 0 }
{{exercise_number}}. Which physical quantities do the following equations represent? Identify them by rearranging the formulas.
  1. \latex{ \varrho=\frac{m}{V} }
  1. \latex{ p=\frac{F_{ny}}{A} }
  1. \latex{ W= F \times s}
  1. \latex{ Q = c \times m \times\Delta t}
Quiz
Solve the equation  \latex{ x+1+2x+2+3x+3+...+99x+99 = 100x+100 }.
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