Algebra Notes

Introducing Algebra

  • Algebra uses letters and symbols to represent numbers.
  • Consider an expression with terms, such as x3+2y+7x^3 + 2y + 7.
    • The expression x3+2y+7x^3 + 2y + 7 has three terms.
    • x3x^3 is the first term.
      • The coefficient of x3x^3 is 1.
      • The power of x is 3.
      • x is a variable.
    • 2y2y is the second term.
      • The coefficient of y is 2.
      • The power of y is 1.
      • y is a variable.
    • 77 is the third term (constant term).
      • There are no variables or coefficients in this term.
      • It is called the constant term.

Various Definitions in Algebra

  • An algebraic expression is a collection of terms connected by mathematical operations (+, -, ×, ÷).
    • Example: 3a−2b+83a - 2b + 8 is an expression.
  • An algebraic term is an element of an expression, consisting of a letter and a coefficient.
    • Example: −2b-2b (where b can be substituted by any given value).
    • In the expression 3a−2b+83a - 2b + 8, the coefficient of a is 3 and the coefficient of b is -2.
  • A constant term is an element of an expression consisting of a number only, with no variables.
    • Example: 5 is the constant term in 5−3c5 - 3c.
  • Like terms have the same letter and the same power.
    • Example: 3y3y and −8y-8y are like terms.
  • Unlike terms have different letters or powers.
    • Example: 2x2x, 3y3y, 5xy5xy, and −8y2-8y^2 are unlike terms.

Algebraic Notation

  • An expression does not have an equals sign (=).
    • Example: p+5q2+18r3p + 5q^2 + 18r^3 is an expression.
  • Some expressions can be simplified by collecting like terms.
    • Example: 3x−7x3x - 7x can be simplified by adding the coefficients to get −4x-4x.
  • An equation has an equals sign (=).
    • Example: 3v−4w=203v - 4w = 20 is an equation.
  • When an equation involves one variable only, we can solve the equation to find the value of the unknown variable.
    • Example: x+2=5x + 2 = 5 can be solved to get x=3x = 3.
  • In an algebraic expression, terms are usually written in the following order:
    • Positive terms first (+ve).
    • Remaining terms alphabetically.
    • Constant terms at the end.

Key Rules of Algebra

  • In algebra, the letter x is written as x to avoid confusion with the multiplication sign (×).
  • a+a+aa + a + a means 3 bags of a, which simplifies to 3a3a.
  • 3×a3 × a means 3 bags of a, which simplifies to 3a3a.
  • b+b+4b + b + 4 means 2 bags of b and 4, which simplifies to 2b+42b + 4.
  • 2×b+42 × b + 4 means 2 bags of b and 4, which simplifies to 2b+42b + 4.
  • xyz+xyz+xyzxyz + xyz + xyz means 3 bags of xyz, which simplifies to 3xyz3xyz.
  • 3×xyz3 × xyz means 3 bags of xyz, which simplifies to 3xyz3xyz.
  • Note:
    • a+2a + 2 cannot be simplified because no terms have the same letter and power.
    • a−ba - b cannot be simplified because no terms have the same letter and power.
    • a+ba + b has no like terms (1 bag of apples + 1 bag of bananas).
  • 1÷21 ÷ 2 is written as 12\frac{1}{2}.
  • a÷2a ÷ 2 is written as a2\frac{a}{2}.
  • 9a÷12b9a ÷ 12b is written as 9a12b\frac{9a}{12b}, which simplifies to 3a4b\frac{3a}{4b}.

Overview of Vocabulary in Algebra

  • Operators: Arithmetic operators perform addition, subtraction, multiplication, division, and exponentiation (indices, powers, exponents, or roots).
  • Term: Items on which mathematical operations take place. Can be a constant, a variable, a function, or a combination of these.
  • Expression: A collection of terms put together with mathematical operators.
  • Equation: A statement that shows that two expressions are equal.
  • Constant: A number (can be known or unknown) that does not change in an expression.
  • Variable: A value that changes according to the situation.
  • Coefficient: A constant value that multiplies the variable or function. Any variable without a visible coefficient has a coefficient of 1.
  • Like terms: Terms that are identical except for the coefficients (variables need to be raised to the same power).
  • Unlike terms: Terms that are not like terms.
  • Simplifying: The process of arranging and combining like terms.
  • Solving: The process of finding the numerical value of unknown values.
  • Substituting: The process of changing an unknown to a number or symbol of equal value.
  • Sum (Total): The result of an addition operation when adding terms.
  • Product: The result of a multiplication operation when multiplying terms.
  • Quotient: The result obtained when dividing a quantity by another.

Simplifying Algebraic Expressions

  • Consider the expression 3x+2y−5x+5y3x + 2y - 5x + 5y.
    • 3x3x and −5x-5x are like terms.
    • 2y2y and 5y5y are like terms.
    • 3x3x and 2y2y are unlike terms.
    • −5x-5x and 5y5y are also unlike terms.
  • Note that only terms containing the same letter and power can be added or subtracted.
  • Exercise A - Simplify the following expressions:
    1. 3x+4y−10−5x+3y3x + 4y - 10 - 5x + 3y
    2. 4x−4y+5y+7−4x4x - 4y + 5y + 7 - 4x
    3. 2x+3y−2+3x+6y+72x + 3y - 2 + 3x + 6y + 7
    4. 3b−(4b−6b+2)+b3b - (4b - 6b + 2) + b
    5. 0.3x2+5x+0.7x2+20.3x^2 + 5x + 0.7x^2 + 2
    6. 1+14+14+141 + \frac{1}{4} + \frac{1}{4} + \frac{1}{4}
    7. hij−hij+jih+ihj+jihhij - hij + jih + ihj + jih
    8. v2+v2+v3+v3+v3+v3v^2 + v^2 + v^3 + v^3 + v^3 + v^3
    9. 8d−3.2b+2.5d−4.1d−3.8d+2b8d - 3.2b + 2.5d - 4.1d - 3.8d + 2b
    10. 2x+9−0.74x+2.242x + 9 - 0.74x + 2.24
    11. −2.3x−8.7+1.3x−1.3-2.3x - 8.7 + 1.3x - 1.3
    12. cd+cd+cdcd + cd + cd
    13. 6a2+3ab+b2−12a2+5ba−7ab6a^2 + 3ab + b^2 - 12a^2 + 5ba - 7ab
    14. 4d+14d+14d4d + \frac{1}{4}d + \frac{1}{4}d
    15. 7+12−3+3m−k+6m+3m7 + \frac{1}{2} - 3 + 3m - k + 6m + 3m
    16. 5m+m−12m5m + m - \frac{1}{2}m

Simplifying Expressions Involving Indices

  • Recall the following important index rules:
    • Multiplication of Terms: Addition of Powers
      • xa×xb=xa+bx^a × x^b = x^{a+b}
    • Division of Terms: Subtraction of Powers
      • xaxb=xa−b\frac{x^a}{x^b} = x^{a-b}
    • Powers raised to other Powers: Multiplication of Powers:
      • (xa)b=xab(x^a)^b = x^{ab}
      • (rxayb)c=rcxacybc(r x^a y^b)^c = r^c x^{ac} y^{bc}
  • Exercise B - Simplify the following expressions:
    1. x3×x2x^3 × x^2
    2. y5÷y3y^5 ÷ y^3
    3. 2h3×4h32h^3 × 4h^3
    4. 3ab2×4a33ab^2 × 4a^3
    5. y2×y2×y2×y2y^2 × y^2 × y^2 × y^2
    6. 20p34p\frac{20p^3}{4p}
    7. (3ax2)2(\frac{3ax}{2})^2
    8. (2cpx15p3d)(\frac{2cp x}{15p^3d})
    9. 3a2×4a33a^2 × 4a^3
    10. 5f2×2g35f^2 × 2g^3
    11. (6m)(\frac{6}{m})
    12. 8rm54m2\frac{8rm^5}{4m^2}
    13. 12a43\frac{12a^4}{3}
    14. 5a25a\frac{5a}{25a}
    15. 10ab25a\frac{10ab}{25a}
    16. 28r4s33rs2\frac{28r^4s^3}{3rs^2}
    17. t+t0t + t^0
    18. k2k\frac{k^2}{k}
    19. 7q5×3q9×a47q^5 × 3q^9 × a^4
    20. 80g510gt\frac{80g^5}{10g t}

Expanding Algebraic Products

  • Product Involving One Bracket
    • Given the expression a(x+y)a(x + y), we can expand the product by multiplying x and y by a.
      • a(x+y)=ax+aya(x + y) = ax + ay
  • Exercise C - Expand and Simplify the following expressions:
    1. 3(a+x)3(a + x)
    2. −(2x−1)-(2x - 1)
    3. y(1+ax)y(1 + ax)
    4. 7+5(x+1)7 + 5(x + 1)
    5. 8(5+x)8(5 + x)
    6. −7(2x+3)-7(2x + 3)
    7. 3a2(4+x)3a^2(4 + x)
    8. 1.6(10+0.5a)+0.2a1.6(10 + 0.5a) + 0.2a
    9. −16(94−12)-16(\frac{9}{4} - \frac{1}{2})
    10. x−(7−4x)x - (7 - 4x)
    11. 12(2+61−10k)\frac{1}{2}(2 + 61 - 10k)
    12. 7y−2y(2y−3)+8−4y7y - 2y(2y - 3) + 8 - 4y
    13. 7(2b+3)7(2b + 3)
    14. 0.5(10−4x)0.5(10 - 4x)
    15. 2(3m+4)+3(m−5)2(3m + 4) + 3(m - 5)
    16. 7x2y3(5xy2−4y9z2)7x^2y^3 (5xy^2 - 4y^9z^2)
    17. −3(2x−1)−2(2x−3)-3(2x - 1) - 2(2x - 3)
    18. t(3t2+4)t(3t^2 + 4)

Algebraic Factors

  • Finding the Factors of a Number
  • Exercise D1 - List all the factors of the following terms:
    1. 18pq18pq
    2. 7x37x^3
    3. 5r2t5r^2t
    4. x3y2x^3y^2

Factorising Expressions

  • Common Factor
  • Exercise D2 - When possible, factorise the following expressions:
    1. 12+36x12 + 36x
    2. 18y−9x+318y - 9x + 3
    3. 16+4b16 + 4b
    4. a2+3aa^2 + 3a
    5. 3g+51h3g + 51h
    6. 71k−3n71k - 3n
    7. 2w+4x+6y+8z2w + 4x + 6y + 8z
    8. ab+2a2b+5ab2ab + 2a^2b + 5ab^2
    9. ab+ba2ab + ba^2
    10. 9y2−6y+18xy9y^2 - 6y + 18xy
    11. 12a+412a + 4
    12. 14x−714x - 7
    13. x2+5xx^2 + 5x
    14. b2+14b^2 + 14
    15. 3y2+27y3y^2 + 27y
    16. 5ab−10bc5ab - 10bc
    17. 9y2−6y9y^2 - 6y
    18. 5x3−10x5x^3 - 10x
    19. 27a2−18a327a^2 - 18a^3
    20. 4a2+8ax−4y4a^2 + 8ax - 4y
    21. 5xy+4xz+3x5xy + 4xz + 3x
    22. 3x2z−6xy+93x^2z - 6xy + 9

Grouping Method

  • Exercise E - Factorise the following expressions, when possible:
    1. 3(x+y)+z(x+y)3(x + y) + z(x + y)
    2. 3p(2q−p)−2q(2q−p)3p(2q - p) - 2q(2q - p)
    3. w(x−y)−8(x−y)w(x - y) - 8(x - y)
    4. x3−4x2−5x+20x^3 - 4x^2 - 5x + 20
    5. 3x+3x3−x−13x + 3x^3 - x - 1
    6. ab−ac+bd+cdab - ac + bd + cd
    7. 5x3+25x2+2x+105x^3 + 25x^2 + 2x + 10
    8. x3+2x2+3x+6x^3 + 2x^2 + 3x + 6
    9. 2a2+6a−a−32a^2 + 6a - a - 3
    10. a5+7a4−6a−42a^5 + 7a^4 - 6a - 42

Factorising & Simplifying

  • Simplifying by taking out a Common Factor
    • Factorising is finding and taking out factors common to terms in the expression.
    • To factorise an expression:
      • Identify the Highest Common Factor;
      • Write the common factor outside the bracket;
      • Fill in each term inside the bracket.
  • Exercise F - Factorise and simplify the following expressions:
    1. 3t+93t + 9
    2. 5a−15+10a5a - 15 + 10a
    3. 4x+6y+124x + 6y + 12
    4. 2ab+ad2ab + ad
    5. 2k2+8k2k^2 + 8k
    6. 12k39rt\frac{12k^3}{9rt}
    7. 12ab+4abc−8abk6q\frac{12ab + 4abc - 8abk}{6q}
    8. −4p2q−6pq3−2pq-4p^2q - 6pq^3 - 2pq
    9. 12x2+6x−6xy33x\frac{12x^2 + 6x - 6xy^3}{3x}
    10. 416xy+24xy\frac{4}{16xy + 24xy}
    11. 5k2+25k10k3−30k\frac{5k^2 + 25k}{10k^3 - 30k}
    12. 7m2+28m4m4+16m3\frac{7m^2 + 28m}{4m^4 + 16m^3}
    13. 7x−3514\frac{7x - 35}{14}
    14. 6x2y−12x2yt8xy2−16xy2t\frac{6x^2y - 12x^2yt}{8xy^2 - 16xy^2t}

Solving Equations

  • Solving an equation means finding the number that the letter stands for so that the two sides are equal.
  • When we are solving equations, we proceed as follows:
    • Expand any brackets (products);
    • Collect any like terms on each side;
    • Gather variable terms on one side and number terms on the other;
    • Find the value of the unknown.
  • Exercise G - Solve:
    1. 5x+3=x+115x + 3 = x + 11
    2. 3x−4=4x−73x - 4 = 4x - 7
    3. 9x+4=13x−89x + 4 = 13x - 8
    4. 7x−3=2x+77x - 3 = 2x + 7
    5. 4x−13=2x4x - 13 = 2x
    6. 3(x+1)=243(x + 1) = 24
    7. 2(x+3)+x=212(x + 3) + x = 21
    8. 7(x+2)−6(x−3)=07(x + 2) - 6(x - 3) = 0
    9. 3(x+2)+4(x+1)=4(x+4)−33(x + 2) + 4(x + 1) = 4(x + 4) - 3
    10. 4(x−2)=3134(x - 2) = \frac{3}{13}
    11. 1.4(x−3)+0.2=0.4(2x−1)1.4(x - 3) + 0.2 = 0.4(2x - 1)
    12. 0.6z+0.9=0.4z+1.50.6z + 0.9 = 0.4z + 1.5

Equations Involving Fractions

  • When we have fractions in equations, we account for them as follows:
    • Change any whole terms to a fraction all over 1;
    • Include brackets to the numerators;
    • Find an LCM common to all denominators;
    • Multiply all the terms (on both sides) by the LCM;
    • Solve the equation.
  • Exercise H1 - Solve the following equations:
    1. x8=518\frac{x}{8} = \frac{5}{18}
    2. 17=x11\frac{1}{7} = \frac{x}{11}
    3. 32=8x\frac{3}{2} = \frac{8}{x}
    4. x4−16=5\frac{x}{4} - 16 = 5
    5. 2x7−16=52\frac{2x}{7} - 16 = \frac{5}{2}
    6. (10x−4)5=3\frac{(10x-4)}{5} = 3
  • Exercise H2- Solve the following equations:
    1. 15−b3=5\frac{15-b}{3} = 5
    2. 4c=28\frac{4}{c} = 28
    3. 7=4j+97 = \frac{4}{j} + 9
    4. 5(i4−9)+7=3j5(\frac{i}{4} -9) + 7 = 3j
    5. 20−2f6=12\frac{20-2f}{6} = 12
    6. 3e+25=23\frac{3e+2}{5} = 23
    7. 5h+34+3=7\frac{5h+3}{4} + 3 = 7
    8. i+39=7\frac{i+3}{9} = 7

Solving Equations using the LCM Method

  • Solve the equation: 4x+35−4x+64=4x−3\frac{4x+3}{5} - \frac{4x+6}{4} = 4x-3(4x+3)5−(4x+6)4=(4x−3)1\frac{(4x+3)}{5} - \frac{(4x+6)}{4} = \frac{(4x-3)}{1}
    • LCM of 1, 4 and 5 is 20. We therefore change every term to a common denominator of 20.
      4(4x+3)20−5(4x+6)20=20(4x−3)20\frac{4(4x+3)}{20} - \frac{5(4x+6)}{20} = \frac{20(4x-3)}{20}
    • Since we have AN EQUATION with SAME DENOMINATOR THROUGHOUT, we can CANCEL OUT THE DENOMINATOR.
      4(4x+3)−5(4x+6)=20(4x−3)4(4x+3) - 5(4x+6) = 20(4x-3)
      16x+12−20x−30=80x−6016x+12 - 20x - 30 = 80x - 60
      16x−20x−80x=−12+30−6016x - 20x - 80x = -12 + 30 - 60
      −84x=−42-84x = -42
      x=−42−84x = \frac{-42}{-84}
      x=12x = \frac{1}{2}
  • Exercise I Solve:
    1. k−53=k−15\frac{k-5}{3} = \frac{k-1}{5}
    2. 16+38+1−45=145\frac{1}{6} + \frac{3}{8} + \frac{1-4}{5} = \frac{14}{5}
    3. (m+5)4=(m−4)7−3\frac{(m+5)}{4} = \frac{(m-4)}{7} - 3
    4. 15(x−1)=2\frac{1}{5}(x-1) = 2
    5. (x+2)5+(x+5)10=11\frac{(x+2)}{5} + \frac{(x+5)}{10} = 11
    6. (x−2)5=710\frac{(x-2)}{5}= \frac{7}{10}
    7. (x+12)4=(x+7)3\frac{(x + 12)}{4} = \frac{(x+7)}{3}
    8. (x−1)4=(x+2)5\frac{(x-1)}{4} = \frac{(x+2)}{5}
    9. 13(x+2)=(x−3)4+2\frac{1}{3}(x+2) = \frac{(x-3)}{4} + 2
    10. 14(2x+10)=x−1\frac{1}{4} (2x + 10) = x − 1
    11. 16(9x+4)=x−1\frac{1}{6} (9x + 4) = x − 1
    12. (20−x4)=17−x(\frac{20-x}{4}) = 17 - x
    13. (30−6x3)−1=7(\frac{30-6x}{3}) - 1 = 7
    14. 1−2x1+2x=0.5\frac{1-2x}{1+2x} = 0.5
    15. 12(x−1)−15(x+1)=1\frac{1}{2}(x-1) - \frac{1}{5}(x + 1) = 1
    16. 53=−3x1−3x\frac{5}{3} = \frac{-3x}{1-3x}
    17. 14(1−2x)+15(1−x)=(x−3)\frac{1}{4}(1-2x) + \frac{1}{5}(1-x) = (x-3)
    18. 16(1−x)−(x−8)=(x−7)\frac{1}{6}(1-x) - (x - 8) = (x - 7)

Algebraic Substitution

  • When we are given the numeric value of a particular letter, we can substitute this number into the expression to simplify it.
  • Example 1:
    • If t = 4 and g = -6, find the value of 2t+3g−52t + 3g - 5.
      • 2t+3g−52t + 3g - 5
      • 2(4)+3(−6)−52(4) + 3(-6) - 5
      • 8−18−58 - 18 - 5
      • −10−5-10 - 5
      • −15-15
  • In substitution, we write a known or given value or symbol equal in value to an unknown in an expression.
  • Example 2:
    • Find the value of x2+3y−z3x^2 + 3y - \frac{z}{3} when x=3x = 3, y=6y = 6, z=9z = 9
  • Example 3:
    • Find the value of m+3np−3\frac{m + 3n}{p} - 3 when m=2m = 2, n=3n = 3, and p=12p = 12
  • Exercise J- Evaluate by substituting using the given information:
    1. What is the value of 3a−16x−49\sqrt{3a - 16x - 49} when a=6a = 6 and x=−5x = -5 ?
    2. What is the value of 7y2+4y+57y^2 + 4y + 5 when y=−2y = -2?
    3. Find the value of x2+y2+5\sqrt{x^2 + y^2 + 5} when:
      • a) x=−4,y=−3x = -4, y = -3
      • b) x=14,y=14x = \frac{1}{4}, y = \frac{1}{4}
    4. Find the value of a+b3\sqrt{\sqrt{a} + b^3} when a=2254a = \frac{225}{4} and b=25b = \frac{2}{5}

Further Algebraic Substitution

  • When we are given a formula involving letters with a specific value, we can find the missing (unknown) by substituting the given numbers into the letters.
  • Exercise K - Answer the following questions:
    1. Evaluate 3x+53x + 5 when:
      • a) x=1x = 1
      • b) x=−1x = -1
      • c) x=0x = 0
    2. Find the value of 2x−3y2x - 3y when:
      • a) x=1.5,y=5x = 1.5, y = 5
      • b) x=−1,y=−1x = -1, y = -1
      • c) x=−2,y=0x = -2, y = 0
    3. The formula used to calculate the perimeter of a square is p=4lp = 4l. Find p when l=6l = 6 cm.
    4. The formula used to calculate the area of a square is A=l2A = l^2. Find A when l=6l = 6 cm.
    5. The formula used to calculate the perimeter of a rectangle is p=2(l+w)p = 2(l + w). Find p when l=7l = 7 cm and w=3w = 3 cm.
    6. The formula used to calculate the area of a rectangle is A=l×wA = l × w. Find A when l=7l = 7 cm and w=3w = 3 cm.
    7. The perimeter of square A is 60 cm and the area of square B is 81 cm².
      • a) Use the formula p=4lp = 4l to find the length of one side of square A.
      • b) Use the formula A=l2A = l^2 to find the length of one side of square B.
    8. The formula to find the perimeter of a rectangle is p=2(l+w)p = 2(l + w). Find the length of a rectangle with perimeter 120 m and width 10 m.
    9. The formula to find the area of a rectangle is A=l×wA = l × w. Find the width, given that the area is 24 cm² and the length is 8 cm.

Subject of the Formula

  • The subject of the formula is a variable positioned alone on one side of an equation, preferably on the left.
  • When we have a subject of the formula, we say that term is expressed in terms of the other terms in the formula.
  • To make a variable the subject, perform mathematical operations on the terms to isolate the required letter on one side.
  • To do this, undo all the former operations on that variable, using BIDMAS in reverse order.
  • Suppose we need to make x the subject of the formula. Then consider the following steps (success criteria) to help you make x the subject:
    • Expand any brackets and take terms containing x on one side of the equation;
    • If the term containing x is positive, then we do not move that term to the other side of the equation;
    • If the term containing x is negative, then we need to move that term to the other side of the equation;
    • We then start to carry out reverse operations until the letter x is the subject of the formula.
  • Example: Make the letter x the subject of the formula in 5−2x=3y+45 - 2x = 3y + 4

Rearranging Formulae

  • To make x the subject of the formula, rearrange it so it starts with x followed by an equals (=) sign.
  • Exercise L - Make the symbol x the subject in each of the following formulae:
    1. y=x+2y = x + 2
    2. y=x−3y = x - 3
    3. y=3xy = 3x
    4. y=x4y = \frac{x}{4}
    5. y=2x+1y = 2x + 1
    6. y=3x−1y = 3x - 1
    7. 5y+3=05y + 3 = 0
    8. y=7x+15y = \sqrt{7x + 15}
    9. x2+y2=r2x^2 + y^2 = r^2
    10. y=(x−2)3×180y = \frac{(x - 2)}{3} × 180

Multiple Occurrences

  • To make x subject of the formula, we have to rearrange the formula such that the formula will start with x and is followed by an equals (=) sign.
  • Example :
    • Make x the subject of the formula in:
      ax=bc+dx\frac{a}{x} = \frac{b}{c} + \frac{d}{x}

Exercise M - Make the letter x the subject of the formula:

1.  d=vx−uxRd = \frac{vx - ux}{R}
2.  10ax−c=d+7x10ax - c = \sqrt{d + 7x}
3.  6(kx+a)=7(x−a)6(kx + a) = 7(x - a)
4.  K=tx2+yx2−ty3K = \frac{tx^2 + y}{x^2 - ty^3}

Exercise N - Change the subject of the formula to the letter indicated in the bracket:

1.  y=x−5y = x - 5    (x)
2.  V=4−yV = 4 - y    (y)
3.  5c=7+P4k5c = \frac{7 + P}{4k}    (k)
4.  r=t−11r = \frac{t}{-11}    (t)
5.  47=74−94\sqrt{7} = \frac{7}{4 - 9}    (t)
6.  m+x−d=5m + \sqrt{x - d} = 5    (x)
7.  x+n3+k=an+xx + \sqrt{n^3 + k} = an + x    (n)
8.  n−x+3=5n - \sqrt{x + 3} = 5    (x)
9.  x2+52=7k\frac{x^2 + 5}{2} = 7k    (x)
10. (r−3)23=x+y6\frac{(r - 3)^2}{3} = \frac{x + y}{6}    (r)
11. s=ut+12at2s = ut + \frac{1}{2}at^2    (a)
12. h2=p2+b2h^2 = p^2 + b^2    (b)
13. y=mx+cy = mx + c    (x)
14. A=πr2A = \pi r^2    (r)
15. A=P(1+R100)2A = P(1 + \frac{R}{100})^2    (R)
16. xy+z=yy+1\frac{x}{y + z} = \frac{y}{y + 1}    (y)

Converting Phrases into Algebra

  • Using Letters to Represent Unknown Values in Expressions
    • Example 1: Use the given letters to write expressions representing unknown values for the following situations:
      • I had €w, but then I lost €10. I now have €(w - 10).
      • John is k years old. In 5 years, he will be (k + 5) years old.
      • Susannah cycles j km to meet Maria. They then cycle m km together to get to school. After school Lara cycles n km to the grocery shop. How many kilometers has Susannah cycled altogether? (j + m + n) km.
      • A test which is out of r marks, consists in section A and section B. In section A, Savannah scored p marks, whereas in section B, she scored q marks. An expression representing Savannah's mark as a percentage is: (p+q)r×100\frac{(p+q)}{r} \times 100
    • In the following example we are not given a pre-defined letter or symbol, so we need to choose and define a letter ourselves. Afterwards, we need to use the chosen letter to write expressions which represent the values in the given situation.
    • Example 2: Angela is relatively tall. Yet, Isabelle is 10 cm taller than Angela, and Reuben is 2 cm shorter than Isabelle. Write down a simplified expression for their mean height.

Miscellaneous Algebraic Problems

  • Exercise O - Answer the following questions:
    1. Let n be the unknown numeric quantity (number) used in this problem. For each phrase, form an expression in terms of n:
      • a. Seven more than twice the quantity: 2n + 7
      • b. Two less than five times the number: 5n - 2
      • c. Result of decreasing the quantity by half of itself: n - (n/2) = n/2
      • d. Twenty percent of the number as a fraction: (20/100)n = n/5
      • e. Result of subtracting the root of the number from the square of the number: *n² - √n