Expressions and Equations
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These are the solutions to the GCSE past questions on Expressions and Equations.
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English to Math: Something About Word Problems
(1.) Word problems are written in standard British English.
(2.) Some word problems are very lengthy and some are unnecessary.
Those ones are meant to discourage you from even trying.
(3.) Word problems in Mathematics demonstrate the real-world applications of mathematical concepts.
(4.) Embrace word problems. See it as writing from English to Math.
Take time to:
(a.) Read to understand. Paraphrase and shorten long sentences as necessary.
(b.) Re-read and note/underline the vocabulary "math" terms written in English.
(c.) Translate/Write "important" sentences one at a time.
(d.) Review what you wrote to ensure correctness.
(e.) Solve the math, and check your solution in the word problem.
Does that solution makes sense? If it does, you are probably correct. If it does not, please re-do it.
| English | Math |
|---|---|
| $7$ more than a number, $p$ | $7 + p$ |
| $7$ increased by a number, $q$ | $7 + q$ |
| $7$ added to a number, $r$ | $7 + r$ |
| A number, $g$ added to $7$ | $g + 7$ |
| The sum of $7$ and a number $t$ | $7 + t$ |
| $7$ less than a number, $p$ | $p - 7$ |
| $7$ decreased by a number, $q$ | $7 - q$ |
| A number, $r$ decreased by $7$ | $r - 7$ |
| $7$ subtracted from a number, $e$ | $e - 7$ |
| A number, $t$ subtracted from $7$ | $7 - t$ |
| The difference of $7$ and a number, $u$ | $7 - u$ |
| The difference of a number, $v$ and $7$ | $v - 7$ |
| $7$ times a number, $p$ | $7 * p$ = $7p$ |
| $7$ multiplied by a number, $q$ | $7 * q$ = $7q$ |
| A number, $r$ multiplied by $7$ | $r * 7$ = $7r$ |
| The product of a $7$ and a number, $h$ | $7 * h$ = $7h$ |
| The product of a number $t$ and $7$ | $t * 7$ = $7t$ |
| $7$ divided by a number, $p$ | $7 \div p$ |
| A number, $q$ divided by $7$ | $q \div 7$ |
| $7$ divides a number, $r$ | $r \div 7$ |
| A number, $m$ divides $7$ | $7 \div m$ |
| The quotient of $7$ and a number, $t$ | $7 \div t$ |
| The quotient of a number, $u$ and $7$ | $u \div 7$ |
| A number, $d$ decreased by the sum of the number and five | $d - (d + 5)$ |
| Three times the product of negative two and a number, $c$ | $3(-2c)$ |
| "is" or "equal to" or "equals" | $=$ |
| $3$ consecutive integers beginning with $3$ | $3, 4, 5$ |
| $3$ consecutive integers beginning with $-3$ | $-3, -2, -1$ |
| $3$ consecutive integers beginning with $p$ | $p, (p + 1), (p + 2)$ |
| $3$ consecutive even integers beginning with $8$ | $8, 10, 12$ |
| $3$ consecutive even integers beginning with $-8$ | $-8, -6, -4$ |
| $3$ consecutive even integers beginning with $q$ | $q, (q + 2), (q + 4)$ |
| $3$ consecutive odd integers beginning with $7$ | $7, 9, 11$ |
| $3$ consecutive odd integers beginning with $-7$ | $-7, -5, -3$ |
| $3$ consecutive odd integers beginning with $r$ | $r, (r + 2), (r + 4)$ |
| The smaller of three consecutive integers beginning with $p$ | $p$ |
| The middle of three consecutive integers beginning with $p$ | $p + 1$ |
| The larger of three consecutive integers beginning with $p$ | $p + 2$ |
| The median of five consecutive odd integers beginning with $p$ | $p + 4$ |
| The larger of five consecutive even integers beginning with $p$ | $p + 8$ |
| The sum of three consecutive integers is $12$ |
Let the first integer = $p$ $p + (p + 1) + (p + 2) = 12$ |
| The sum of three consecutive even integers is $12$ |
Let the first even integer = $p$ $p + (p + 2) + (p + 4) = 12$ |
| The sum of three consecutive odd integers is $12$ |
Let the first odd integer = $k$ $k + (k + 2) + (k + 4) = 12$ |
| Twice a number, $p$ | $2 * p$ = $2p$ |
| Double a number, $d$ | $d * 2$ = $2d$ |
| Thrice a number, $r$ | $3 * r$ = $3r$ |
| Triple a number, $c$ | $c * 3$ = $3c$ |
| Quadruple a number, $p$ | $4 * p$ = $4p$ |
| Multiplicative inverse or Reciprocal of a number, $p$ | $\dfrac{1}{p}$ |
| $5$ less than twice a number, $c$ |
Twice a number, $c$ = $2c$ $5$ less than $2c$ = $2c -5$ $2c - 5$ |
| $4$ more than twice as many tickets, $t$ |
Twice as many tickets, $t$ = $2t$ $4$ more than $2t$ = $4 + 2t$ $4 + 2t$ |
| The difference between the product of seven and a number, $k$; and twice the number. |
Product of seven and $k$ = $7k$ Twice $k$ = $2 * k = 2k$ Difference between $7k$ and $2k$ = $7k - 2k$ $7k - 2k$ |
| The product of seven less than the temperature, $t$ and three |
Seven less than $t$ = $t - 7$ Product of $t - 7$ and three = $3(t - 7)$ $3(t - 7)$ |
| The difference between the profit, $P$ divided by four and nine |
$P$ divided by four = $\dfrac{P}{4}$ Difference between $\dfrac{P}{4}$ and nine = $\dfrac{P}{4} - 9$ $\dfrac{P}{4} - 9$ |
| The difference between twelve times a number, $d$; and five more than three times the number. |
Twelve times $d$ = $12d$ Three times $d$ = $3 * d = 3d$ Five more than $3d$ = $5 + 3d$ Difference between $12d$ and $5 + 3d$ = $12d - (5 + 3d)$ $12d - (5 + 3d)$ |
| A number, $p$ decreased by the difference between ten and the number. |
Difference between ten and $p$ = $10 - p$ $p$ decreased by $10 - p$ = $p - (10 - p)$ $p - (10 - p)$ |
| The difference between the square of a number, $m$ and six times the number. |
Square of $m$ = $m^2$ Six times $m$ = $6 * m = 6m$ Difference between $m^2$ and $6m$ = $m^2 - 6m$ $m^2 - 6m$ |
| The cube of the sum of a number, $p$ and seven. |
Sum of $p$ and seven = $p + 7$ Cube of $p + 7$ = $(p + 7)^3$ $(p + 7)^3$ |
| The sum of a number, $k$ increased by nine and a number, $d$ decreased by seven. |
$k$ increased by nine = $k + 9$ $d$ decreased by seven = $d - 7$ Sum of $k + 9$ and $d - 7$ = $(k + 9) + (d - 7)$ $(k + 9) + (d - 7)$ |
| Twelve less than the quotient of the square of a number, $e$ and eight. |
Square of $e$ = $e^2$ Quotient of $e^2$ and eight = $\dfrac{e^2}{8}$ Twelve less than $\dfrac{e^2}{8}$ = $\dfrac{e^2}{8} - 12$ $\dfrac{e^2}{8} - 12$ |
| Complement of an angle, $\theta$ | $90 - \theta$ |
| Suppleement of an angle, $\theta$ | $180 - \theta$ |
These are the notable notes regarding factoring
Factoring Formulas
$ \underline{Difference\;\;of\;\;Two\;\;Squares} \\[3ex] (1.)\;\;x^2 - y^2 = (x + y)(x - y) \\[5ex] \underline{Difference\;\;of\;\;Two\;\;Cubes} \\[3ex] (2.)\;\; x^3 - y^3 = (x - y)(x^2 + xy + y^2) \\[5ex] \underline{Sum\;\;of\;\;Two\;\;Cubes} \\[3ex] (3.)\;\; x^3 + y^3 = (x + y)(x^2 - xy + y^2) \\[4ex] $Formulas Relating to Quadratic Expressions and Equations
$ (1.)\;\; Discriminant = b^2 - 4ac \\[5ex] (2.)\;\; \text{Quadratic Formula}:\;\; x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[6ex] (3.)\;\; \text{Sum of roots} = -\dfrac{b}{a} \\[5ex] (4.)\;\; \text{Product of roots} = \dfrac{c}{a} $
(1.) Solve the equation without a calculator: $\dfrac{6}{2^x} - 5^x = 50^x$
Leave your answer in the exact form.
You may check your solution with a calculator using the exact form of the solution.
$ \dfrac{6}{2^x} - 5^x = 50^x \\[5ex] LCD = 2^x \\[3ex] 2^x\left(\dfrac{6}{2^x} - 5^x\right) = 2^x * 50^x \\[5ex] 2^x\left(\dfrac{6}{2^x}\right) - (2^x * 5^x) = 2^x * (2 * 25)^x \\[5ex] 6 - (2^x * 5^x) = 2^x * (2 * 5^2)^x \\[3ex] 6 - (2^x * 5^x) = 2^x * 2^x * 5^{2x}...Law\;5...Exp \\[3ex] 6 - (2^x * 5^x) = 2^{x + x} * 5^{2x}...Law\;1...Exp \\[3ex] 6 - (2^x * 5^x) = 2^{2x} * 5^{2x} \\[3ex] 2^{2x} * 5^{2x} + (2^x * 5^x) = 6 \\[3ex] (2^x * 5^x)^2 + (2^x * 5^x) = 6 ...Law\;5...Exp \\[3ex] Let\;\; 2^x * 5^x = p \\[3ex] \text{Make the substitution in the exponential equation} \\[3ex] \text{This turns into a quadratic equation} \\[3ex] p^2 + p = 6 \\[3ex] \text{Solve the quadratic equation for p} \\[3ex] p^2 + p - 6 = 0 \\[3ex] (p + 3)(p - 2) = 0 \\[3ex] p + 3 = 0 \hspace{2em}OR\hspace{2em} p - 2 = 0 \\[3ex] p = -3 \hspace{2em}OR\hspace{2em} p = 2 \\[3ex] \text{Substitute back for p to find x} \\[3ex] When\;\;p = -3 \\[3ex] 2^x * 5^x = -3 \\[3ex] \text{Introduce log to both sides} \\[3ex] \log(2^x * 5^x) = \log(-3) \\[3ex] \log(-3) \text{ does not exist. Discard this value.} \\[5ex] When\;\;p = 2 \\[3ex] 2^x * 5^x = 2 \\[3ex] \text{Introduce log to both sides} \\[3ex] \log(2^x * 5^x) = \log(2) \\[3ex] \log 2^x + \log 5^x = \log 2...Law\;1...Log \\[3ex] x\log 2 + x\log 5 = \log 2 ...Law\;5...Log \\[3ex] x(\log 2 + \log 5) = \log 2 \\[3ex] x = \dfrac{\log 2}{\log 2 + \log 5} $
$ \dfrac{6}{2^x} - 5^x = 50^x \\[5ex] x = \dfrac{\log 2}{\log 2 + \log 5} \\[5ex] LHS = RHS \\[3ex] $
Leave your answer in the exact form.
You may check your solution with a calculator using the exact form of the solution.
$ \dfrac{6}{2^x} - 5^x = 50^x \\[5ex] LCD = 2^x \\[3ex] 2^x\left(\dfrac{6}{2^x} - 5^x\right) = 2^x * 50^x \\[5ex] 2^x\left(\dfrac{6}{2^x}\right) - (2^x * 5^x) = 2^x * (2 * 25)^x \\[5ex] 6 - (2^x * 5^x) = 2^x * (2 * 5^2)^x \\[3ex] 6 - (2^x * 5^x) = 2^x * 2^x * 5^{2x}...Law\;5...Exp \\[3ex] 6 - (2^x * 5^x) = 2^{x + x} * 5^{2x}...Law\;1...Exp \\[3ex] 6 - (2^x * 5^x) = 2^{2x} * 5^{2x} \\[3ex] 2^{2x} * 5^{2x} + (2^x * 5^x) = 6 \\[3ex] (2^x * 5^x)^2 + (2^x * 5^x) = 6 ...Law\;5...Exp \\[3ex] Let\;\; 2^x * 5^x = p \\[3ex] \text{Make the substitution in the exponential equation} \\[3ex] \text{This turns into a quadratic equation} \\[3ex] p^2 + p = 6 \\[3ex] \text{Solve the quadratic equation for p} \\[3ex] p^2 + p - 6 = 0 \\[3ex] (p + 3)(p - 2) = 0 \\[3ex] p + 3 = 0 \hspace{2em}OR\hspace{2em} p - 2 = 0 \\[3ex] p = -3 \hspace{2em}OR\hspace{2em} p = 2 \\[3ex] \text{Substitute back for p to find x} \\[3ex] When\;\;p = -3 \\[3ex] 2^x * 5^x = -3 \\[3ex] \text{Introduce log to both sides} \\[3ex] \log(2^x * 5^x) = \log(-3) \\[3ex] \log(-3) \text{ does not exist. Discard this value.} \\[5ex] When\;\;p = 2 \\[3ex] 2^x * 5^x = 2 \\[3ex] \text{Introduce log to both sides} \\[3ex] \log(2^x * 5^x) = \log(2) \\[3ex] \log 2^x + \log 5^x = \log 2...Law\;1...Log \\[3ex] x\log 2 + x\log 5 = \log 2 ...Law\;5...Log \\[3ex] x(\log 2 + \log 5) = \log 2 \\[3ex] x = \dfrac{\log 2}{\log 2 + \log 5} $
$ \dfrac{6}{2^x} - 5^x = 50^x \\[5ex] x = \dfrac{\log 2}{\log 2 + \log 5} \\[5ex] LHS = RHS \\[3ex] $
(2.) For real numbers, x, y; solve the system of equations for x and y
Check your solution.
$ x^2 - xy + y^2 = 1 \\[3ex] x^2 + xy + 2y^2 = 4 \\[3ex] $
Sometimes, when confronted with these kinds of equations: quadratic in x, quadratic in y, and linear in xy:
Would it be interesting to know that you can solve a system of equations by ensuring that the RHS of both equations are the same?
Let's make sure that 4 = 4
So, let's multiply eqn.(1) by 4
$ x^2 - xy + y^2 = 1 ...eqn.(1) \\[3ex] x^2 + xy + 2y^2 = 4 ...eqn.(2) \\[3ex] 4 * eqn.(1) \implies \\[3ex] 4(x^2 - xy + y^2) = 4(1) \\[3ex] 4x^2 - 4xy + 4y^2 = 4 ...eqn.(3) \\[3ex] 4 = 4 \implies RHS = RHS \\[3ex] \text{Equate the LHS of } eqn.(3) \text{ to the LHS of } eqn.(1) \\[3ex] 4x^2 - 4xy + 4y^2 = x^2 + xy + 2y^2 \\[3ex] 4x^2 - 4xy + 4y^2 - x^2 - xy - 2y^2 = 0 \\[3ex] 3x^2 - 5xy + 2y^2 = 0 \\[3ex] 3x^2 - 3xy - 2xy + 2y^2 = 0 \\[3ex] 3x(x - y) - 2y(x - y) = 0 \\[3ex] (x - y)(3x - 2y) = 0 \\[3ex] x - y = 0 \hspace{2em}OR\hspace{2em} 3x - 2y = 0 \\[3ex] When: \\[3ex] x - y = 0 \\[3ex] x = y ...eqn.(4) \\[5ex] When: \\[3ex] 3x - 2y = 0 \\[3ex] 3x = 2y \\[3ex] x = \dfrac{2y}{3}...eqn.(5) \\[5ex] \text{Let us find the real numbers} \\[3ex] \text{Substitute eqn.(4) into eqn.(1) for } y \\[3ex] x^2 - x(x) + x^2 = 1 \\[3ex] x^2 - x^2 + x^2 = 1 \\[3ex] x^2 = 1 \\[3ex] x = \pm \sqrt{1} \\[3ex] x = \pm 1 \\[3ex] y = x = \pm 1 \\[5ex] \text{Substitute eqn.(5) into eqn.(1) for } x \\[3ex] x^2 - xy + y^2 = 1 \\[3ex] \left(\dfrac{2y}{3}\right)^2 - \dfrac{2y}{3} * y + y^2 = 1 \\[5ex] \dfrac{4y^2}{9} - \dfrac{2y^2}{3} + y^2 = 1 \\[5ex] LCD = 9 \\[3ex] 9 * \dfrac{4y^2}{9} - 9 * \dfrac{2y^2}{3} + 9 * y^2 = 9 * 1 \\[5ex] 4y^2 - 6y^2 + 9y^2 = 9 \\[3ex] 7y^2 = 9 \\[3ex] y^2 = \dfrac{9}{7} \\[5ex] y = \pm\sqrt{\dfrac{9}{7}} \\[5ex] y = \pm\dfrac{\sqrt{9}}{\sqrt{7}} \\[5ex] y = \pm \dfrac{3}{\sqrt{7}} * \dfrac{\sqrt{7}}{\sqrt{7}} \\[5ex] y = \pm \dfrac{3\sqrt{7}}{7} \\[5ex] \text{Substitute these values of } y \text{ in eqn.(5) for } x \\[3ex] x = 2 * y \div 3 \\[3ex] x = 2 * \dfrac{3\sqrt{7}}{7} \div 3 \\[5ex] x = \dfrac{6\sqrt{7}}{7} * \dfrac{1}{3} \\[5ex] x = \pm\dfrac{2\sqrt{7}}{7} \\[5ex] $
Check your solution.
$ x^2 - xy + y^2 = 1 \\[3ex] x^2 + xy + 2y^2 = 4 \\[3ex] $
Sometimes, when confronted with these kinds of equations: quadratic in x, quadratic in y, and linear in xy:
Would it be interesting to know that you can solve a system of equations by ensuring that the RHS of both equations are the same?
Let's make sure that 4 = 4
So, let's multiply eqn.(1) by 4
$ x^2 - xy + y^2 = 1 ...eqn.(1) \\[3ex] x^2 + xy + 2y^2 = 4 ...eqn.(2) \\[3ex] 4 * eqn.(1) \implies \\[3ex] 4(x^2 - xy + y^2) = 4(1) \\[3ex] 4x^2 - 4xy + 4y^2 = 4 ...eqn.(3) \\[3ex] 4 = 4 \implies RHS = RHS \\[3ex] \text{Equate the LHS of } eqn.(3) \text{ to the LHS of } eqn.(1) \\[3ex] 4x^2 - 4xy + 4y^2 = x^2 + xy + 2y^2 \\[3ex] 4x^2 - 4xy + 4y^2 - x^2 - xy - 2y^2 = 0 \\[3ex] 3x^2 - 5xy + 2y^2 = 0 \\[3ex] 3x^2 - 3xy - 2xy + 2y^2 = 0 \\[3ex] 3x(x - y) - 2y(x - y) = 0 \\[3ex] (x - y)(3x - 2y) = 0 \\[3ex] x - y = 0 \hspace{2em}OR\hspace{2em} 3x - 2y = 0 \\[3ex] When: \\[3ex] x - y = 0 \\[3ex] x = y ...eqn.(4) \\[5ex] When: \\[3ex] 3x - 2y = 0 \\[3ex] 3x = 2y \\[3ex] x = \dfrac{2y}{3}...eqn.(5) \\[5ex] \text{Let us find the real numbers} \\[3ex] \text{Substitute eqn.(4) into eqn.(1) for } y \\[3ex] x^2 - x(x) + x^2 = 1 \\[3ex] x^2 - x^2 + x^2 = 1 \\[3ex] x^2 = 1 \\[3ex] x = \pm \sqrt{1} \\[3ex] x = \pm 1 \\[3ex] y = x = \pm 1 \\[5ex] \text{Substitute eqn.(5) into eqn.(1) for } x \\[3ex] x^2 - xy + y^2 = 1 \\[3ex] \left(\dfrac{2y}{3}\right)^2 - \dfrac{2y}{3} * y + y^2 = 1 \\[5ex] \dfrac{4y^2}{9} - \dfrac{2y^2}{3} + y^2 = 1 \\[5ex] LCD = 9 \\[3ex] 9 * \dfrac{4y^2}{9} - 9 * \dfrac{2y^2}{3} + 9 * y^2 = 9 * 1 \\[5ex] 4y^2 - 6y^2 + 9y^2 = 9 \\[3ex] 7y^2 = 9 \\[3ex] y^2 = \dfrac{9}{7} \\[5ex] y = \pm\sqrt{\dfrac{9}{7}} \\[5ex] y = \pm\dfrac{\sqrt{9}}{\sqrt{7}} \\[5ex] y = \pm \dfrac{3}{\sqrt{7}} * \dfrac{\sqrt{7}}{\sqrt{7}} \\[5ex] y = \pm \dfrac{3\sqrt{7}}{7} \\[5ex] \text{Substitute these values of } y \text{ in eqn.(5) for } x \\[3ex] x = 2 * y \div 3 \\[3ex] x = 2 * \dfrac{3\sqrt{7}}{7} \div 3 \\[5ex] x = \dfrac{6\sqrt{7}}{7} * \dfrac{1}{3} \\[5ex] x = \pm\dfrac{2\sqrt{7}}{7} \\[5ex] $
| LHS | RHS |
|---|---|
|
$
x^2 - xy + y^2 \\[3ex]
(x, y) = (1, 1) \\[3ex]
1^2 - 1(1) + 1^2 \\[3ex]
1 - 1 + 1 \\[3ex]
1
$
$ (x, y) = (-1, -1) \\[3ex] (-1)^2 - (-1)(-1) + (-1)^2 \\[3ex] 1 - 1 + 1 \\[3ex] 1 $ $ (x, y) = \left(\dfrac{2\sqrt{7}}{7}, \dfrac{3\sqrt{7}}{7}\right) \\[5ex] \left(\dfrac{2\sqrt{7}}{7}\right)^2 - \left(\dfrac{2\sqrt{7}}{7}\right)\left(\dfrac{3\sqrt{7}}{7}\right) + \left(\dfrac{3\sqrt{7}}{7}\right)^2 \\[5ex] \dfrac{4(7)}{7(7)} - \dfrac{6(7)}{7(7)} + \dfrac{9(7)}{7(7)} \\[5ex] \dfrac{4}{7} - \dfrac{6}{7} + \dfrac{9}{7} \\[5ex] \dfrac{7}{7} \\[5ex] 1 $ $ (x, y) = \left(-\dfrac{2\sqrt{7}}{7}, -\dfrac{3\sqrt{7}}{7}\right) \\[5ex] \left(-\dfrac{2\sqrt{7}}{7}\right)^2 - \left(-\dfrac{2\sqrt{7}}{7}\right)\left(-\dfrac{3\sqrt{7}}{7}\right) + \left(-\dfrac{3\sqrt{7}}{7}\right)^2 \\[5ex] \dfrac{4(7)}{7(7)} - \dfrac{6(7)}{7(7)} + \dfrac{9(7)}{7(7)} \\[5ex] \dfrac{4}{7} - \dfrac{6}{7} + \dfrac{9}{7} \\[5ex] \dfrac{7}{7} \\[5ex] 1 $ |
1 |
|
$
x^2 + xy + 2y^2 \\[3ex]
(x, y) = (1, 1) \\[3ex]
1^2 + 1(1) + 2(1)^2 \\[3ex]
1 + 1 + 2(1) \\[3ex]
2 + 2 \\[3ex]
4
$
$ (x, y) = (-1, -1) \\[3ex] (-1)^2 + (-1)(-1) + 2(-1)^2 \\[3ex] 1 + 1 + 2(1) \\[3ex] 2 + 2 \\[3ex] 4 $ $ (x, y) = \left(\dfrac{2\sqrt{7}}{7}, \dfrac{3\sqrt{7}}{7}\right) \\[5ex] \left(\dfrac{2\sqrt{7}}{7}\right)^2 + \left(\dfrac{2\sqrt{7}}{7}\right)\left(\dfrac{3\sqrt{7}}{7}\right) + 2\left(\dfrac{3\sqrt{7}}{7}\right)^2 \\[5ex] \dfrac{4(7)}{7(7)} + \dfrac{6(7)}{7(7)} + \dfrac{2(9)(7)}{7(7)} \\[5ex] \dfrac{4(7)}{7(7)} + \dfrac{6(7)}{7(7)} + \dfrac{18(7)}{7(7)} \\[5ex] \dfrac{4}{7} + \dfrac{6}{7} + \dfrac{18}{7} \\[5ex] \dfrac{28}{7} \\[5ex] 4 $ $ (x, y) = \left(-\dfrac{2\sqrt{7}}{7}, -\dfrac{3\sqrt{7}}{7}\right) \\[5ex] \left(-\dfrac{2\sqrt{7}}{7}\right)^2 + \left(-\dfrac{2\sqrt{7}}{7}\right)\left(-\dfrac{3\sqrt{7}}{7}\right) + 2\left(-\dfrac{3\sqrt{7}}{7}\right)^2 \\[5ex] \dfrac{4(7)}{7(7)} + \dfrac{6(7)}{7(7)} + \dfrac{2(9)(7)}{7(7)} \\[5ex] \dfrac{4(7)}{7(7)} + \dfrac{6(7)}{7(7)} + \dfrac{18(7)}{7(7)} \\[5ex] \dfrac{4}{7} + \dfrac{6}{7} + \dfrac{18}{7} \\[5ex] \dfrac{28}{7} \\[5ex] 4 $ |
4 |
(5.) Solve the equation: $2^x = 4x$
This equation, often referred to as a transcendal equation because it is not solved by the usual algebraic methods is solved by Inspection (Trial-and-Error/Guessing), Graphical, and/or Numerical methods.
A quick inspection (guessing) gives us one solution: $x = 4$ because $2^4 = 4(4)$
If we decide to solve it numerically, we shall get the other solution.
A graphical approach gives us the two solutions.
$ x = 4 \text{ and } x = 0.3099069310441 $
This equation, often referred to as a transcendal equation because it is not solved by the usual algebraic methods is solved by Inspection (Trial-and-Error/Guessing), Graphical, and/or Numerical methods.
A quick inspection (guessing) gives us one solution: $x = 4$ because $2^4 = 4(4)$
If we decide to solve it numerically, we shall get the other solution.
A graphical approach gives us the two solutions.
$ x = 4 \text{ and } x = 0.3099069310441 $
(6.) $\dfrac{1}{(x)(x - 1)} + \dfrac{2}{(x - 1)(x - 3)} + \dfrac{3}{(x - 3)(x - 6)} + \dfrac{4}{(x - 6)(x -
10)} = -0.4$
Did you notice any pattern in the fractions?
Let us resolve each term into partial fractions and find out if we can see more patterns.
This will most likely lead to a telescoping sum: a series where the intermediate values will cancel out, leaving us with only the first and last terms.
$ \dfrac{1}{(x)(x - 1)} + \dfrac{2}{(x - 1)(x - 3)} + \dfrac{3}{(x - 3)(x - 6)} + \dfrac{4}{(x - 6)(x - 10)} = -0.4 \\[5ex] ..................................................................... \\[3ex] \underline{\text{1st term}} \\[3ex] \dfrac{1}{(x)(x - 1)} = \dfrac{A}{x} + \dfrac{B}{x - 1} \\[5ex] A(x - 1) + Bx = 1 \\[3ex] Ax - A + Bx = 1 \\[3ex] Ax + Bx - A = 1 \\[3ex] x(A + B) - A = 0x + 1 \\[3ex] -A = 1 \\[3ex] A = -1 \\[3ex] A + B = 0 \\[3ex] B = -A \\[3ex] B = -(-1) \\[3ex] B = 1 \\[3ex] \implies \\[3ex] \dfrac{1}{(x)(x - 1)} = \dfrac{-1}{x} + \dfrac{1}{x - 1} \\[5ex] ..................................................................... \\[3ex] \underline{\text{2nd term}} \\[3ex] \dfrac{2}{(x - 1)(x - 3)} = \dfrac{C}{x - 1} \dfrac{D}{x - 3} \\[5ex] C(x - 3) + D(x - 1) = 2 \\[3ex] Cx - 3C + Dx - D = 2 \\[3ex] Cx + Dx - 3C - D = 2 \\[3ex] x(C + D) - 3C - D = 0x + 2 \\[3ex] C + D = 0 ...eqn.(1) \\[3ex] -3C - D = 2 ...eqn.(2) \\[3ex] eqn.(1) + eqn.(2) \implies \\[3ex] -2C = 2 \\[3ex] C = -\dfrac{2}{2} \\[5ex] C = -1 \\[3ex] \text{Substitute for c in eqn.(1)} \\[3ex] D = 0 - C \\[3ex] D = 0 - (-1) \\[3ex] D = 1 \\[3ex] \implies \\[3ex] \dfrac{2}{(x - 1)(x - 3)} = \dfrac{-1}{x - 1} + \dfrac{1}{x - 3} \\[5ex] ..................................................................... \\[3ex] \underline{\text{3rd term}} \\[3ex] \dfrac{3}{(x - 3)(x - 6)} = \dfrac{E}{x - 3} + \dfrac{F}{x - 6} \\[5ex] E(x - 6) + F(x - 3) = 3 \\[3ex] Ex - 6E + Fx - 3F = 3 \\[3ex] Ex + Fx - 6E - 3F = 3 \\[3ex] x(E + F) - 6E - 3F = 0x + 3 \\[3ex] E + F = 0 ...eqn.(3) \\[3ex] -6E - 3F = 3...eqn.(4) \\[3ex] 3 * eqn.(3) \implies \\[3ex] 3E + 3F = 0 ...eqn.(5) \\[3ex] eqn.(4) + eqn.(5) \implies \\[3ex] -3E = 3 \\[3ex] E = -\dfrac{3}{3} \\[5ex] E = -1 \\[3ex] \text{Substitute for E in eqn.(3)} \\[3ex] F = 0 - E \\[3ex] F = 0 - (-1) \\[3ex] F = 1 \\[3ex] \implies \\[3ex] \dfrac{3}{(x - 3)(x - 6)} = \dfrac{-1}{x - 3} + \dfrac{1}{x - 6} \\[5ex] ..................................................................... \\[3ex] $ Teacher: Without working out the last term, do you know what the partial fractions will be?
Student: I can guess...I think so...
Is it going to be:
$\dfrac{4}{(x - 6)(x - 10)} = \dfrac{-1}{x - 6} + \dfrac{1}{x - 10}$
Teacher: Yes, that is correct.
You noticed the pattern.
This leads us to a telescoping series sum
Student: You mentioned it earlier.
But, what does it mean?
Teacher: "telescoping" is analogous to how a collapsible telescope extends and retracts.
A telescoping sum (also known as a telescoping series) is a series in which the intermediate values cancels out, with the first and last terms as the remaining terms.
Student: Can we complete the 4th term...just to verify?
Teacher: Sure, let's do it.
$ ..................................................................... \\[3ex] \underline{\text{4th term}} \\[3ex] \dfrac{4}{(x - 6)(x -10)} = \dfrac{G}{x - 6} + \dfrac{H}{x - 10} \\[5ex] G(x - 10) + H(x - 6) = 4 \\[3ex] Gx - 10G + Hx - 6H = 4 \\[3ex] Gx + Hx - 10G - 6H = 4 \\[3ex] x(G + H) - 10G - 6H = 0x + 4 \\[3ex] G + H = 0 ...eqn.(6) \\[3ex] -10G - 6H = 4...eqn.(7) \\[3ex] 10 * eqn.(1) \implies \\[3ex] 10G + 10H = 0 ...eqn.(8) \\[3ex] eqn.(7) + eqn.(8) \implies \\[3ex] 4H = 4 \\[3ex] H = \dfrac{4}{4} \\[5ex] H = 1 \\[3ex] \text{Substitute for H in eqn.(6)} \\[3ex] G = 0 - H \\[3ex] G = 0 - 1 \\[3ex] G = -1 \\[3ex] \implies \\[3ex] \dfrac{4}{(x - 6)(x - 10)} = \dfrac{-1}{x - 6} + \dfrac{1}{x - 10} \\[5ex] ..................................................................... \\[3ex] \text{So, we have}: \\[3ex] \dfrac{-1}{x} + \dfrac{1}{x - 1} + \dfrac{-1}{x - 1} + \dfrac{1}{x - 3} + \dfrac{-1}{x - 3} + \dfrac{1}{x - 6} + \dfrac{-1}{x - 6} + \dfrac{1}{x - 10} = -0.4 \\[5ex] \dfrac{-1}{x} + \dfrac{1}{x - 1} - \dfrac{1}{x - 1} + \dfrac{1}{x - 3} - \dfrac{1}{x - 3} + \dfrac{1}{x - 6} - \dfrac{1}{x - 6} + \dfrac{1}{x - 10} = -0.4 \\[5ex] $ This is a telescoping sum
The intermediate values cancel out.
The first and last terms remain.
We have:
$ \dfrac{-1}{x} + \dfrac{1}{x - 10} = -0.4 \\[5ex] \dfrac{-1(x - 10) + 1(x)}{x(x - 10)} = -0.4 \\[5ex] -x + 10 + x = -0.4[x(x - 10)] \\[3ex] -0.4[x(x - 10)] = 10 \\[3ex] x(x - 10) = -\dfrac{10}{0.4} \\[5ex] x^2 - 10x = - 25 \\[3ex] x^2 - 10x + 25 = 0 \\[3ex] (x - 5)(x - 5) = 0 \\[3ex] x - 5 = 0 ...\text{twice} \\[3ex] x = 5...\text{twice} $
Did you notice any pattern in the fractions?
Let us resolve each term into partial fractions and find out if we can see more patterns.
This will most likely lead to a telescoping sum: a series where the intermediate values will cancel out, leaving us with only the first and last terms.
$ \dfrac{1}{(x)(x - 1)} + \dfrac{2}{(x - 1)(x - 3)} + \dfrac{3}{(x - 3)(x - 6)} + \dfrac{4}{(x - 6)(x - 10)} = -0.4 \\[5ex] ..................................................................... \\[3ex] \underline{\text{1st term}} \\[3ex] \dfrac{1}{(x)(x - 1)} = \dfrac{A}{x} + \dfrac{B}{x - 1} \\[5ex] A(x - 1) + Bx = 1 \\[3ex] Ax - A + Bx = 1 \\[3ex] Ax + Bx - A = 1 \\[3ex] x(A + B) - A = 0x + 1 \\[3ex] -A = 1 \\[3ex] A = -1 \\[3ex] A + B = 0 \\[3ex] B = -A \\[3ex] B = -(-1) \\[3ex] B = 1 \\[3ex] \implies \\[3ex] \dfrac{1}{(x)(x - 1)} = \dfrac{-1}{x} + \dfrac{1}{x - 1} \\[5ex] ..................................................................... \\[3ex] \underline{\text{2nd term}} \\[3ex] \dfrac{2}{(x - 1)(x - 3)} = \dfrac{C}{x - 1} \dfrac{D}{x - 3} \\[5ex] C(x - 3) + D(x - 1) = 2 \\[3ex] Cx - 3C + Dx - D = 2 \\[3ex] Cx + Dx - 3C - D = 2 \\[3ex] x(C + D) - 3C - D = 0x + 2 \\[3ex] C + D = 0 ...eqn.(1) \\[3ex] -3C - D = 2 ...eqn.(2) \\[3ex] eqn.(1) + eqn.(2) \implies \\[3ex] -2C = 2 \\[3ex] C = -\dfrac{2}{2} \\[5ex] C = -1 \\[3ex] \text{Substitute for c in eqn.(1)} \\[3ex] D = 0 - C \\[3ex] D = 0 - (-1) \\[3ex] D = 1 \\[3ex] \implies \\[3ex] \dfrac{2}{(x - 1)(x - 3)} = \dfrac{-1}{x - 1} + \dfrac{1}{x - 3} \\[5ex] ..................................................................... \\[3ex] \underline{\text{3rd term}} \\[3ex] \dfrac{3}{(x - 3)(x - 6)} = \dfrac{E}{x - 3} + \dfrac{F}{x - 6} \\[5ex] E(x - 6) + F(x - 3) = 3 \\[3ex] Ex - 6E + Fx - 3F = 3 \\[3ex] Ex + Fx - 6E - 3F = 3 \\[3ex] x(E + F) - 6E - 3F = 0x + 3 \\[3ex] E + F = 0 ...eqn.(3) \\[3ex] -6E - 3F = 3...eqn.(4) \\[3ex] 3 * eqn.(3) \implies \\[3ex] 3E + 3F = 0 ...eqn.(5) \\[3ex] eqn.(4) + eqn.(5) \implies \\[3ex] -3E = 3 \\[3ex] E = -\dfrac{3}{3} \\[5ex] E = -1 \\[3ex] \text{Substitute for E in eqn.(3)} \\[3ex] F = 0 - E \\[3ex] F = 0 - (-1) \\[3ex] F = 1 \\[3ex] \implies \\[3ex] \dfrac{3}{(x - 3)(x - 6)} = \dfrac{-1}{x - 3} + \dfrac{1}{x - 6} \\[5ex] ..................................................................... \\[3ex] $ Teacher: Without working out the last term, do you know what the partial fractions will be?
Student: I can guess...I think so...
Is it going to be:
$\dfrac{4}{(x - 6)(x - 10)} = \dfrac{-1}{x - 6} + \dfrac{1}{x - 10}$
Teacher: Yes, that is correct.
You noticed the pattern.
This leads us to a telescoping series sum
Student: You mentioned it earlier.
But, what does it mean?
Teacher: "telescoping" is analogous to how a collapsible telescope extends and retracts.
A telescoping sum (also known as a telescoping series) is a series in which the intermediate values cancels out, with the first and last terms as the remaining terms.
Student: Can we complete the 4th term...just to verify?
Teacher: Sure, let's do it.
$ ..................................................................... \\[3ex] \underline{\text{4th term}} \\[3ex] \dfrac{4}{(x - 6)(x -10)} = \dfrac{G}{x - 6} + \dfrac{H}{x - 10} \\[5ex] G(x - 10) + H(x - 6) = 4 \\[3ex] Gx - 10G + Hx - 6H = 4 \\[3ex] Gx + Hx - 10G - 6H = 4 \\[3ex] x(G + H) - 10G - 6H = 0x + 4 \\[3ex] G + H = 0 ...eqn.(6) \\[3ex] -10G - 6H = 4...eqn.(7) \\[3ex] 10 * eqn.(1) \implies \\[3ex] 10G + 10H = 0 ...eqn.(8) \\[3ex] eqn.(7) + eqn.(8) \implies \\[3ex] 4H = 4 \\[3ex] H = \dfrac{4}{4} \\[5ex] H = 1 \\[3ex] \text{Substitute for H in eqn.(6)} \\[3ex] G = 0 - H \\[3ex] G = 0 - 1 \\[3ex] G = -1 \\[3ex] \implies \\[3ex] \dfrac{4}{(x - 6)(x - 10)} = \dfrac{-1}{x - 6} + \dfrac{1}{x - 10} \\[5ex] ..................................................................... \\[3ex] \text{So, we have}: \\[3ex] \dfrac{-1}{x} + \dfrac{1}{x - 1} + \dfrac{-1}{x - 1} + \dfrac{1}{x - 3} + \dfrac{-1}{x - 3} + \dfrac{1}{x - 6} + \dfrac{-1}{x - 6} + \dfrac{1}{x - 10} = -0.4 \\[5ex] \dfrac{-1}{x} + \dfrac{1}{x - 1} - \dfrac{1}{x - 1} + \dfrac{1}{x - 3} - \dfrac{1}{x - 3} + \dfrac{1}{x - 6} - \dfrac{1}{x - 6} + \dfrac{1}{x - 10} = -0.4 \\[5ex] $ This is a telescoping sum
The intermediate values cancel out.
The first and last terms remain.
We have:
$ \dfrac{-1}{x} + \dfrac{1}{x - 10} = -0.4 \\[5ex] \dfrac{-1(x - 10) + 1(x)}{x(x - 10)} = -0.4 \\[5ex] -x + 10 + x = -0.4[x(x - 10)] \\[3ex] -0.4[x(x - 10)] = 10 \\[3ex] x(x - 10) = -\dfrac{10}{0.4} \\[5ex] x^2 - 10x = - 25 \\[3ex] x^2 - 10x + 25 = 0 \\[3ex] (x - 5)(x - 5) = 0 \\[3ex] x - 5 = 0 ...\text{twice} \\[3ex] x = 5...\text{twice} $
