Comprehensive Notes on Algebraic Methods, Direct Proportion, and Inverse Proportion

Principles of Algebraic Fractions and Quadratic Equations

Simplifying algebraic fractions requires finding common denominators and factoring out shared variables or constants. Consider the expression where two fraction terms are subtracted:

$\frac{2x+3}{(3x+1)(3x-1)} - \frac{7}{3x-1}$

To combine these terms into a single fraction, express both over the common denominator $(3x+1)(3x-1)$. Multiplying the numerator and denominator of the second term by $(3x+1)$ yields:

$\frac{2x+3 - 7(3x+1)}{(3x+1)(3x-1)}$

Expanding the numerator gives $2x + 3 - 21x - 7$, which simplifies to:

$\frac{-19x-4}{(3x+1)(3x-1)}$

Similarly, consider combining fractions with linear denominators:

$\frac{x-1}{2(2x-1)} + \frac{5-2x}{3(2x-1)}$

Using the common denominator $6(2x-1)$, scale each numerator accordingly:

$\frac{3(x-1) + 2(5-2x)}{6(2x-1)}$

Expanding the numerator gives $3x - 3 + 10 - 4x$, which simplifies to:

$\frac{7-x}{6(2x-1)}$

Rearranging algebraic equations to isolate a variable requires systematically clearing fractions and factoring. Given the equation:

$5a = \frac{3x-4}{2y-3x}$

To express $x$ in terms of $a$ and $y$, first multiply both sides by $(2y-3x)$, resulting in:

$5a(2y-3x) = 3x - 4$

Expanding the left side yields:

$10ay - 15ax = 3x - 4$

Rearrange terms to collect all terms containing $x$ on one side and constant/other terms on the opposite side:

$10ay + 4 = 3x + 15ax$

Factor out $x$ on the right side:

$10ay + 4 = x(3 + 15a)$

Dividing both sides by $(3 + 15a)$ yields the isolated expression for $x$:

$x = \frac{10ay+4}{3+15a}$

To find the specific numerical value of $x$ when $a = 1$ and $y = 5$, substitute these values directly into the derived formula:

$x = \frac{10(1)(5) + 4}{3 + 15(1)} = \frac{50 + 4}{3 + 15} = \frac{54}{18} = 3$

Solving quadratic equations involved in proportion contexts can be accomplished via expansion and factorization. Consider the non-standard quadratic form:

$(2x-1)(3-4x) = 2(1-2x)$

Expanding both sides yields:

$6x - 8x^2 - 3 + 4x = 2 - 4x$

$-8x^2 + 10x - 3 = 2 - 4x$

Rearranging all terms to one side to set the equation to zero produces:

$-8x^2 + 14x - 5 = 0$

Multiplying by $-1$ converts this to a standard quadratic with a positive leading coefficient:

$8x^2 - 14x + 5 = 0$

Factoring the quadratic expression gives:

$(2x-1)(4x-5) = 0$

Setting each factor to zero provides the two solutions:

$x = \frac{1}{2} \quad \text{or} \quad x = \frac{5}{4}$

When given a known root of a quadratic polynomial, the unknown coefficients can be determined directly. Given the polynomial equation $x^2 + px + 15 = 0$ with a known solution $x = 3$, substituting $x = 3$ yields:

$3^2 + p(3) + 15 = 0$

$9 + 3p + 15 = 0$

$3p = -24 \implies p = -8$

Substituting $p = -8$ back into the original quadratic equation gives $x^2 - 8x + 15 = 0$. Factoring this equation yields $(x-3)(x-5) = 0$, which confirms $x = 3$ and identifies the other solution as $x = 5$.

Simplifying multi-variable algebraic expressions involves identifying common factors across terms. For the algebraic fraction:

$\frac{4a^2b^2 - 8a^3b - 14ab^3}{4ab}$

Factor out the greatest common term $2ab$ from the numerator:

$\frac{2ab(2ab - 4a^2 - 7b^2)}{4ab}$

Canceling $2ab$ from numerator and denominator simplifies the expression to:

$\frac{2ab - 4a^2 - 7b^2}{2}$

For an expression with negative signs in the denominator, such as:

$\frac{6c^3d + 12cd^2 - 9c^2d^3}{-3c^2d^2}$

Factoring out $3cd$ from the numerator gives:

$\frac{3cd(2c^2 + 4d - 3cd)}{-3c^2d^2}$

Dividing each individual term by the denominator $-3c^2d^2$ simplifies the fraction to:

$\frac{3d - 2cd^2 - 4c^2}{cd}$

Direct Proportion and Linear Relationships

Two variables $x$ and $y$ are directly proportional if an increase in $x$ causes a proportional scale increase in $y$. Mathematically, direct proportion between $y$ and $x$ is written as $y \propto x$, which is expressed in equation form as:

$y = kx$

In this relationship, $k$ represents a constant of proportionality. To determine the exact value of $k$, any given pair of corresponding values for $x$ and $y$ can be substituted into the equation. For instance, if $y = 5$ when $x = 2$, substituting these values yields $5 = k(2)$, which gives $k = 2.5$, producing the governing equation $y = 2.5x$. Under direct proportion, if $x$ is doubled, $y$ is also doubled.

Practical problems in direct proportion can be resolved using either the unitary method or the proportion method. Consider determining the cost of $380\,\text{g}$ of sweets when $50\,\text{g}$ costs $\$2.10$.

Using the unitary method, first find the cost of $1\,\text{g}$ of sweets:

$\text{Cost per gram} = \frac{\$2.10}{50} = \$0.042$

Multiplying by the target weight gives:

$\text{Cost of } 380\,\text{g} = 0.042 \times 380 = \$15.96$

Using the proportion method, let $\$x$ be the cost of $380\,\text{g}$ of sweets:

$\frac{x}{2.10} = \frac{380}{50}$

$x = \frac{2.10}{50} \times 380 = \$15.96$

Consider another proportional problem where $\frac{3}{4}$ of a piece of metal has a mass of $15\,\text{kg}$, and the goal is to find the mass of $\frac{2}{5}$ of the same piece of metal.

By the unitary method, the total mass of 1 whole piece of metal is calculated as:

$\text{Total mass} = 15 \div \frac{3}{4} = 20\,\text{kg}$

The mass of $\frac{2}{5}$ of the metal piece is then:

$\text{Mass} = 20 \times \frac{2}{5} = 8\,\text{kg}$

By the proportion method, set $x\,\text{kg}$ as the mass of $\frac{2}{5}$ of the metal piece:

$\frac{x}{15} = \frac{2/5}{3/4}$

$x = 15 \times \frac{2/5}{3/4} = 8\,\text{kg}$

Graphical representations of direct proportion always yield a straight line passing directly through the origin $(0,0)$. For example, consider an overdue library fine system where $x$ represents the number of days a book is overdue and $y$ represents the fine in cents. The data points are defined as follows: for $x = 0$, $y = 0$; for $x = 1$, $y = 15$; for $x = 2$, $y = 30$; for $x = 3$, $y = 45$; for $x = 4$, $y = 60$; for $x = 5$, $y = 75$; for $x = 6$, $y = 90$; for $x = 7$, $y = 105$; for $x = 8$, $y = 120$; for $x = 9$, $y = 135$; and for $x = 10$, $y = 150$.

Plotting $y$ against $x$ yields a straight line with a y-intercept of $0$. The gradient of this graph represents the rate of the fine per overdue day, which is $15\,\text{cents/day}$.

When establishing equations for proportional variables, given that $y \propto x$ and $y = 12$ when $x = 4$, the constant $k$ is calculated via:

$12 = k(4) \implies k = 3$

This gives the governing equation $y = 3x$. Using this equation, when $x = 8$, $y = 3(8) = 24$. Conversely, when $y = 21$, solving $21 = 3x$ gives x = 7$.\n\nSimilarly, if y \propto x$and$y = 10$when$x = 2$, the constant$k = \frac{10}{2} = 5$, producing the equation$y = 5x$. When$x = 10$,$y = 5(10) = 50$. When$y = 60$,$60 = 5x \implies x = 12$.

When variable values are presented in tabular form for proportional quantities $q$ and $p$ such that $q \propto p$, the proportionality relationship $q = kp$ is determined from known pairs. Given $p = 5$ and $q = 30$, the constant $k = \frac{30}{5} = 6$, establishing $q = 6p$. The remaining values are calculated as:

When $p = 4$, $q = 6(4) = 24$

When $p = 7$, $q = 6(7) = 42$

When $q = 48$, $48 = 6p \implies p = 8$

When $q = 57$, $57 = 6p \implies p = 9.5$

For two variables $Q$ and $P$ where $Q \propto P$ and $Q = 28$ when $P = 4$, the relation is expressed as $Q = kP$. Substituting gives $28 = k(4) \implies k = 7$, leading to $Q = 7P$. Evaluating for $P = 5$ yields $Q = 7(5) = 35$. Calculating $P$ when $Q = 42$ gives 42 = 7P \implies P = 6$.\n\n# Real-World Applications of Direct Proportion\n\nDirect proportion applies to physical systems such as transport costs, mechanical forces, and electrical circuits.\n\nThe cost \$C$of transporting goods is directly proportional to the distance covered$d\,\text{km}$. If transporting goods over a distance of$60\,\text{km}$costs$\$100:\n\nC = kd\n\n100 = k(60) \implies k = \frac{100}{60} = \frac{5}{3}\n\nThus, the governing equation connecting C$and$d is:\n\nC = \frac{5}{3}d\n\nTo find the cost of transporting goods over 45\,\text{km}:\n\nC = \frac{5}{3}(45) = \$75\n\nIf the cost of transporting goods is \$120$, the distance covered$d is computed as:\n\n120 = \frac{5}{3}d \implies d = \frac{120 \times 3}{5} = 72\,\text{km}\n\nThe graph of C$against$d$is a straight line starting at origin$(0,0)$and passing through$(60, 100).\n\nThe net force F\,\text{N}$required to push a block along a horizontal surface is directly proportional to the block's mass$m\,\text{kg}$. Given that$F = 49$when$m = 5:\n\nF = km\n\n49 = k(5) \implies k = \frac{49}{5} = 9.8\n\nThe equation connecting F$and$m is:\n\nF = \frac{49}{5}m \quad (\text{or } F = 9.8m)\n\nEvaluating the force F$required for a mass$m = 14\,\text{kg} yields:\n\nF = 9.8(14) = 137.2\,\text{N}\n\nCalculating the mass m$when the required force$F = 215.6\,\text{N} yields:\n\n215.6 = 9.8m \implies m = \frac{215.6}{9.8} = 22\,\text{kg}\n\nThe graph of F$against$m$forms a straight line passing through the origin$(0,0).\n\nThe voltage V\,\text{V}$needed to drive a fixed current through a wire is directly proportional to its resistance$R\,\Omega$. Given$V = 9$when$R = 6:\n\nV = kR\n\n9 = k(6) \implies k = \frac{9}{6} = \frac{3}{2}\n\nV = \frac{3}{2}R\n\nFor a resistance R = 15\,\Omega, the voltage required is:\n\nV = \frac{3}{2}(15) = 22.5\,\text{V}\n\nIf the voltage V = 15\,\text{V}$, the corresponding resistance$R is:\n\n15 = \frac{3}{2}R \implies R = \frac{15 \times 2}{3} = 10\,\Omega\n\nThe graph of V$against$R$is a straight line through the origin$(0,0).\n\n# Non-Examples of Direct Proportion and Fixed-Cost Models\n\nA relationship is not directly proportional if its graph does not pass through the origin (0,0), even if it forms a straight line. This occurs in systems governed by a fixed linear equation containing a non-zero y-intercept.\n\nConsider the total monthly cost \$C$of operating a kindergarten, which comprises a fixed base cost of$\$5000$plus a variable cost of$\$41$per enrolled child$n:\n\nC = 5000 + 41n\n\nIf student enrolment n = 80, the total monthly cost is calculated as:\n\nC = 5000 + 41(80) = 5000 + 3280 = \$8280\n\nIf the total monthly running cost is \$7378$, the number of enrolled children$n is found by solving:\n\n7378 = 5000 + 41n\n\n41n = 2378 \implies n = 58\n\nPlotting C$against$n$produces a straight-line graph starting at the y-intercept$(0, 5000)$. Because the line does not pass through the origin$(0,0)$,$C$is not directly proportional to$n.\n\n# Direct Proportion with Powers, Polynomials, and Roots\n\nDirect proportion can exist between a variable and a non-linear expression of another variable, such as squares, cubes, square roots, or polynomials.\n\nIf y$is directly proportional to$x^2$and$y = 18$when$x = 3:\n\ny = kx^2\n\n18 = k(3^2) \implies 18 = 9k \implies k = 2\n\ny = 2x^2\n\nEvaluating y$when$x = 5$gives$y = 2(5^2) = 50$. Evaluating$x$when$y = 32 gives:\n\n32 = 2x^2 \implies x^2 = 16 \implies x = \pm 4\n\nA graph of y$against$x^2$produces a straight line through the origin, whereas a graph of$y$against$x produces a non-linear parabolic curve.\n\nConsider another squared proportion where k \propto h^2$for positive real numbers$h$. Given$k = 81$when$h = 3:\n\nk = ch^2\n\n81 = c(3^2) \implies c = 9 \implies k = 9h^2\n\nUsing this relation to complete missing data:\n\nWhen h = 2$,$k = 9(2^2) = 36\n\nWhen k = 56.25$,$56.25 = 9h^2 \implies h^2 = 6.25 \implies h = 2.5\n\nWhen h = 5$,$k = 9(5^2) = 225\n\nWhen k = 441$,$441 = 9h^2 \implies h^2 = 49 \implies h = 7\n\nIf x$is directly proportional to$y^3$and$x = 32$when$y = 2:\n\nx = ky^3\n\n32 = k(2^3) \implies 8k = 32 \implies k = 4\n\nx = 4y^3\n\nWhen y = 6$,$x = 4(6^3) = 4(216) = 864$. When$x = 108$, solving for$y gives:\n\n108 = 4y^3 \implies y^3 = 27 \implies y = 3\n\nIn geometric scaling laws expressed as y = kx^n:\n\nFor the area y\,\text{m}^2$of a square with side length$x\,\text{m}$,$y = x^2$, so$n = 2.\n\nFor the volume y\,\text{cm}^3$of a cube with side length$x\,\text{cm}$,$y = x^3$, so$n = 3$.

During a specific biological growth phase, the length $L\,\text{cm}$ of an earthworm is directly proportional to the square root of hours since birth $N$. Given $L = 2.5\,\text{cm}$ when $N = 1\,\text{hour}$:

$L = k\sqrt{N}$

$2.5 = k\sqrt{1} \implies k = 2.5$

$L = 2.5\sqrt{N}$

To find the earthworm's length at $N = 4\,\text{hours}$:

$L = 2.5\sqrt{4} = 2.5(2) = 5\,\text{cm}$

To calculate how long it takes for the earthworm to reach a length of $15\,\text{cm}$:

$15 = 2.5\sqrt{N} \implies \sqrt{N} = 6 \implies N = 36\,\text{hours}$

Consider a system where $y \propto x^2$ for all positive values of $y$, and the difference in $y$ when $x = 1$ and $x = 3$ is $32$. Express $y = kx^2$:

$\text{When } x = 1, \quad y_1 = k(1^2) = k$

$\text{When } x = 3, \quad y_2 = k(3^2) = 9k$

$\text{Difference } y_2 - y_1 = 9k - k = 8k$

$8k = 32 \implies k = 4$

Thus, the formula is $y = 4x^2$. Evaluating $y$ when $x = -2$ gives:

$y = 4(-2)^2 = 16$

For a system where $y \propto (2x+1)^2$ and $y = 48$ when $x = 1.5$:

$y = k(2x+1)^2$

$48 = k(2(1.5)+1)^2 = k(3+1)^2 = 16k \implies k = 3$

$y = 3(2x+1)^2$

Evaluating $y$ when $x = 7$ yields:

$y = 3(2(7)+1)^2 = 3(15)^2 = 3(225) = 675$

For a system where $y \propto \sqrt{x}$ and $y = 6$ when $x = 9$:

$y = k\sqrt{x}$

$6 = k\sqrt{9} \implies 3k = 6 \implies k = 2$

$y = 2\sqrt{x}$

When $y = 24$, solving for $x$ gives $24 = 2\sqrt{x} \implies \sqrt{x} = 12 \implies x = 144$. When $x = 49$, y = 2\sqrt{49} = 14$.\n\nIf y \propto (x+1)^2$and$y = 9$when$x = 1:\n\ny = k(x+1)^2\n\n9 = k(1+1)^2 \implies 4k = 9 \implies k = \frac{9}{4}\n\ny = \frac{9}{4}(x+1)^2\n\nEvaluating y$when$x = 3 yields:\n\ny = \frac{9}{4}(3+1)^2 = \frac{9}{4}(16) = 36\n\nFinding all possible values of x$when$y = 81 yields:\n\n81 = \frac{9}{4}(x+1)^2 \implies (x+1)^2 = \frac{81 \times 4}{9} = 36\n\nx + 1 = \pm 6 \implies x = 5 \quad \text{or} \quad x = -7\n\nIf y \propto (x+2)^2$and$y = 32$when$x = 2:\n\ny = k(x+2)^2\n\n32 = k(2+2)^2 \implies 16k = 32 \implies k = 2\n\ny = 2(x+2)^2\n\nEvaluating y$when$x = 5 gives:\n\ny = 2(5+2)^2 = 2(7^2) = 2(49) = 98\n\n# Fundamental Principles and Applications of Inverse Proportion\n\nTwo variables x$and$y$are in inverse proportion if an increase in$x$causes a proportional decrease in$y$, such that their product remains constant. Mathematically, this is expressed as$y \propto \frac{1}{x} or:\n\ny = \frac{k}{x} \quad (\text{or } xy = k)\n\nConsider an electrical copper wire of fixed length whose resistance R\,\Omega$is inversely proportional to the square of its diameter$d\,\text{mm}$. Given$R = 20\,\Omega$when$d = 2.5\,\text{mm}:\n\nR = \frac{k}{d^2}\n\n20 = \frac{k}{2.5^2} \implies 20 = \frac{k}{6.25} \implies k = 125\n\nR = \frac{125}{d^2}\n\nTo find the diameter d$when the resistance$R = 31.25\,\Omega:\n\n31.25 = \frac{125}{d^2} \implies d^2 = \frac{125}{31.25} = 4 \implies d = 2\,\text{mm}\n\nConsider a case where y$is inversely proportional to$2x^2 + 5$, and$y = 7$when$x = 2:\n\ny = \frac{k}{2x^2 + 5}\n\n7 = \frac{k}{2(2)^2 + 5} = \frac{k}{13} \implies k = 91\n\ny = \frac{91}{2x^2 + 5}\n\nEvaluating y$when$x = 8 yields:\n\ny = \frac{91}{2(8)^2 + 5} = \frac{91}{2(64) + 5} = \frac{91}{133}\n\nCalculating the values of x$when$y = 5.2 yields:\n\n5.2 = \frac{91}{2x^2 + 5} \implies 2x^2 + 5 = \frac{91}{5.2} = 17.5\n\n2x^2 = 12.5 \implies x^2 = 6.25 \implies x = \pm 2.5\n\nFor two quantities x$and$y$in inverse proportion where the difference in$y$when$x = 5$and$x = 10$is$100:\n\ny = \frac{k}{x}\n\n\text{When } x = 5, \quad y_1 = \frac{k}{5}\n\n\text{When } x = 10, \quad y_2 = \frac{k}{10}\n\n\frac{k}{5} - \frac{k}{10} = 100 \implies \frac{k}{10} = 100 \implies k = 1000\n\nThe connecting equation is xy = 1000$or$y = \frac{1000}{x}$. When$x = 25$,$y = \frac{1000}{25} = 40$. When$y = 20$,$x = \frac{1000}{20} = 50$.

If $m$ is inversely proportional to the cube of $n$ ($m = \frac{k}{n^3}$) and $m = 36$ for a particular value of $n$, scaling $n$ by halving it ($n_{\text{new}} = \frac{n}{2}$) changes $m$ to:

$m_{\text{new}} = \frac{k}{(n/2)^3} = \frac{k}{n^3/8} = 8 \left(\frac{k}{n^3}\right) = 8 \times 36 = 288$

If $p$ is inversely proportional to $q^2$ ($p = \frac{k}{q^2}$) and $p = 8$ for a specific $q$, increasing $q$ by $100\%$ doubles $q$ ($q_{\text{new}} = 2q$):

$p_{\text{new}} = \frac{k}{(2q)^2} = \frac{k}{4q^2} = \frac{1}{4}(8) = 2$

The percentage change in $p$ is a decrease calculated by:

$\text{Percentage decrease} = \frac{8 - 2}{8} \times 100\% = 75\%$

Advanced Proportion and Work-Rate Problems

Percentage changes in proportional quantities can be analyzed using general structural formulas.

If stored energy $E$ in a stretched spring is directly proportional to the square of its extension $d$ ($E = kd^2$), and extension decreases by $25\%$, the new extension becomes $d_{\text{new}} = 0.75d = \frac{3}{4}d$. The new energy stored is:

$E_{\text{new}} = k\left(\frac{3}{4}d\right)^2 = \frac{9}{16}kd^2$

The percentage decrease in stored energy is:

$\text{Percentage decrease} = \frac{kd^2 - \frac{9}{16}kd^2}{kd^2} \times 100\% = \left(1 - \frac{9}{16}\right) \times 100\% = \frac{7}{16} \times 100\% = 43.75\%$

If the braking distance $d$ of a truck is directly proportional to the square of its speed $x$ ($d = kx^2$), and a speed of $x\,\text{m/s}$ gives a braking distance of $16\,\text{m}$, decreasing speed $x$ by $50\%$ ($x_{\text{new}} = 0.5x$) alters the braking distance to:

$d_{\text{new}} = k(0.5x)^2 = 0.25 kx^2 = 0.25 \times 16 = 4\,\text{m}$

The percentage decrease in braking distance is:

$\text{Percentage decrease} = \frac{16 - 4}{16} \times 100\% = 75\%$

Work-rate problems represent inverse proportional relationships where total worker-time remains constant.

If 5 painters paint a house in 12 days, total workload is $5 \times 12 = 60\,\text{painter-days}$. Hiring 6 painters alters total time taken to:

$\text{Time} = \frac{60}{6} = 10\,\text{days}$

$\text{Time saved} = 12 - 10 = 2\,\text{days}$

If a house must be painted in $d$ days, the required number of painters is expressed as:

$\text{Painters required} = \frac{60}{d}$

For road repair operations, if repairing a $4\,\text{km}$ road takes 8 men working for 12 days, total effort required for $4\,\text{km}$ is $8 \times 12 = 96\,\text{man-days}$, which equals $24\,\text{man-days per kilometer}$.

To repair a $3\,\text{km}$ road, total work required is:

$\text{Work required} = 3 \times 24 = 72\,\text{man-days}$

If 6 men perform the repair, days required $d$ is calculated as:

$d = \frac{72}{6} = 12\,\text{days}$