Optional algebra refresher

Use the entry diagnostic first. If the first five tasks are unfamiliar, work through this refresher. If only one operation is uncertain, study that section and its practice. The estimated six hours include writing your own calculations, drawing graphs, correcting mistakes, and explaining why each operation is allowed.

Work on paper before using a calculator. A numerical answer is useful, but the purpose is to recover the relationship that produces it. Keep a short error log: record the incorrect step, the rule it violated, and a corrected example. Revisit that example at the start of your next session.

A fraction $p/q$ describes division by a nonzero denominator. Multiplying its numerator and denominator by the same nonzero value preserves the ratio: $1/2=2/4$. To add fractions, express them using the same denominator; adding denominators does not perform the required operation.

Multiplication gives $(p/q)(r/s)=pr/(qs)$. Division by a nonzero fraction multiplies by its reciprocal: $(3/4)/(2/3)=(3/4)(3/2)=9/8$. The divisor cannot be zero. Cancel factors in products, rather than cancelling terms in sums: $(2x)/(2y)=x/y$ when $y\ne0$, but $(x+2)/(y+2)$ generally does not simplify to $x/y$.

The negative of a negative value is positive. Subtracting a negative adds the corresponding positive amount: $5-(-2)=7$. Parentheses determine whether the minus belongs to a squared base: $-3^2=-9$, but $(-3)^2=9$. For several operations, handle parentheses first, then powers, then multiplication/division, then addition/subtraction. Operations of equal precedence are read in their appropriate order; add parentheses whenever a formula could be misread.

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The results are $5/6$ and $6$. Dividing each numerator term by two gives $x/2+2/2=x/2+1$. The numerator is a sum; deleting the two would not preserve it.

A percentage is a ratio per hundred. Twenty-five percent is $25/100=1/4$. A percentage change compares a difference with the chosen starting value: increasing from eighty to one hundred gives $(100-80)/80=25\%$, while falling from one hundred to eighty is a $20\%$ decrease. The different denominators explain the difference.

Ratios need units and a reference. Three milliseconds per request multiplied by ten requests gives thirty milliseconds under a simple additive timing model. The same arithmetic does not establish a concurrent system’s elapsed time; the model tells us when adding durations is appropriate.

Scientific notation writes a nonzero value as $a\times10^k$, normally with $1\le|a|<10$. Thus $0.00032=3.2\times10^{-4}$ and $45000=4.5\times10^4$. When multiplying, multiply the leading factors and add the powers: $(2\times10^3)(3\times10^{-2})=6\times10^1=60$. This keeps small and large scales readable.

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$4.5\times10^{-3}$; the increase is $(60-50)/50=0.2=20\%$. State which original value supplies the denominator.

Solving an equation means finding values that make both sides equal. Apply the same reversible operation to both sides to preserve its solutions. Adding or subtracting a fixed value is reversible. Multiplying or dividing by a nonzero fixed value is reversible. Dividing by an expression that might be zero needs a separate case; otherwise you may discard a valid solution.

For inequalities, adding the same value preserves the comparison direction. Multiplying or dividing by a positive value also preserves it, but a negative value reverses it. From $-2x<6$, division by $-2$ gives $x>-3$. Check one included value such as zero, and the excluded boundary $-3$.

Squaring is not reversible on all reals: both two and minus two square to four. If a rearrangement squares both sides, verify any candidate in the original equation. For example, $\sqrt{x}=x-2$ requires both $x\ge0$ and $x-2\ge0$; squaring alone can produce a candidate that violates these restrictions.

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$x=6$; $x\le-3$. The product equation permits $x=0$, so dividing by $x$ would remove that valid case. Its full solution set is $\{0,1\}$.

Distributivity is the rule $a(b+c)=ab+ac$. Applying it to each term gives $(x+2)(x-3)=x^2-3x+2x-6=x^2-x-6$. Combine like terms only: $x$ and $x^2$ represent different powers and cannot be added into a single coefficient of $x$.

Factoring reverses expansion. A common factor gives $3x+6=3(x+2)$. The polynomial $x^2-x-6$ factors as $(x-3)(x+2)$ because minus three and two have sum minus one and product minus six. Expand your factorisation to check it.

If a product of real numbers is zero, at least one factor is zero. Thus $(x-3)(x+2)=0$ has solutions $x=3$ and $x=-2$. A factor can be cancelled in a fraction only where it is nonzero. The expression $(x^2-1)/(x-1)$ agrees with $x+1$ for $x\ne1$; it remains undefined at one unless a new function is deliberately defined there.

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Expansion gives $x^2+2x+1$. At one the original expression is four while $x^2+1$ is two; the missing cross term causes the discrepancy.

For nonzero $a$, $a^0=1$ and $a^{-k}=1/a^k$. With positive bases, $a^r a^s=a^{r+s}$ and $(a^r)^s=a^{rs}$. Distinguish $x^2$ from $2^x$: the variable occupies a different role. Real square roots require nonnegative inputs; a root symbol gives the nonnegative square root, so $\sqrt9=3$, while solving $x^2=9$ gives two answers.

For $a>0$, $a\ne1$, and $x>0$, $\log_a x$ is the exponent that makes $a$ produce $x$. Thus $\log_2 8=3$ because $2^3=8$. Solving $2^x=8$ gives $x=3$. Logarithms turn positive products into sums; they do not generally distribute over addition. Module 01 explains these domains and identities in detail.

Trigonometry is a preview for later geometry and calculus. An angle in radians measures arc length divided by the circle’s radius. A half-turn is $\pi$ radians, or $180$ degrees. On the unit circle, an angle $\theta$ gives coordinates $(\cos\theta,\sin\theta)$, satisfying $\cos^2\theta+\sin^2\theta=1$. At zero the point is $(1,0)$; at $\pi/2$ it is $(0,1)$. Python’s standard trigonometric functions use radians. No trigonometric calculation is required for Module 01’s labs.

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$1/8$; the equation has $x=3$ and $x=-3$. The logarithm requires $x-2>0$, hence $x>2$.

An ordered pair $(x,y)$ names a point: horizontal coordinate first, vertical second. A function graph places $(x,f(x))$ for allowed inputs. To read a value, find the input on the horizontal axis and follow to the curve, then read the vertical coordinate. Check the axis units and displayed range before interpreting a shape.

The slope through points $(x_1,y_1)$ and $(x_2,y_2)$ with distinct horizontal coordinates is $(y_2-y_1)/(x_2-x_1)$. Through $(1,2)$ and $(3,6)$ it is $(6-2)/(3-1)=2$. A corresponding line is $y=2x$. For $y=2x+1$, the slope is still two, but the vertical intercept is one.

Draw a line by calculating two or three points. For $y=2x+1$, use $(0,1),(1,3),(2,5)$. Draw the square using $(-2,4),(-1,1),(0,0),(1,1),(2,4)$; it is curved, so joining just two points with a straight line would misrepresent it. A vertical line has no finite slope under this formula because its horizontal difference is zero.

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The slope is $(5-1)/(2-0)=2$. Changing units on one axis changes the numerical ratio; state the units before interpreting it physically.

The first six tasks are hand calculations, the next two concern graphs, and the final two repair invalid manipulations. Rework missed tasks with new values until you can justify each step.

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$9/12-8/12=1/12$; multiplying by the reciprocal gives $9/8$.

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$-9,9,7$; the parentheses determine whether the negative sign is squared.

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Subtract five, divide by three: $x=4$; substituting gives seventeen.

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$x^2-x-6=(x+2)(x-3)$. The cross terms are $-3x+2x=-x$.

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$x=3$; $3.2\times10^{-4}$. Moving the decimal four places left corresponds to a negative power of ten.

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$x>-3$, since dividing by a negative reverses the inequality. The increase is $20/80=25\%$.

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The slope is two and the line is $y=2x$; both points satisfy it.

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The points are $(-2,4),(-1,1),(0,0),(1,1),(2,4)$. The value at zero is zero. Opposite inputs square to the same value, giving symmetry about the vertical axis.

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Consider $x=0$ first; it is valid. For $x\ne0$, division gives $x=1$. Together the solutions are zero and one.

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Factor the numerator as $(x-1)(x+1)$; cancellation is valid only for $x\ne1$. The original expression is undefined at one. The simplified expression describes equal values only on the original domain.

Repeat the first five diagnostic tasks without consulting their answers. Explain the denominator in fraction addition, the placement of a negative sign, each reversible equation step, the cross terms in an expansion, and the inverse relationship defining a logarithm. Then start Module 01.

For extra practice, use the single-variable calculus course’s preparation materials selectively; its later calculus is not required yet. The refresher’s worked examples and tasks are original.