from fractions import Fraction
result = -Fraction(3, 4) + Fraction(5, 6) / Fraction(-10, 9)
resultFraction(-3, 2)
Advanced problems can fail for a basic reason. A correct plan for a parameter problem may collapse because of a sign error; a calculus derivation may fail because fractions were combined incorrectly; a model may be numerically correct but meaningless because units were ignored. The remedy is not to repeat an entire elementary course. It is to identify the exact weak link, repair it, and verify the repair on a parallel problem.
This chapter uses a diagnostic cycle:
\[ \text{attempt}\longrightarrow\text{classify}\longrightarrow \text{repair}\longrightarrow\text{reattempt}\longrightarrow\text{reflect}. \]
A diagnostic is not an entrance barrier and its total score is not a measure of mathematical potential. It is evidence about present fluency. Honest evidence lets you spend time where it produces the largest gain.
After completing the chapter, you should be able to:
Complete How to read and write an algebraic argument first. In particular, use \(=\) only for equality, preserve exact values, and mark steps that produce candidates rather than equivalent equations.
0.3333333333, what exact number might have produced it, and what can the display alone prove?Set aside 45–60 uninterrupted minutes. Use paper, but do not use notes, calculator, search, or computer algebra on the first attempt. Write enough work to reveal your method. If you are unsure, record a best attempt and mark your confidence as high, medium, or low.
The structured diagnostic near the end contains 24 questions in six strands:
| Questions | Strand | Evidence sought |
|---|---|---|
| 1–4 | exact rational arithmetic | grouping, signs, common denominators, exact form |
| 5–8 | powers, roots, factors | laws with conditions, square roots, divisibility |
| 9–12 | ratios, percentages, units | multiplicative comparison and dimensional meaning |
| 13–16 | elementary equations and order | inverse operations, inequalities, checks |
| 17–20 | coordinates and graphs | scale, slope, intercepts, qualitative reading |
| 21–24 | estimation and interpretation | magnitude, plausibility, exact versus approximate |
After submitting, score each strand separately. A result of 3 or 4 correct means “continue, then retrieve again this week.” A result of 2 means “complete the strand repair below.” A result of 0 or 1 means “complete the repair slowly, practice five additional examples, and reassess before Unit 1.” Confidence also matters: a correct low-confidence answer belongs in retrieval practice.
Use one of these categories for every missed or guessed problem.
Do not classify from the final wrong line. Mark the first line that would need to change for the remainder to become valid.
A learner computes
\[ \frac23-\frac58=\frac{-3}{5}. \]
This is not merely an arithmetic slip. The learner subtracted numerators and denominators as if fraction bars were independent components. That is an algebra/representation error: unlike units were combined. The repair is to rename both fractions in fifteenths:
\[ \frac23-\frac58=\frac{16}{24}-\frac{15}{24}=\frac1{24}. \]
A useful reattempt changes the surface form, for example \(5/6-7/10\), while testing the same common-unit idea.
A fraction \(a/b\) with \(b\ne0\) is a number, not a pair of independent digits. Addition compares common-sized parts:
\[ \frac ab+\frac cd=\frac{ad+bc}{bd},\qquad b,d\ne0. \]
Division asks how many copies of one number fit in another, so multiplying by a reciprocal is derived from the inverse property:
\[ \frac ab\div\frac cd=\frac ab\cdot\frac dc, \qquad b,c,d\ne0. \]
Evaluate
\[ -\frac34+\frac56\div\left(-\frac{10}{9}\right). \]
Division comes before addition:
\[ \frac56\div\left(-\frac{10}{9}\right) =\frac56\left(-\frac9{10}\right) =-\frac{45}{60}=-\frac34. \]
Therefore the total is \(-3/4-3/4=-3/2\). A decimal check gives \(-1.5\), consistent with adding two negative numbers of magnitude \(0.75\).
Compute \(\frac7{12}-\frac58\) exactly, then estimate to check the sign.
The least common denominator is \(24\): \(14/24-15/24=-1/24\). Since \(7/12\approx0.583\) and \(5/8=0.625\), a small negative answer is plausible.
Mastery signal: four mixed fraction problems correct in succession, with unsimplified denominators and signs checked rather than guessed.
A leading negative sign means multiplication by \(-1\). Thus \(-(a-b)=-a+b\), not \(-a-b\). On a number line, \(-7<-2\) because \(-7\) lies farther left, even though its absolute value is larger.
Multiplying an inequality by a negative reverses order. If \(a<b\), then \(b-a>0\); multiplying by \(-1\) gives \(a-b<0\). This is the reason, not a memorized symbol trick.
Simplify \(-2[3-(x-4)]\).
First preserve the inner grouping:
\[ 3-(x-4)=3-x+4=7-x. \]
Then distribute \(-2\) to the entire bracket:
\[ -2(7-x)=-14+2x. \]
At \(x=0\), both the original and final expressions equal \(-14\), a quick check that catches the common answer \(-14-2x\).
Solve \(-3(2x-1)\le9\) and test one allowed and one excluded value.
Expand: \(-6x+3\le9\), so \(-6x\le6\). Division by \(-6\) reverses order, giving \(x\ge-1\). The boundary \(x=-1\) yields \(9\le9\); the excluded value \(x=-2\) yields \(15\le9\), false.
For nonzero \(a\) and integers \(m,n\),
\[ a^ma^n=a^{m+n},\qquad \frac{a^m}{a^n}=a^{m-n},\qquad a^{-n}=\frac1{a^n}. \]
The product and quotient laws require a common base. Powers do not distribute over addition: \((a+b)^2=a^2+2ab+b^2\). Over the reals, \(\sqrt{u}\) denotes the nonnegative number whose square is \(u\), so \(u\ge0\) and \(\sqrt{x^2}=|x|\).
Scientific notation writes a nonzero number as \(c\times10^n\) with \(1\le|c|<10\). It separates significant size from decimal placement.
Compute
\[ \frac{(3\times10^{-4})(8\times10^7)}{6\times10^2}. \]
Separate coefficients and powers:
\[ \frac{3\cdot8}{6}\times10^{-4+7-2} =4\times10^1=40. \]
The exponent estimate \(-4+7-2=1\) predicts tens, so \(40\) has a plausible scale.
Simplify \((12x^5y^{-2})/(18x^{-1}y)\) for nonzero \(x,y\).
Reduce coefficients and subtract exponents:
\[ \frac{12}{18}x^{5-(-1)}y^{-2-1} =\frac23x^6y^{-3} =\frac{2x^6}{3y^3}. \]
The conditions \(x,y\ne0\) justify the negative exponents and original quotient.
A ratio compares quantities multiplicatively. A rate carries units, such as kilometers per hour. A percentage change is measured relative to the original value:
\[ \text{relative change}=\frac{\text{new}-\text{old}}{\text{old}}. \]
An increase of \(20\%\) multiplies by \(1.20\); a later decrease of \(20\%\) multiplies by \(0.80\). The combined factor is \(0.96\), not \(1\).
Dimensional analysis treats units as algebraic factors. It catches impossible additions and missing conversions.
A vehicle travels \(180\) kilometers in \(2.5\) hours. Its average speed is
\[ \frac{180\ \mathrm{km}}{2.5\ \mathrm h}=72\ \mathrm{km/h}. \]
Converting to meters per second,
\[ 72\frac{\mathrm{km}}{\mathrm h} \cdot\frac{1000\ \mathrm m}{1\ \mathrm{km}} \cdot\frac{1\ \mathrm h}{3600\ \mathrm s} =20\ \mathrm{m/s}. \]
The cancelled units are part of the reasoning.
A price rises from \(80\) to \(92\), then receives a \(10\%\) discount. Find the final price and the overall percentage change from the original.
The discount multiplier is \(0.90\), so the final price is \(92(0.90)=82.80\). The overall change is \(2.80/80=0.035=3.5\%\) increase. Adding \(15\%-10\%=5\%\) would be wrong because the changes use different bases.
Solving means describing every value in the stated universe that makes the original statement true. Reversible operations preserve the solution set; nonreversible operations create candidates that need checks.
For a proportion \(a/b=c/d\), cross multiplication is shorthand for multiplying by \(bd\). It requires \(b,d\ne0\). For a linear equation with a parameter, isolate the coefficient before dividing and split the zero case.
Solve \(k(x-2)=3k\) for real \(x\).
If \(k\ne0\), division by \(k\) gives \(x-2=3\), so \(x=5\). If \(k=0\), the equation is \(0=0\) for every real \(x\). Thus
\[ S_k=\begin{cases} \mathbb R,&k=0,\\ \{5\},&k\ne0. \end{cases} \]
The classification is more complete than the unqualified answer \(x=5\).
Solve \(\frac{x-1}{3}-\frac{x+2}{4}=2\).
Multiply both sides by \(12\): \(4(x-1)-3(x+2)=24\). Expanding gives \(4x-4-3x-6=24\), so \(x=34\). Check: \(33/3-36/4=11-9=2\).
A point \((x,y)\) records an input and output. The slope between distinct points \((x_1,y_1)\) and \((x_2,y_2)\) is
\[ m=\frac{y_2-y_1}{x_2-x_1}. \]
Read axis labels and scale before estimating. A steep-looking line can have a small numerical slope if the axes use different units. A graph suggests behavior on its window; it does not prove exact roots or global behavior.
A tank contains \(120\) liters at minute \(2\) and \(84\) liters at minute \(8\). The average rate of change is
\[ \frac{84-120}{8-2}=-6\ \mathrm{liters/minute}. \]
The negative sign means volume decreased as time increased. A linear extrapolation would predict \(132\) liters at minute \(0\), but that conclusion is a model assumption, not a fact supplied by two data points.
A line passes through \((-2,5)\) and \((4,-7)\). Find its slope and equation, then check both points.
The slope is \((-7-5)/(4-(-2))=-12/6=-2\). Using \(y=mx+b\) with \((-2,5)\) gives \(5=4+b\), so \(b=1\) and \(y=-2x+1\). Substitution gives \(5\) and \(-7\) at the stated inputs.
Two learners with 16 correct answers may need entirely different repairs. One may miss all unit questions while another misses sign and order questions. Record strand scores and first causes; the total is only a summary.
Rereading -13/36 does not repair fraction addition. Close the solution and perform the same operation on a new pair of fractions. A transfer problem—not recognition of the old answer—is the evidence of repair.
An immediate parallel problem can measure short-term imitation. Insert another activity or a delay, then retrieve the method without the model in view. Repeat again later in mixed practice.
A decimal output may conceal a wrong denominator, a lost exact form, or an incorrect unit. Write the mathematical operation first, estimate its scale, and then use a tool for independent verification.
Early repair should prioritize a correct, explainable method. Add time pressure only after several accurate attempts. A fast invalid rule becomes harder—not easier—to unlearn through repetition.
This diagnostic samples prerequisites; it cannot prove mastery of every possible school-algebra task. Low confidence, a long pause, or an unexplained guess is useful evidence even when the selected answer is correct.
Record each answer and confidence before submitting. If website controls are not available, the generated notebook provides the same questions and collapsed solutions.
For every missed or low-confidence item, use this six-column record:
| Field | Required entry |
|---|---|
| Problem ID | stable question identifier |
| First attempt | enough work to reveal the original method |
| First invalid step | exact line or decision where validity was lost |
| Error class | one category from Section 2 |
| Corrected argument | complete work, not only the corrected answer |
| Transfer check | a new problem testing the same idea |
A repair is complete only when the transfer check is solved without copying the worked example. Schedule another one-question retrieval after two or three days. This delay tests learning rather than short-term imitation.
Problem. Decrease \(200\) by \(15\%\).
First attempt. \(200-15=185\).
First invalid step. The percent sign was treated as 15 units rather than \(15/100\) of the original quantity.
Class. Representation error.
Correction. \(200(1-0.15)=200(0.85)=170\).
Transfer check. Increase \(240\) by \(12\%\): \(240(1.12)=268.8\).
Reflection. Translate a percentage to a multiplier before performing the arithmetic.
Complete these after the targeted repairs, preferably on a later day.
Compute \(-\frac79+\frac5{12}\) exactly.
Simplify \(-[2x-(3-x)]\).
Simplify \((6a^{-2}b^4)(3a^5b^{-1})\) for nonzero \(a,b\).
Convert \(3.6\times10^5\) millimeters to kilometers.
A quantity falls from \(250\) to \(215\). Find its percentage decrease.
Solve \(-4<2x+6\le12\) and write interval notation.
Solve \(m(x+1)=m^2\) for \(x\), classifying all real \(m\).
A line through \((1,8)\) has slope \(-3\). Find where it crosses each axis and interpret the sign of the slope.
Without a calculator, place \(\sqrt{70}\) between consecutive tenths.
A value is increased by \(p\%\) and then decreased by \(p\%\). Prove that the result is below the original for every \(0<p<100\), and find the percentage loss in terms of \(p\).
Let the original positive value be \(V\). The increase multiplies it by \(1+p/100\) and the decrease multiplies the result by \(1-p/100\). Thus
\[ V\left(1+\frac p{100}\right)\left(1-\frac p{100}\right) =V\left(1-\frac{p^2}{10000}\right). \]
For \(0<p<100\), we have \(0<p^2/10000<1\), so the multiplier is positive and strictly below \(1\). The loss is \(V p^2/10000\), which is \(p^2/100\) percent of the original. The proof explains why equal percentage changes do not cancel: the second percentage uses a different base.
Positive numbers \(a\) and \(b\) satisfy \(a/b=3/5\). Without assigning arbitrary values, determine
\[ \frac{2a+3b}{4a-b} \]
and state why the denominator is nonzero.
Plan. A ratio determines a common scale. Write \(a=3t\) and \(b=5t\) for some \(t>0\).
Execution. Substitution gives
\[ \frac{2(3t)+3(5t)}{4(3t)-5t} =\frac{21t}{7t}=3. \]
Because \(t>0\), the denominator \(7t\) is nonzero and cancellation is valid.
Why this method. Choosing \(a=3,b=5\) would produce the right number but would not explain why every positive pair with the ratio works. Introducing the scale \(t\) converts an example into a general argument.
Use these if a strand score remains below three. Select at least two from the weak strand and one mixed problem.
Evaluate \(5-2[3-(4-9)]\).
Compute \(7/15\div(-14/25)\).
Write \(0.000072\) in scientific notation.
Explain why \((-3)^4\) and \(-3^4\) differ.
A recipe ratio is \(3\) cups flour to \(2\) cups liquid. Scale it to \(7.5\) cups flour.
Convert \(54\) kilometers per hour to meters per second.
Solve \(3(x-4)-2(x+1)=5\) and check.
Find the slope through \((-3,-2)\) and \((5,10)\).
Find the number that becomes \(126\) after a \(12\%\) increase.
Determine all real \(q\) for which \(q/(q-2)>0\) using sign reasoning.
A graph’s horizontal axis is seconds and vertical axis is meters. Explain the units and meaning of a secant slope of \(-4.2\).
Prove that if \(0<a<b\), then \(1/a>1/b\), without relying on decimal examples.
Return to argument language. For each claim below, state whether the displayed relation should be \(\Rightarrow\) or \(\Leftrightarrow\), and name conditions.
Proceed to Sets, intervals, statements, and quantifiers. If one strand remains below mastery, interleave its repair exercises with the next chapter rather than postponing all progress.
Use Python to verify a completed correction, not to skip the first attempt. The first cell preserves exact fractions; the second compares an exact value with a floating approximation.
Fraction(-3, 2)
The output Fraction(-3, 2) is exact. A binary floating-point result would be an approximation even if its printed digits looked simple.
(sqrt(70), 8.36660026534, (6889/100, 1764/25))
The squared endpoints \(68.89\) and \(70.56\) prove \(8.3<\sqrt{70}<8.4\); the decimal is a convenient report after the bound has already explained its plausibility.
For a portfolio, store only problem IDs, classifications, and your own written corrections. Do not build code that exposes solutions before an honest attempt or uploads raw answers without informed consent.