Section 9.2

Subsection 9.2.1 Chapter 2 Solutions”

Exercises Exercises

1.

Solution.

Simplify both sides to the same expression, naming the property used at each step.

\begin{align*} x \amp= 2a + 7 - a\\ \amp= (2a - a) + 7 \quad \text{(commutativity and associativity of addition)}\\ \amp= a + 7 \quad \text{(combine like terms: }2a-a=a\text{)} \end{align*}

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\begin{align*} y \amp= 3a - 8 - 2a + 15\\ \amp= (3a - 2a) + (-8 + 15) \quad \text{(commutativity and associativity of addition)}\\ \amp= a + 7 \quad \text{(combine like terms and integer addition }-8+15=7\text{)} \end{align*}

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Therefore, $x = a + 7$ and $y = a + 7\text{,}$ so $x = y$ by the substitution (transitivity) property of equality.

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2.

Solution.

True: $2^0=1=63^0$ (with the standard convention that $a^0=1$ for any nonzero base $a$).

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3.

Solution.

Throughout, use the given definitions $x=2\text{,}$ $y=x+b\text{,}$ $z=c-y+a\text{.}$

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4.

Solution.

\begin{align*} d \amp = x - y + z \quad \text{(given definition of } d\text{)}\\ \amp = 2 - (x + b) + c - y + a \quad \text{(substitute } x=2,\ y=x+b,\ z=c-y+a\text{)}\\ \amp = 2 - x - b + c - y + a \quad \text{(distribute and remove parentheses)}\\ \amp = 2 - 2 - b + c - (x + b) + a \quad \text{(substitute } x=2,\ y=x+b\text{ again)}\\ \amp = -b + c - x - b + a \quad \text{(additive inverses } 2-2=0\text{, then add identity and regroup)}\\ \amp = a - 2b + c - x \quad \text{(commutativity/associativity of addition)}\\ \amp = a - 2b + c - 2 \quad \text{(substitute } x=2\text{)} \end{align*}

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5.

Solution.

$b=15-3(2)=9\text{,}$ $a=3b-4=23\text{,}$ $c=a+14=37\text{.}$ From above, $d=a-2b+c-2=23-18+37-2=40\text{.}$

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6.

Solution.

Let the relation be “logical equivalence” $\leftrightarrow\text{.}$

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Reflexive: $A\leftrightarrow A$ is a tautology.

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Symmetric: $A\leftrightarrow B$ is logically equivalent to $B\leftrightarrow A\text{.}$

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Transitive: From $A\leftrightarrow B$ and $B\leftrightarrow C$ we get $A\leftrightarrow C$ (e.g., by truth table or chaining equivalences).

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7.

Solution.

By definition, $A=B$ iff $A\subseteq B$ and $B\subseteq A\text{.}$

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Reflexive: $A\subseteq A\text{,}$ so $A=A\text{.}$

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Symmetric: If $A=B\text{,}$ then trivially $B=A\text{.}$

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Transitive: If $A=B$ and $B=C\text{,}$ then $A=C$ (subset inclusions chain).

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8.

Solution.

$<\text{:}$ not reflexive (never $a<a$), not symmetric, transitive (always).

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$>\text{:}$ not reflexive, not symmetric, transitive (always).

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$\le\text{:}$ reflexive and transitive (always); symmetry holds only in the special case $a=b$ (so not in general).

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$\ge\text{:}$ reflexive and transitive (always); symmetry only when $a=b$ (so not in general).

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9.

Solution.

Reflexive only: On $\mathbb{R}\text{,}$ define $a\,R\,b$ iff $a=b$ or $(a,b)\in\{(0,1),(1,2)\}\text{.}$ Then reflexive holds; $0R1$ but not $1R0$ (not symmetric); and $0R1\text{,}$ $1R2$ but not $0R2$ (not transitive).

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Transitive only: On $\mathbb{R}\text{,}$ $a\,R\,b$ iff $a<b\text{.}$ Transitive; not reflexive; not symmetric.

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Symmetric only: On $\mathbb{R}\text{,}$ $a\,R\,b$ iff $a\ne b\text{.}$ Symmetric; not reflexive; not transitive (e.g., $0R1$ and $1R2$ but not $0R2$).

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10.

Solution.

False. Choose $a=-3\text{,}$ $b=-2\text{,}$ then $a<b$ but $a/b=1.5$ which is not in $\{x\in\mathbb{Q}\mid x<1,\ |x|<1\}\text{.}$

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True.

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True.

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True.

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False; left set is $\{-2<x<2\}\text{,}$ not just the endpoints.

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True.

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True: the primes less than 10 are $\{2,3,5,7\}\text{.}$

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Subsection 9.2.2 Solutions to “Problem Solving Questions”

Exercises Exercises

1.

Solution.

Equations (for reference):

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\begin{align*} (1)\; x+4-2(3x-5) \amp = \dfrac{x+4}{3}\\ (2)\; axy+bx+4+7y \amp = 46\\ (3)\; 3 \amp = 24 - c\\ (4)\; x^2 + 33x + 33 \amp = 1\\ (5)\; 3 \amp = 4 - 1 \end{align*}

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2.

Solution.

Terms: LHS $x,\,4,\,-2(3x-5)\text{;}$ RHS $\dfrac{x}{3},\,\dfrac{4}{3}\text{.}$

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Terms: LHS $axy,\,bx,\,4,\,7y\text{;}$ RHS $46\text{.}$

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Terms: LHS $3\text{;}$ RHS $24,\,-c\text{.}$

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Terms: LHS $x^2,\,33x,\,33\text{;}$ RHS $1\text{.}$

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Terms: LHS $3\text{;}$ RHS $4,\,-1\text{.}$

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3.

Solution.

Variable term coefficients: $x$ has coefficient $1$ on LHS and $\tfrac{1}{3}$ on RHS; inside $-2(3x-5)$ the $x$-term has coefficient $-6$ after distribution.

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$axy$ has variables $x,y$ with (constant) coefficient $a\text{;}$ $bx$ has variable $x$ with coefficient $b\text{;}$ $7y$ has variable $y$ with coefficient $7\text{.}$

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$-c$ has variable $c$ with coefficient $-1\text{.}$

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$x^2$ has coefficient $1\text{;}$ $x$ has coefficient $33\text{.}$

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No variable terms.

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4.

Solution.

Constants: $4$ and $-2(-5)=10$ after distribution; also $\tfrac{4}{3}$ on RHS.

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Constants: $a,b,4\text{,}$ and $46$ (with $a,b$ declared constants).

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Constants: $3,24\text{.}$

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Constants: $33$ (LHS) and $1$ (RHS).

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Constants: $3,4,1\text{.}$

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5.

Solution.

Linear in $x$ (only first power; distribution preserves linearity).

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Not linear in $x,y$ because of the product $xy\text{.}$

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Linear in $c\text{.}$

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Not linear (quadratic in $x$).

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No variables (a true numeric identity: $3=3$).

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6.

Solution.

Let $w$ be weeks. Equation: $7+4w=63$ so $w=14\text{.}$

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Let $n$ be number of pizzas. Equation: $7.85\,n=219.80$ so $n=28\text{.}$

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Let $r$=bags of Rory’s, $p$=bags of Pete’s. Equations: $47r+39p=493\text{,}$ $380r+400p=4240\text{.}$ Solution: $(r,p)=(8,3)\text{.}$

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7.

Solution.

Equation: $2x+3y=27$ with $x,y\in\mathbb{Z}_{\ge 0}\text{.}$ Modulo $3\text{,}$ we need $2x\equiv 0\pmod 3\text{,}$ so $x\equiv 0\pmod 3\text{.}$ With $0\le x\le 13\text{,}$ $x\in\{0,3,6,9,12\}\text{,}$ giving

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$(x,y)\in\{(0,9),(3,7),(6,5),(9,3),(12,1)\}$ (five ordered pairs). Completeness follows from the congruence condition.

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8.

Solution.

“A straight angle measures $180^\circ\text{.}$ If one angle is $x\text{,}$ the other supplementary angle is $36^\circ\text{:}$ $x+36=180\text{.}$”

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“A base fare of $\$50$ plus $\$3$ per mile totals $\$200\text{:}$ $3x+50=200\text{.}$”

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“Two products are blended: profit or resource constraints yield $18y-3x=90$ and $10y+5x=72$ (e.g., liters or dollars of two goods).”

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9.

Solution.

$2x+y=47\text{,}$ $2x-3y=19$ ⇒ subtract to get $4y=28$ ⇒ $y=7\text{,}$ then $x=20\text{.}$

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$10x-3y=15\text{,}$ $-8x+14y=2$ ⇒ combine to get $58y=70$ ⇒ $y=\tfrac{35}{29}\text{,}$ $x=\tfrac{54}{29}\text{.}$

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From $2y-3x=-44$ get $2y=3x-44\text{.}$ Substitute into $13x+2y=228$ ⇒ $16x=272$ ⇒ $x=17\text{,}$ $y=\tfrac{7}{2}\text{.}$

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$3x=149-10y\text{,}$ $11y=142+4x$ ⇒ $y=14\text{,}$ $x=3\text{.}$

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10.

Solution.

Substitution: the first equation is solved for $x$ in terms of $y\text{,}$ then that expression is substituted into the second equation.

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11.

Solution.

Unique solution (nonzero determinant: $3\cdot7-2\cdot8=5\ne 0$).

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Infinitely many real solutions (two equations in three variables; e.g., pick one free parameter, solve the others consistently).

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Two real solutions: substitute $y=4x-30$ into $3xy+12y=10$ to get $6x^2-21x-185=0$ with positive discriminant $4881\text{;}$ hence two real $x\text{,}$ and then $y=4x-30\text{.}$

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12.

Solution.

Check by substitution. Correct solution: from $19y=7x+27$ get $y=\dfrac{7x+27}{19}\text{.}$ Substitute into $3x-4y=5$ to obtain $29x=203\text{,}$ so $x=7\text{,}$ $y=4\text{.}$ Jonah’s pair is incorrect.

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13.

Solution.

No algebra error: from $14x=2y+20$ we get $y=7x-10\text{;}$ then $4x+3(7x-10)=45$ gives $x=3\text{,}$ $y=11\text{.}$

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Work is correct: from $z=37-4w$ and $3z-6w=3$ we get $w=6\text{,}$ $z=13\text{.}$

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14.

Solution.

Both lines have the same slope and different intercepts: each is of the form $4x-3y=c\text{.}$ Parallel distinct lines do not intersect, so there is no solution.

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Subsection 9.2.3 Solutions to “Explanatory and Critical Thinking Questions”

Exercises Exercises

1.

Solution.

Disagree. Graphing is useful for visual insight and approximate solutions, but exact intersections can be hard to read (non-integer, steep, or nearly parallel lines). Substitution/elimination are more precise and scale to systems with more variables.

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2.

Solution.

Counterexample: $2x+4y=10$ and $x+2y=5$ look different but have exactly the same solution set (the first is twice the second). Different forms can represent the same line/equation after scaling.

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3.

Solution.

The equations are multiples: $6x+4y=12$ is $2\cdot(3x+2y=6)\text{.}$ They represent the same line, so there are infinitely many solutions (all points on that line), not a unique one.

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4.

Solution.

If algebraic steps produce an unexpected identity or contradiction at the wrong time, an error likely occurred. Substituting the found values back into both original equations verifies correctness; any failure indicates a mistake in the manipulation.

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5.

Solution.

“No solution” (inconsistent): e.g., $x+1=x$ or parallel distinct lines; algebra gives a contradiction like $0=1\text{.}$ “Infinitely many”: identical equations (same line/plane), giving an identity like $0=0\text{.}$ “Exactly one”: intersecting, nonparallel lines (nonzero determinant in a $2\times2$ linear system).

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6.

Solution.

Subtracting $2x$ from both sides yields $5=7\text{,}$ a contradiction; there is no $x$ that satisfies it. An inconsistent equation/system has no solutions because its conditions are mutually incompatible.

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7.

Solution.

Let the numbers be $x$ and $y$ with $x+y=30$ and $x=y+4\text{.}$ Substituting gives $y+4+y=30\text{,}$ so $y=13$ and $x=17\text{.}$ Checking: $17+13=30$ and the difference is $4\text{,}$ matching the description.

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