2.6: Exercise 2
- Page ID
- 13972
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)State the property or indicate that no property is applicable.
- \(5+x=x+5\)
- \(7(a-2)=7a-7(2)\)
- \(5(x\cdot 2)=(5x)(2)\)
- \((y+2)+4=4+(y+2)\)
- \(6\cdot (t\cdot 2)=(6\cdot t)\cdot (6\cdot 2)\)
- \(w-5=5-w\)
- \(xy=yx\)
- \(a\div \pi=\pi \div a\)
- \((-4.7)+[(6.5)+(\sqrt{2})]=[(-4.7)+(6.5)]+(\sqrt{2})\)
- \(x-4=4-x\)
- Factor \(12x+18\)
- Factor \(24x^2+12x\)
The solutions follow:
Keep them covered up till you have worked out each of the problems above.
- \(5+x=x+5\)
Solution:
\(5+x=x+5\) illustrates the commutative property of addition. The terms commuted to each other’s position. - \(7(a-2)=7a-7(2)\)
Solution:
\(7(a-2)=7a-7(2)\) illustrates the distributive property of multiplication over addition. - \(5(x\cdot 2)=(5x)(2)\)
Solution:
\(5(x\cdot 2)=(5x)(2)\) illustrates the associative property of multiplication. - \((y+2)+4=4+(y+2)\)
Solution:
\((y+2)+4=4+(y+2)\) illustrates the commutative property of addition. \((y+2)\) and 4 have commuted to each other’s position. \(y\) remains grouped with 2.N regrouping took place.
- \(6\cdot (t\cdot 2)=(6\cdot t)\cdot (6\cdot 2)\)
Solution:
\(6\cdot (t\cdot 2)=(6t)\cdot (6\cdot 2)\) does not illustrate any of the mentioned properties. You cannot distribute multiplication over multiplication. - \(w-5=5-w\)
Solution:
\(w-5=5-w\) does not illustrate any of the mentioned properties. Subtraction is not commutative.
We shall learn later that \(w-5=w+(-5)=(-5)+w\) because addition is commutative. - \(xy=yx\)
Solution:
\(xy=yx\) illustrates the commutative property of multiplication. The terms commuted to each other’s position. - \(a\div \pi=\pi \div a\)
Solution:
\(a\div \pi=\pi \div a\) does not illustrate any of the mentioned properties. Division is not commutative. - \((-4.7)+[(6.5)+(\sqrt{2})]=[(-4.7)+(6.5)]+(\sqrt{2})\)
Solution:
\((-4.7)+[(6.5)+(\sqrt{2})]=[(-4.7)+(6.5)]+(\sqrt{2})\) illustrates the associative property of addition. \(6.5\) is associated (grouped) with \(\sqrt{2}\) on the left and \(-4.7\) on the right. - \(x-4=4-x\)
Solution:
\(x-4=4-x\) does not illustrate any of the mentioned properties. Subtraction is not commutative. - Factor \(12x+18\)
Solution:
\(\begin{array}{rcl lll} 12x+18&=&(3)(4)x+(3)(6)\\ &=&3(4x+6)\\ \end{array}\)A quick check: \(3(4x+6)=3(4x)+3(6)=12x+18\)Gthe result grees wiht the iginal statement.
- Factor \(24x^2+12x\)
Solution:
\(\begin{array}{rcl lll} 24x^2+12x&=&2(12)x\ x+1(12)x\\ &=&12x(2x+1)\\ \end{array}\)
Note the \(1\) besides the \(12\).A quick check: \(12x(2x+1)=12x(2x)+12x(1)=24x^2+12x\)