0.4: Properties of Algebra
- Page ID
- 45024
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In algebra, we will often need to simplify an expression. There are three basic forms of simplifying which we will discuss in this section.
The term “Algebra” comes from the Arabic word al-jabr which means “reunion.” It was first used in Iraq in 830 AD by Mohammad ibn-Musa al-Khwarizmi.
An algebraic expression consists of coefficients, variables, and terms. Given an algebraic expression, a
- coefficient is the number in front of the variable.
- variable is a letter representing any number.
- term is a product of a coefficient and variable(s).
For example, \[t \qquad 2x \qquad 3st \qquad 7x^2 \qquad 5ab^3c \nonumber\] are all examples of terms because each is a product of a coefficient and variable(s).
Evaluating expressions
The first form of simplifying expressions is evaluating expressions. Given particular values for each variable, we can simplify the expression by replacing the variables with its corresponding values.
Evaluate \(p(q + 6)\) when \(p = 3\) and \(q = 5\).
Solution
\[\begin{array}{rl} p (q + 6) & \text{Replace }p\text{ with }3\text{ and }q\text{ with }5 \\ (3) ((5) + 6) & \text{Evaluate parenthesis} \\ (3) (11) & \text{Multiply} \\ 33 & \text{Result} \end{array}\nonumber\]
Whenever we replace a variable, we will put the new number inside a set of parenthesis. Notice the 3 and 5 in Example \(\PageIndex{1}\) are in parenthesis. This is to preserve operations that are sometimes lost in a simple replacement. Sometimes the parenthesis won’t make a difference, but it is a good habit to always use them to prevent potential future arithmetic errors.
Evaluate \(x + z x (3 - z) \left( \dfrac{x}{3} \right)\) when \(x = - 6\) and \(z = - 2\).
Solution
\[\begin{array}{rl} x + z x (3 - z) \left(\dfrac{x}{3} \right) & \text{Replace }x\text{ with }6\text{ and }z\text{ with }2 \\ && \\ (- 6) + (- 2) (- 6) (3 - (- 2)) \left( \dfrac{(- 6)}{3} \right) & \text{Evaluate parenthesis} \\ && \\ - 6 + (5) (- 2) & \text{Multiply left to right} \\ - 6 + {12 (5)} (- 2) & \text{Multiply left to right} \\ - 6 + {60 (- 2)} & \text{Multiply} \\ {- 6 - 120} & \text{Subtract} \\ - 126 & \text{Result} \end{array}\nonumber\]
Like terms
It is common in the study of Algebra that the values of the variables are unknown. In this case, we simplify by combining like terms.
Two terms are like terms if the base variable(s) and exponent on each variable are identical.
For example, \(3 x^2 y \text{and} - 7 x^2 y\) are like terms because they both contain the same base variables, \(x\) and \(y\), and the exponents on \(x\) (the \(x\) is squared on both terms) and \(y\) are the same.
If two terms are like terms, we add (or subtract) the coefficients, then keep the variables (and exponents on the corresponding variable) the same.
Simplify: \(5 x - 2 y - 8 x + 7 y\)
Solution
\[\begin{array}{rl} 5 x - 2 y - 8 x + 7 y & \text{Combine like terms }5x-8x\text{ and }-2y+7y \\ - 3 x + 5 y & \text{Result} \end{array}\nonumber\]
Simplify: \(8 x^2 - 3 x + 7 - 2 x^2 + 4 x - 3\)
Solution
\[\begin{array}{rl} 8 x^2 - 3 x + 7 - 2 x^2 + 4 x - 3 & \text{Combine like terms }8x^2-2x^2\text{ and }-3x+4x\text{ and} \\ &7-3 \\ 6 x^2 + x + 4 & \text{Result} \end{array}\nonumber\]
As we combine like terms, we interpret subtraction signs as part of the following term. Hence, if we see a subtraction sign, we treat the following term as a negative term.
Notice, when we write the simplified result, it is common practice to write the expression in standard form, terms written with descending exponents. E.g., looking at the result in Example \(\PageIndex{4}\), we wrote \(6 x^2 + x + 4\), where the \(x^2\) term is written first since it is the largest exponent and then the \(x\) term. We always write the term with just the coefficient at the end, e.g., \(4\).
Distribution
The final method for simplifying algebraic expressions is distribution. Many times we are given algebraic expressions with sets of parenthesis and terms directly in front of the expressions (as product). By using the distributive property, we can rewrite the expression without parenthesis.
Property. The distributive property is a product between one term and a sum or difference of two or more terms: \[a(b + d) = a\cdot b + a \cdot d\nonumber \]
Simplify: \(4 (2 x - 7)\)
Solution
\[\begin{array}{rl} 4 (2 x - 7) & \text{Multiply each term by }4 \\ \textcolor{blue}{4} \cdot 2x - \textcolor{blue}{4} \cdot 7 & \text{Simplify} \\ 8 x - 28 & \text{Result} \end{array}\nonumber\]
Simplify: \(- 7 (5 x - 6)\)
Solution
\[\begin{array}{rl} - 7 (5 x - 6) & \text{Multiply each term by }-7 \\ \textcolor{blue}{(-7)} \cdot 5x - \textcolor{blue}{(-7)} \cdot 6 & \text{Simplify} \\ - 35x+ 42 & \text{Result} \end{array}\nonumber\]
In the previous example, we use the fact that the sign is attached with the number, i.e., we treat the \(- 6\) as a negative number: \((-7)(-6) = 42\), a positive number. The most common error in using the distributive property is a sign (negatives) error. Be very careful with your signs!
It is possible to distribute a negative through parenthesis. When there is a negative in front of parenthesis, we can think of the negative as a \(- 1\). We don’t always write it, but we know it’s there. Then we distribute the \(- 1\) as usual.
Simplify \(- (4 x - 5 y + 6)\)
Solution
\[\begin{array}{rl} - (4 x - 5 y + 6) & \text{Negative can be thought of as }-1 \\ \textcolor{blue}{-1} (4 x - 5 y + 6) & \text{Multiply each term by } \textcolor{blue}{- 1} \\ \textcolor{blue}{(-1)}4x - \textcolor{blue}{(- 1)} 5y + \textcolor{blue}{(- 1)} 6 & \text{Simplify} \\ \textcolor{blue}{-} 4 x \textcolor{blue}{+} 5 y \textcolor{blue}{-} 6 & \text{Result} \end{array}\nonumber\]
Putting it all together
Distributing through parenthesis and combining like terms can be combined into one problem. Order of operations implies multiplication (distribute) first, then add or subtract (combine like terms). Thus, we first distribute and then combine like terms.
Simplify: \(5 + 3 (2 x - 4)\)
Solution
\[\begin{array}{rl} 5 + 3 (2 x - 4) & \text{Distribute} \\ 5 + 6 x - 12 & \text{Combine like terms} \\ - 7 + 6 x & \text{Rewrite in standard form} \\ 6x - 7 & \text{Result} \\ \end{array}\nonumber\]
Simplify: \(3 x - 2 (4 x - 5)\)
Solution
\[\begin{array}{rl} 3 x - 2 (4 x - 5) & \text{Distribute} \\ 3 x - 8 x + 10 & \text{Combine like terms} \\ - 5 x + 10 & \text{Result} \end{array}\nonumber\]
Simplify: \(2 (5 x - 8) - 6 (4 x + 3)\)
Solution
\[\begin{array}{rl} 2 (5 x - 8) - 6 (4 x + 3) & \text{Distribute} \\ 10 x - 16 - 24 x - 18 & \text{Combine like terms} \\ - 14 x - 34 & \text{Result} \end{array}\nonumber\]
Simplify: \(4 (3 x - 8) - (2 x - 7)\)
Solution
\[\begin{array}{rl} 4 (3 x - 8) - (2 x - 7) & \text{Treat the negative as a } \textcolor{blue}{- 1} \\ 4 (3 x - 8) \textcolor{blue}{- 1} (2 x - 7) & \text{Distribute} \\ 12 x - 32 - 2 x + 7 & \text{Combine like terms} \\ 10 x - 25 & \text{Result} \end{array}\nonumber\]
Properties of Algebra Homework
Evaluate each expression given the values for each variable.
\(p + 1 + q − m;\quad m = 1, p = 3, q = 4\)
\(p − \frac{pq}{6};\quad p = 6\text{ and }q = 5\)
\(c^2 − (a − 1);\quad a = 3\text{ and }c = 5\)
\(5j + \frac{kh}{2};\quad h = 5, j = 4, k = 2\)
\(\frac{4 − (p − m)}{2} + q;\quad m = 4, p = 6, q = 6\)
\(m + n + m + \frac{n}{2};\quad m = 1\text{ and }n = 2\)
\(q − p − (q − 1 − 3);\quad p = 3, q = 6\)
\(y^2 + y − z;\quad y = 5, z = 1\)
\(\frac{6 + z − y}{3};\quad y = 1, z = 4\)
\(x + 6z − 4y;\quad x = 6, y = 4, z = 4\)
\(5(b + a) + 1 + c;\quad a = 2, b = 6, c = 5\)
\(z + x − (1^2)^3;\quad x = 5, z = 4\)
\(3 + z − 1 + y − 1;\quad y = 5, z = 4\)
\(p + (q − r)(6 − p);\quad p = 6, q = 5, r = 5\)
\(y − [4 − y − (z − x)];\quad x = 3, y = 1, z = 6\)
\(4z − (x + x − (z − z))\quad x = 3, z = 2\)
\(k \times 3^2 − (j + k) − 5;\quad j = 4, k = 5\)
\(zx −\left(z-\frac{4+x}{6}\right);\quad x = 2, z = 6\)
\(a^3 (c^2 − c);\quad a = 3, c = 2\)
\(5 + qp + pq − q;\quad p = 6, q =3\)
Simplify.
\(r − 9 + 10\)
\(n+n\)
\(8v + 7v\)
\(−7x − 2x\)
\(k − 2 + 7\)
\(x − 10 − 6x + 1\)
\(m − 2m\)
\(9n − 1 + n + 4\)
\(−4x + 2 − 4\)
\(4b + 6 + 1 + 7b\)
\(−x + 8x\)
\(−7a − 6 + 5\)
\(−8p + 5p\)
\(1 − 10n − 10\)
\(1 − r − 6\)
\(−4b + 9b\)
\(−8(x − 4)\)
\(8n(n + 9)\)
\(7k(−k + 6)\)
\(−6(1 + 6x)\)
\(8m(5 − m)\)
\(−9x(4 − x)\)
\(−9b(b − 10)\)
\(−8n(5 + 10n)\)
\(3(8v + 9)\)
\(−(−5 + 9a)\)
\(10x(1 + 2x)\)
\(−2(n + 1)\)
\(−2p(9p − 1)\)
\(4(8n − 2)\)
\(−4(1 + 7r)\)
\(2x(8x − 10)\)
\(9(b + 10) + 5b\)
\(−3x(1 − 4x) − 4x^2\)
\(−4k^2 − 8k(8k + 1)\)
\(1 − 7(5 + 7p)\)
\(−10 − 4(n − 5)\)
\(4(x + 7) + 8(x + 4)\)
\(−8(n + 6) − 8n(n + 8)\)
\(7(7 + 3v) + 10(3 − 10v)\)
\(2n(−10n + 5) − 7(6 − 10n)\)
\(5(1 − 6k) + 10(k − 8)\)
\((8n^2 − 3n) − (5 + 4n^2)\)
\((5p − 6) + (1 − p)\)
\((2 − 4v^2) + (3v^2 + 2v)\)
\((4 − 2k^2) + (8 − 2k^2)\)
\((x^2 − 8) + (2x^2 − 7)\)
\(4v − 7(1 − 8v)\)
\(−8x + 9(−9x + 9)\)
\(−9 − 10(1 + 9a)\)
\(−10(x − 2) − 3\)
\(−6(5 − m) + 3m\)
\(−2r(1 + 4r) + 8r(−r + 4)\)
\(9(6b + 5) − 4b(b + 3)\)
\(−7(4x − 6) + 2(10x − 10)\)
\(−3(4 + a) + 6a(9a + 10)\)
\(−7(4x + 3) − 10(10x + 10)\)
\((7x^2 − 3) − (5x^2 + 6x)\)
\((3x^2 − x) − (7 − 8x)\)
\((2b − 8) + (b − 7b^2)\)
\((7a^2 + 7a) − (6a^2 + 4a)\)
\((3 − 7n^2) + (6n^2 + 3)\)