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1.5: Algebraic Expressions

  • Page ID
    19853
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    The associative property of multiplication is valid for all numbers.

    Associative Property of Multiplication

    Let \(a\), \(b\), and \(c\) be any numbers. Then: \[a \cdot(b \cdot c)=(a \cdot b) \cdot c \nonumber \]

    The associative property of multiplication is useful in a number of situations.

    Example \(\PageIndex{1}\)

    Simplify: \(-3(4y)\).

    Solution

    Currently, the grouping \(-3(4y)\) demands that we first multiply \(4\) and \(y\). However, we can use the associative property of multiplication to regroup, first multiplying \(-3\) and \(4\).

    \[\begin{aligned} -3(4 y)=&(-3 \cdot 4) y \quad \color{Red} \text{ The associative property of multiplication.}\\ &=-12 y \quad \color{Red} \text { Multiply: }-3 \cdot 4=-12 \end{aligned} \nonumber \]

    Thus, \(-3(4y)=-12y\).

    Exercise \(\PageIndex{1}\)

    Simplify: \(2(3 x)\).

    Answer

    \(6x\)

    Let’s look at another example.

    Example \(\PageIndex{2}\)

    Simplify: \(-2(-4 x y)\).

    Solution

    Currently, the grouping \(-2(-4xy)\) demands that we first multiply \(-4\) and \(xy\). However, we can use the associative property of multiplication to regroup, first multiplying \(-2\) and \(-4\).

    \[\begin{aligned}-2(-4 x y) &=(-2 \cdot(-4)) x y \quad \color{Red} \text { The associative property of multiplication. } \\ &=8 x y \quad \color{Red} \text { Multiply: } -2 \cdot (-4)=8 \end{aligned} \nonumber \]

    Thus, \(-2(-4xy)=8xy\).

    Exercise \(\PageIndex{2}\)

    Simplify: \(-3\left(-8 u^{2}\right)\).

    Answer

    \(24 u^{2}\)

    In practice, we can move quicker if we perform the regrouping mentally, then simply write down the answer. For example:

    \[-2(-4 t)=8 t \quad \text { and } \quad 2\left(-5 z^{2}\right)=-10 z^{2} \quad \text { and } \quad-3\left(4 u^{3}\right)=-12 u^{3} \nonumber \]

    The Distributive Property

    We now discuss a property that couples addition and multiplication. Consider the expression \(2 \cdot(3+5)\). The Rules Guiding Order of Operations require that we first simplify the expression inside the parentheses.

    \[\begin{aligned} 2 \cdot(3+5) &=2 \cdot 8 \quad \color{Red} \text { Add: } 3+5=8 \\ &=16 \quad \color{Red} \text { Multiply: } 2 \cdot 8=16 \end{aligned} \nonumber \]

    Alternatively, we can instead distribute the \(2\) times each term in the parentheses. That is, we will first multiply the \(3\) by \(2\), then multiply the \(5\) by \(2\). Then we add the results.

    \[\begin{aligned} 2 \cdot(3+5) &=2 \cdot 3+2 \cdot 5 \quad \color{Red} \text { Distribute the 2. }\\ &=6+10 \quad \color{Red} \text { Multiply: } 2 \cdot 3=6 \text { and } 2 \cdot 5=10\\ &=16 \quad \color{Red} \text { Add: } 6+10=16 \end{aligned} \nonumber \]

    Note that both methods produce the same result, namely 16. This example demonstrates an extremely important property of numbers called the distributive property.

    The Distributive Property

    Let \(a\), \(b\), and \(c\) be any numbers. Then: \[a \cdot(b+c)=a \cdot b+a \cdot c \nonumber \] That is, multiplication is distributive with respect to addition.

    Example \(\PageIndex{3}\)

    Use the distributive property to expand \(2(3x + 7)\).

    Solution

    First distribute the \(2\) times each term in the parentheses. Then simplify.

    \[\begin{aligned} 2(3 x+7)&=2(3 x)+2(7) \quad \color{Red} \text { Use the distributive property. } \\ &=6 x+14 \quad \color{Red} \text { Multiply: } 2(3 x)=6 x \text { and } 2(7)=14 \end{aligned} \nonumber \]

    Thus, \(2(3 x+7)=6 x+14\).

    Exercise \(\PageIndex{3}\)

    Expand: \(5(2 y+7)\).

    Answer

    \(10y+35\)

    Multiplication is also distributive with respect to subtraction.

    Example \(\PageIndex{4}\)

    Use the distributive property to expand \(-2(5y-6)\).

    Solution

    Change to addition by adding the opposite, then apply the distributive property.

    \[\begin{aligned} -2(5 y-6)&=-2(5 y+(-6)) \quad \color{Red} \text { Add the opposite. } \\ &=-2(5 y)+(-2)(-6) \quad \color{Red} \text { Use the distributive property. } \\ &=-10 y+12 \quad \color{Red} \text { Multiply: } -2(5 y)=-10 y \text { and }(-2)(-6)=12 \end{aligned} \nonumber \]

    Thus, \(-2(5 y-6)=-10 y+12\)

    Exercise \(\PageIndex{4}\)

    Expand: \(-3(2z-7)\).

    Answer

    \(-6 z+21\)

    Speeding Things Up a Bit

    In Example \(\PageIndex{4}\), we changed the subtraction to addition, applied the distributive property, then several steps later we were finished. However, if you understand that subtraction is really the same as adding the opposite, and if you are willing to do a few steps in your head, you should be able to simply write down the answer immediately following the given problem.

    If you look at the expression \(-2(5y-6)\) from Example \(\PageIndex{4}\) again, only this time think “multiply \(-2\) times \(5y\), then multiply \(-2\) times \(-6\), then the result is immediate. \[-2(5y-6) = -10y + 12 \nonumber \]

    Let’s try this “speeding it up” technique in a couple more examples.

    Example \(\PageIndex{5}\)

    Use the distributive property to expand \(-3(-2 x+5 y-12)\).

    Solution

    To distribute the \(-3\), we simply think as follows: “\(-3(-2x)=6x\), \(-3(5y)=-15y\), and \(-3(-12) = 36\).” This sort of thinking allows us to write down the answer immediately without any additional steps. \[-3(-2 x+5 y-12)=6 x-15 y+36 \nonumber \]

    Exercise \(\PageIndex{5}\)

    Expand: \(-3(-2 a+3 b-7)\).

    Answer

    \(6 a-9 b+21\)

    Example \(\PageIndex{6}\)

    Use the distributive property to expand \(-5(-2 a-5 b+8)\).

    Solution

    To distribute the \(-5\), we simply think as follows: “\(-5(-2a) = 10a\), \(-5(-5b) = 25 b\), and \(-5(8) = -40\).” This sort of thinking allows us to write down the answer immediately without any additional steps. \[-5(-2 a-5 b+8)=10 a+25 b-40 \nonumber \]

    Exercise \(\PageIndex{6}\)

    Expand: \(-4(-x-2 y-7)\).

    Answer

    \(4 x+8 y+28\)

    Distributing a Negative Sign

    Recall that negating a number is equivalent to multiplying the number by \(-1\).

    Multiplicative Property of Minus One

    If \(a\) is any number, then: \[(-1)a = -a \nonumber \]

    This means that if we negate an expression, it is equivalent to multiplying the expression by \(-1\).

    Example \(\PageIndex{7}\)

    Expand \(-(7 x-8 y-10)\).

    Solution

    First, negating is equivalent to multiplying by \(-1\). Then we can change subtraction to addition by “adding the opposite” and use the distributive property to finish the expansion.

    \[\begin{aligned} -(7 x-8 y-10) &=-1(7 x-8 y-10) \quad \color{Red} \text { Negating is equivalent to multiplying by } -1 \\ &=-1(7 x+(-8 y)+(-10)) \quad \color{Red} \text { Add the opposite. } \\ &=-1(7 x)+(-1)(-8 y)+(-1)(-10) \quad \color{Red} \text { Distribute the } -1 \\ &=-7 x+8 y+10 \quad \color{Red} \text { Multiply.} \end{aligned} \nonumber \]

    Thus, \(-(7 x-8 y-10)=-7 x+8 y+10\)

    Exercise \(\PageIndex{7}\)

    Expand: \(-(-a-2 b+11)\).

    Answer

    \(a+2 b-11\)

    While being mathematically precise, the technique of Example \(\PageIndex{7}\) can be simplified by noting that negating an expression surrounded by parentheses simply changes the sign of each term inside the parentheses to the opposite sign.

    Once we understand this, we can simply “distribute the minus sign” and write:

    \[-(7 x-8 y-10)=-7 x+8 y+10 \nonumber \]

    In similar fashion,

    \[-(-3 a+5 b-c)=3 a-5 b+c \nonumber \]

    and,

    \[-(-3 x-8 y+11)=3 x+8 y-11 \nonumber \]

    Combining Like Terms

    We can use the distributive property to distribute a number times a sum. \[a(b+c)=a b+a c \nonumber \]

    However, the distributive property can also be used in reverse, to “unmultiply” or factor an expression. Thus, we can start with the expression \(ab + ac\) and “factor out” the common factor a as follows:

    \[a b+a c=a(b+c) \nonumber \]

    You can also factor out the common factor on the right.

    \[a c+b c=(a+b) c \nonumber \]

    We can use this latter technique to combine like terms.

    Example \(\PageIndex{8}\)

    Simplify: \(7 x+5 x\).

    Solution

    Use the distributive property to factor out the common factor \(x\) from each term, then simplify the result.

    \[\begin{aligned} 7x+5x&=(7+5)x \quad \color{Red} \text { Factor out an } x \text { using the distributive property. } \\ &=12x \quad \color{Red} \text { Simplify: } 7+5=12 \end{aligned} \nonumber \]

    Thus, \(7x +5x = 12x\).

    Exercise \(\PageIndex{8}\)

    Simplify: \(3 y+8 y\).

    Answer

    \(11y\)

    Example \(\PageIndex{9}\)

    Simplify: \(-8 a^{2}+5 a^{2}\).

    Solution

    Use the distributive property to factor out the common factor \(a^2\) from each term, then simplify the result.

    \[\begin{aligned} -8 a^{2}+5 a^{2}&=(-8+5) a^{2} \quad \color{Red} \text { Factor out an } a^{2} \text { using the distributive property. } \\ &=-3 a^{2} \quad \color{Red} \text { Simplify: }-8+5=-3 \end{aligned} \nonumber \]

    Thus, \(-8 a^{2}+5 a^{2}=-3 a^{2}\).

    Exercise \(\PageIndex{9}\)

    Simplify: \(-5 z^{3}+9 z^{3}\).

    Answer

    \(4z^3\)

    Examples \(\PageIndex{8}\) and \(\PageIndex{9}\) combine what are known as “like terms.” Examples \(\PageIndex{8}\) and \(\PageIndex{9}\) also suggest a possible shortcut for combining like terms.

    Like Terms

    Two terms are called like terms if they have identical variable parts, which means that the terms must contain the same variables raised to the same exponents.

    For example, \(2x^2y\) and \(11x^2y\) are like terms because they contain identical variables raised to the same exponents. On the other hand, \(-3st^2\) and \(4s^2t\) are not like terms. They contain the same variables, but the variables are not raised to the same exponents.

    Consider the like terms \(2x^2y\) and \(11x^2y\). The numbers \(2\) and \(11\) are called the coefficients of the like terms. We can use the distributive property to combine these like terms as we did in Examples \(\PageIndex{8}\) and \(\PageIndex{9}\), factoring out the common factor \(x^2y\).

    \[\begin{aligned} 2 x^{2} y+11 x^{2} y &=(2+11) x^{2} y \\ &=13 x^{2} y \end{aligned} \nonumber \]

    However, a much quicker approach is simply to add the coefficients of the like terms, keeping the same variable part. That is, \(2 + 11 = 13\), so:

    \[2 x^{2} y+11 x^{2} y=13 x^{2} y \nonumber \]

    This is the procedure we will follow from now on.

    Example \(\PageIndex{10}\)

    Simplify: \(-8 w^{2}+17 w^{2}\).

    Solution

    These are like terms. If we add the coefficients \(-8\) and \(17\), we get \(9\). Thus:

    \[-8 w^{2}+17 w^{2}=9 w^{2} \quad \color{Red} \text{Add the coefficients and repeat the variable part.} \nonumber \]

    Exercise \(\PageIndex{10}\)

    Simplify: \(4 a b-15 a b\).

    Answer

    \(-11ab\)

    Example \(\PageIndex{11}\)

    Simplify: \(-4 u v-9 u v\).

    Solution

    These are like terms. If we add \(-4\) and \(-9\), we get \(-13\). Thus:

    \[-4 u v-9 u v=-13 u v \quad \color{Red} \text{Add the coefficients and repeat the variable part.} \nonumber \]

    Exercise \(\PageIndex{11}\)

    Simplify: \(-3 x y-8 x y\).

    Answer

    \(-11xy\)

    Example \(\PageIndex{12}\)

    Simplify: \(-3 x^{2} y+2 x y^{2}\)

    Solution

    These are not like terms. They do not have the same variable parts. They do have the same variables, but the variables are not raised to the same exponents. Consequently, this expression is already simplified as much as possible.

    \[-3 x^{2} y+2 x y^{2} \quad \color{Red} \text{Unlike terms. Already simplified.} \nonumber \]

    Exercise \(\PageIndex{12}\)

    Simplify: \(5ab+11bc\).

    Answer

    \(5ab+11bc\)

    Sometimes we have more than just a single pair of like terms. In that case, we want to group together the like terms and combine them.

    Example \(\PageIndex{13}\)

    Simplify: \(-8 u-4 v-12 u+9 v\).

    Solution

    Use the associative and commutative property of addition to change the order and regroup, then combine line terms.

    \[\begin{aligned} -8u-4v-12u+9v &=(-8u-12u)+(-4v+9v) \quad \color{Red} \text { Reorder and regroup. } \\ &=-20u+5v \quad \color{Red} \text { Combine like terms. } \end{aligned} \nonumber \]

    Note that \(-8u-12u =-20u\) and \(-4v +9v =5 v\).

    Alternate solution

    You may skip the reordering and regrouping step if you wish, simply combining like terms mentally. That is, it is entirely possible to order your work as follows:

    \[-8 u-4 v-12 u+9 v=-20 u+5 v \quad \color{Red} \text {Combine like terms.} \nonumber \]

    Exercise \(\PageIndex{13}\)

    Simplify: \(-3 z^{2}+4 z-8 z^{2}-9 z\).

    Answer

    \(-11 z^{2}-5 z\)

    In Example \(\PageIndex{13}\), the “Alternate solution” allows us to move more quickly and will be the technique we follow from here on, grouping and combining terms mentally.

    Order of Operations

    Now that we know how to combine like terms, let’s tackle some more complicated expressions that require the Rules Guiding Order of Operations.

    Rules Guiding Order of Operations

    When evaluating expressions, proceed in the following order.

    1. Evaluate expressions contained in grouping symbols first. If grouping symbols are nested, evaluate the expression in the innermost pair of grouping symbols first.
    2. Evaluate all exponents that appear in the expression.
    3. Perform all multiplications and divisions in the order that they appear in the expression, moving left to right.
    4. Perform all additions and subtractions in the order that they appear in the expression, moving left to right.

    Example \(\PageIndex{14}\)

    Simplify: \(4(-3 a+2 b)-3(4 a-5 b)\).

    Solution

    Use the distributive property to distribute the \(4\) and the \(-3\), then combine like terms.

    \[\begin{aligned} 4(-3a+2b)-3(4a-5b) &=-12a+8b-12a+15b \quad \color{Red} \text { Distribute. } \\ &=-24a+23b \quad \color{Red} \text { Combine like terms. } \end{aligned} \nonumber \]

    Note that \(-12a-12a =-24a\) and \(8b + 15b=23b\)

    Exercise \(\PageIndex{14}\)

    Simplify: \(-2x-3(5-2x)\).

    Answer

    \(4 x-15\)

    Example \(\PageIndex{15}\)

    Simplify: \(-2(3 x-4 y)-(5 x-2 y)\).

    Solution

    Use the distributive property to multiply \(-2\) times \(3x-4y\), then distribute the minus sign times each term of the expression \(5x-2y\). After that, combine like terms.

    \[\begin{aligned} -2(3x-4y)-(5x-2y) &=-6x+8y-5x+2y \quad \color{Red} \text { Distribute. } \\ &=-11x+10y \quad \color{Red} \text { Combine like terms. } \end{aligned} \nonumber \]

    Note that \(-6x-5x =-11x\) and \(8y +2y = 10 y\).

    Exercise \(\PageIndex{15}\)

    Simplify: \(-3(u+v)-(u-5 v)\).

    Answer

    \(-4 u+2 v\)

    Example \(\PageIndex{16}\)

    Simplify: \(-2\left(x^{2} y-3 x y^{2}\right)-4\left(-x^{2} y+3 x y^{2}\right)\).

    Solution

    Use the distributive property to multiply \(-2\) times \(x^2y-3xy^2\) and \(-4\) times \(-x^2y +3xy^2\). After that, combine like terms.

    \[\begin{aligned}-2\left(x^{2} y-3 x y^{2}\right)-4\left(-x^{2} y+3 x y^{2}\right) &=-2 x^{2} y+6 x y^{2}+4 x^{2} y-12 x y^{2} \\ &=2 x^{2} y-6 x y^{2} \end{aligned} \nonumber \]

    Note that \(-2 x^{2} y+4 x^{2} y=2 x^{2} y\) and \(6 x y^{2}-12 x y^{2}=-6 x y^{2}\).

    Exercise \(\PageIndex{16}\)

    Simplify: \( 8 u^{2} v-3\left(u^{2} v+4 u v^{2}\right)\).

    Answer

    \(5 u^{2} v-12 u v^{2}\)

    When grouping symbols are nested, evaluate the expression inside the innermost pair of grouping symbols first.

    Example \(\PageIndex{17}\)

    Simplify: \(-2 x-2(-2 x-2[-2 x-2])\).

    Solution

    Inside the parentheses, we have the expression \(-2 x-2[-2 x-2]\). The Rules Guiding Order of Operations dictate that we should multiply first, expanding \(-2[-2 x-2]\) and combining like terms.

    \[\begin{aligned} -2x-2({\color{Red}-2x-2[-2x-2]}) &=-2x-2({\color{Red}-2x+4x+2}) \\ &=-2x-2({\color{Red}2x+2}) \end{aligned} \nonumber \]

    In the remaining expression, we again multiply first, expanding \(-2(2x+2)\) and combining like terms.

    \[\begin{aligned} &=-2x-4x-4\\ &=-6x-4 \end{aligned} \nonumber \]

    Exercise \(\PageIndex{17}\)

    Simplify: \(x-2[-x+4(x+1)]\).

    Answer

    \(-5 x-8\)

    Contributors


    This page titled 1.5: Algebraic Expressions is shared under a CC BY-NC-ND 3.0 license and was authored, remixed, and/or curated by David Arnold via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.