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Remove the brackets from algebraic expressions. Part of HELM Book 1.3

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Removing Brackets from expressions $a(b+c)$ and $a(b-c)$

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Removing brackets means multiplying out. For example

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\\[5(2+4) = 5\\times 2+5\\times 4 = 10+20 = 30.\\]

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In this simple example we could alternatively get the same result as follows:

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\\[5(2 + 4) = 5 \\times 6 = 30.\\]

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That is:

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\\[5(2 + 4) = 5 \\times 2 + 5 \\times 4\\]

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In an expression such as $5(x+y)$ it is intended that the $5$ multiplies both $x$ and $y$ to produce $5x+5y$.

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Thus the expressions $5(x + y)$ and $5x + 5y$ are equivalent. In general we have the following rules known as distributive laws:

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Key Point 14

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\\(a(b+c)=ab+ac\\)

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\\(a(b-c)=ab-ac\\)

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Note that when the brackets are removed both terms in the brackets are multiplied by $a$.

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As we have noted above, if you insert numbers instead of letters into these expressions you will see that both left and right hand sides are equivalent. For example

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$4(3 + 5)$ has the same value as $4(3) + 4(5)$; that is $32$

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and

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$7(8 - 3)$ has the same value as $7(8) - 7(3)$; that is $35$

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Example 31

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Remove the brackets from (a) $9(2+y),\\quad$ (b) $9(2y)$

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Solution

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(a) In the expression $9(2 + y)$ the $9$ must multiply both terms in the brackets:

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\\[ \\begin{align*} 9(2+y)&=9(2)+9(y) \\\\ &=18+9y \\end{align*} \\]

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(b) Recall that $9(2y)$ means $9 \\times (2 \\times y)$ and that when multiplying numbers together the presence of brackets is irrelevant. Thus $9(2y) = 9 \\times 2 \\times y = 18y$

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The crucial distinction between the role of the factor $9$ in the two expressions $9(2 + y)$ and $9(2y)$ in Example 31 should be noted.

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Example 32

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Remove the brackets from $9(x+2y)$.

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Solution

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In the expression $9(x+2y)$ the $9$ must multiply both the $x$ and the $2y$ in the brackets. Thus

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\\[ \\begin{align*} 9(x+2y) &= 9x + 9(2y)\\\\ &= 9x + 18 y \\end{align*} \\]

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", "advice": "

Task 1:

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$\\var{q1expr}=\\var{q1ans}$

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Task 2:

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$\\var{q2expr}=\\var{q2ans}$

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Task 3:

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$\\var{q3expr}=\\var{q3ans}$

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Task 4:

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$\\var{q4expr}=\\var{q4ans}$

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Remove the brackets from $\\var{q1expr}$

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Example 33

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Remove the brackets from $-3(5x-z)$.

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Solution

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The number $-3$ must multiply both the $5x$ and the $z$.

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\\[ \\begin{align*} -3(-5x-z) &= (-3)(5x) - (-3)(z)\\\\ &= -15x+3z \\end{align*} \\]

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Remove the brackets from $\\var{q2expr}$.

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Remember to type * when two variables are multipled together.

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Example 34

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Remove the brackets from $-(3x+1)$.

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Solution

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Although the $1$ is unwritten, the minus sign outside the brackets stands for $-1$. We must therefore consider the expression $-(3x+1)$.

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\\[ \\begin{align*} -1(3x-1) &= (-1)(3x) + (-1)(1)\\\\ &= -3x + (-1) \\\\ &= -3x -1 \\end{align*} \\]

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Example 35

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Remove the brackets from the expression $5x-(3x+1)$ and simplify the result by collecting like terms.

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Solution

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The brackets in $-(3x+1)$ were removed in Example 34.

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\\[ \\begin{align*} 5x-(3x+1) &= 5x -1(3x+1)\\\\ &= 5x-3x-1\\\\&= 2x-1 \\end{align*} \\]

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Example 36

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Show that $\\displaystyle{\\frac{-x-1}{4}}$, $\\displaystyle{\\frac{-(x+1)}{4}}$ and $\\displaystyle {-\\frac{x+1}{4}}$ are all equivalent expressions.

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Solution

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Consider $-(x+1)$. Removing the brackets we obtain $-x-1$ and so \\[\\frac{-x-1}{4} = \\frac{-(x+1)}{4}\\]

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A negative quantity divided by a positive quantity will be negative. Hence \\[ \\frac{-(x+1)}{4} = -\\frac{x+1}{4} =\\]

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You should study all three expressions carefully to recognise the variety of equivalent ways in which we can write an algebraic expression.

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Sometimes the bracketed expression can appear on the left, as in (a + b)c. To remove the brackets here we use the following rules:

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Key Point 15

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\\[(a+b)c = ac + bc\\]

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\\[(a-b)c = ac-bc\\]

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\n

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Note that when the brackets are removed both the terms in the brackets multiply $c$.

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Example 37

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Remove the brackets from $(2x+3y)x$.

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Solution

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Both terms in the brackets multiply the $x$ outside. Thus

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\\[ \\begin{align*} (2x+3y)x &= 2x(x) + 3y(x) \\\\ &= 2x^2 + 3yx \\end{align*} \\]

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