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Question

In the following exercise, identify which expressions are equivalent to the given expression.

12x+3+12x\frac{1}{2} x+3+\frac{1}{2} x

a. 12(x+3)\frac{1}{2}(x+3)

b. x+3x+3

c. 3x+3x3 x+3-x

Solution

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TRY x=2\underline{TRY\ x= 2}

12x+3+12x=12(2)+3+12(2)=1+3+1=4+1=5{\color{#4257b2} \dfrac{1}{2}x+3+\dfrac{1}{2}x } = \dfrac{1}{2}(2)+3+\dfrac{1}{2}(2) =1+3+1=4+1={\color{#c34632}5}

a) 12(x+3)=12((1)+3)=12(1+3)=12(4)=2a)\ {\color{#4257b2} \dfrac{1}{2}\left(x+3 \right) } = \dfrac{1}{2}\left((1)+3 \right)= \dfrac{1}{2}\left(1+3 \right)=\dfrac{1}{2}\left(4 \right) ={\color{#c34632}2}

b) x+3=(2)+3=2+3=5b)\ {\color{#4257b2} x+3 } = (2)+3=2+3={\color{#c34632}5}

c) 3x+3x=3(2)+3(2)=6+32=92=7c)\ {\color{#4257b2} 3x+3-x } = 3(2)+3-(2)=6+3-2=9-2 ={\color{#c34632}7}

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