Two expressions involving fractions are equivalent if they are the same.

Even if two expressions initially appear different, they might still be equivalent. For example, the expressions \dfrac{1}{2}x and \dfrac{x}{2} both represent the same thing: "half of x. " Consequently, they are equivalent.

We can also see that these expressions are equivalent because \dfrac{1}{2}x can be transformed into \dfrac{x}{2}. Writing x as \dfrac{x}{1}, we can multiply \dfrac 1 2 and \dfrac x 1 using fraction multiplication, as follows:

\begin{align*} \dfrac{1}{2} x &= \dfrac{1}{2} \cdot x \\[5pt] &= \dfrac{1}{2} \cdot \dfrac{x}{1} \\[5pt] &= \dfrac {1 \cdot x}{2\cdot 1}\\[5pt] &= \dfrac{x}{2} \end{align*}

FLAG

Find an expression that's equivalent to \dfrac{2}3 a.

EXPLANATION

Writing a as \dfrac{a}{1}, we have

\begin{align*} \dfrac 23\, a &= \dfrac 23\cdot a\\[5pt] &= \dfrac 23\cdot \dfrac a1\\[5pt] &= \dfrac {2\cdot \color{} a}{3\cdot \color{} 1}\\[5pt] &= \dfrac{2a}{3}. \end{align*}

Thus, the expressions \dfrac{2}{3}a and \dfrac{2a}{3} are equivalent.

FLAG

$\dfrac 13 x$ is equivalent to

a
$\dfrac {1}{3x}$
b
$\dfrac 3x$
c
$\dfrac x3$
d
$3x$
e
$x^3$

$\dfrac 25 y$ is equivalent to

a
$y^2$
b
$y^5$
c
$\dfrac{5}{2}y$
d
$\dfrac{2y}{5}$
e
$\dfrac{2}{5y}$

Find an expression equivalent to -\dfrac{2}9 (-z).

EXPLANATION

First, remember that the product of two negative numbers is a positive number. Thus,

-\dfrac{2}9 (-z) = \dfrac{2}{9} z.

Now, writing z as \dfrac{z}{1}, we have

\begin{align*} \dfrac{2}{9} z &= \dfrac{2}{9} \cdot z \\[5pt] &= \dfrac{2}{9} \cdot \dfrac{z}{1} \\[5pt] &= \dfrac{2 \cdot z}{9 \cdot 1} \\[5pt] &= \dfrac{2z}{9}. \end{align*}

Thus, the expressions -\dfrac{2}9 (-z) and \dfrac{2z}{9} are equivalent.

FLAG

$-\dfrac 65\, {(-h)} $ is equivalent to

a
$\dfrac{6h}{30}$
b
$\dfrac{6h}{5}$
c
$\dfrac{6+h}{5}$
d
$-\dfrac{-6-h}{5}$
e
$-\dfrac{6}{5h}$

$\dfrac 18\, {(-f)} $ is equivalent to

a
$-\dfrac{1}{8f}$
b
$\dfrac{1-f}{8}$
c
$-8f$
d
$\dfrac{f+1}{8}$
e
$-\dfrac{f}{8}$

Find an expression equivalent to \dfrac 27\, {(y-3)}

EXPLANATION

Writing y-3 as \dfrac{y-3}{1}, we have

\begin{align*} \dfrac 27\, {(y-3)} &= \dfrac 27\cdot (y-3) \\[5pt] &= \dfrac 27\cdot \dfrac{y-3}1\\[5pt] &= \dfrac{2\cdot (y-3) }{7\cdot 1}\\[5pt] &= \dfrac{2y-6}{7}\, . \end{align*}

FLAG

$\dfrac 13\, {(k+1)} $ is equivalent to

a
$\dfrac{k}{3}$
b
$\dfrac{k}{4}$
c
$\dfrac{k+3}{3}$
d
$\dfrac{k+2}{3}$
e
$\dfrac{k+1}{3}$

The expression $\dfrac 54\, {(m+1)} $ is equivalent to

a
$\dfrac{m+1}{20}$
b
$\dfrac{5m+1}{4}$
c
$\dfrac{m+5}{4}$
d
$\dfrac{5(m+1)}{4}$
e
$\dfrac{m+1}{9}$

The usual procedure also works in the reverse direction. For example, to find an equivalent expression for \dfrac{2(y+1)}5, we can separate the variable part from the rest of the fraction, as follows:

\begin{align*} \dfrac{2(y+1)}5 &= \dfrac{2\cdot (y+1)}{5\cdot 1}\\[5pt] &= \dfrac{2}{5}\cdot \dfrac{(y+1)}{1} \\[5pt] &= \dfrac{2}5\cdot (y+1)\\[5pt] &= \dfrac{2}5\, (y+1) \end{align*}

FLAG

Find an expression equivalent to \dfrac{2(3-x)}{5}.

EXPLANATION

Pulling out the variable part, we have

\begin{align*} \dfrac {2(3-x)}{5} &= \dfrac{2\cdot (3-x)}{5\cdot 1}\\[5pt] &= \dfrac{2}{5}\cdot\dfrac{3-x}{1}\\[5pt] &= \dfrac{2}5\cdot (3-x)\\[5pt] &= \dfrac{2}5(3-x)\,. \end{align*}

FLAG

Find an expression equivalent to $\dfrac{3(a-7)}{2}.$

a
$\dfrac{1}{2}(a-7)$
b
$\dfrac{3}2(a-7)$
c
$3a - \dfrac{7}{2}$
d
$\dfrac{2}3(a-7)$
e
$\dfrac{3}{2}a-7$

Find an expression equivalent to $\dfrac{k+2}{3}.$

a
$3(k+2)$
b
$\dfrac{3}{k+2}$
c
$\dfrac{1}{3}(k+2)$
d
$\dfrac{3(k+2)}{1}$
e
$\dfrac{1}{3(k+2)}$
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