### Home > CC2 > Chapter Ch1 > Lesson 1.2.6 > Problem1-110

1-110.

Vu needs to add $\frac { 3 } { 8 } + \frac { 5 } { 6 }$.  He knows that he can rewrite each fraction as a portion of $48$ because $6$ and $8$ are each factors of $48$ in his multiplication table.  “Is there a smaller number that could work as a common denominator?” he wonders.

 $\mathbf{1}$ $\mathbf{2}$ $\mathbf{3}$ $\mathbf{4}$ $\mathbf{5}$ $\mathbf{6}$ $\mathbf{7}$ $\mathbf{8}$ $\mathbf{9}$ $\mathbf{10}$ $\mathbf{1}$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\mathbf{2}$ $2$ $4$ $6$ $8$ $10$ $12$ $14$ $16$ $18$ $20$ $\mathbf{3}$ $3$ $6$ $9$ $12$ $15$ $18$ $21$ $24$ $27$ $30$ $\mathbf{4}$ $4$ $8$ $12$ $16$ $20$ $24$ $28$ $32$ $36$ $40$ $\mathbf{5}$ $5$ $10$ $15$ $20$ $25$ $30$ $35$ $40$ $45$ $50$ $\mathbf{6}$ $\mathbf{6}$ $\mathbf{12}$ $\mathbf{18}$ $\mathbf{24}$ $\mathbf{30}$ $\mathbf{36}$ $\mathbf{42}$ $\enclose{circle}{\mathbf{48}}$ $\mathbf{54}$ $\mathbf{60}$ $\mathbf{7}$ $7$ $14$ $21$ $28$ $35$ $42$ $49$ $56$ $63$ $70$ $\mathbf{8}$ $\mathbf{8}$ $\mathbf{16}$ $\mathbf{24}$ $\mathbf{32}$ $\mathbf{40}$ $\enclose{circle}{\mathbf{48}}$ $\mathbf{56}$ $\mathbf{64}$ $\mathbf{72}$ $\mathbf{80}$ $\mathbf{9}$ $9$ $18$ $27$ $36$ $45$ $54$ $63$ $72$ $81$ $90$ $\mathbf{10}$ $10$ $20$ $30$ $40$ $50$ $60$ $70$ $80$ $90$ $100$
1. Where can you look in the multiplication table to see if $6$ and $8$ are each factors of a number less than $48$

Examine the picture and notice how the square is broken up into $25$ square units. If you created a $100$ square unit box and shaded in $40$ of the squares, would that be the same as having four of the above shapes placed together to form a box?

What about a rectangle with $5$ square units and $2$ shaded squares? Can the shape above be rearranged to form 5 rectangles like the one pictured at below?

 tile $\space$ tile $\space$ tile $\space$ shaded tile $\space$ shaded tile $\space$

Each of the fractions are equivalent to each other because of the multiplicative identity as noted in the Math Notes box in this lesson. Reminder: Any number divided by itself is equal to $1$.

$\frac{2}{5}\times\enclose{circle}{\frac{5}{5}}= \frac{10}{25}\rightarrow \frac{10}{25} \times \enclose{circle}{\frac{4}{4}}=\frac{40}{100}$

Consider why using the multiplicative identity changes the numbers of the fractions, but they are all still equivalent.

1. Use the multiplication table on the Lesson 1.2.6 Resource Page to find other number(s) you could use as a common denominator to add $\frac{3}{8}$ and $\frac{5}{6}$.

$\frac{2}{5}\rightarrow\frac{?}{10}$

If the equivalent fraction denominator desired is $10$, then the denominator of $5$ must be multiplied by what number?

Now, consider the multiplicative identity from part (a).
Now that you have solved for the denominator, what must the numerator be?

$\frac{4}{10}$or the Multiplication Table eTool (CPM) to find other number(s) you could use as a common denominator to add $\frac { 3 } { 8 }$ and $\frac { 5 } { 6 }$

1. Vu’s next problem is to add $\frac { 5 } { 4 } + \frac { 7 } { 10 }$ .  Use the multiplication table to find the smallest number you could use as a common denominator.  This number is called the lowest common denominator.  After you find the lowest common denominator for Vu’s problem, find the sum.