How do you write $\frac{5 b + 9 y}{6 b} - \frac{5 b + 6 y}{2 b} + 1$ $\frac { 5b + 9y } { 6b } - \frac { 5b + 6y } { 2b } + 1$ as a single fraction?

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Answer 1 · 2018-03-24T00:45:15

You may want to use $6 b$ $6b$ as a common denominator.

Multiply the second and third term by a "disguised" $1$ $1$ :

2nd term: $\frac{5 b + 6 y}{2 b} \times \frac{3}{3} = \frac{15 b + 18 y}{6 b}$ $(5b+6y)/(2b)xx3/3=(15b+18y)/(6b)$

3rd term: $1 \times \frac{6 b}{6 b} = \frac{6 b}{6 b}$ $1xx(6b)/(6b)=(6b)/(6b)$

Now put it all together:

$= \frac{5 b + 9 y}{6 b} - \frac{15 b + 18 y}{6 b} + \frac{6 b}{6 b}$ $=(5b+9y)/(6b)-(15b+18y)/(6b)+(6b)/(6b)$

$= \frac{5 b + 9 y - (15 b + 18 y) + 6 b}{6 b}$ $=(5b+9y-(15b+18y)+6b)/(6b)$

$= \frac{- 4 b - 9 y}{6 b} = - \frac{4 b + 9 y}{6 b}$ $=(-4b-9y)/(6b)=-(4b+9y)/(6b)$

How do you write 5b+9y6b−5b+6y2b+1\frac { 5b + 9y } { 6b } - \frac { 5b + 6y } { 2b } + 1 as a single fraction?