Question

How do you solve `\frac { 35} { w } = \frac { 63} { 9}`?

Answer 1

See some solution processes below:

Explanation:

Process 1:

First, because both sides of the equation are single fractions we can "flip" the fractions giving:

`w/35 = 9/63`

Now, multiply each side of the equation by `color(red)(35)` to solve for `w` while keeping the equation balanced:

`color(red)(35) xx w/35 = color(red)(35) xx 9/63`

`cancel(color(red)(35)) xx w/color(red)(cancel(color(black)(35))) = color(red)(35) xx 9/(9 xx 7)`

`cancel(color(red)(35)) xx w/color(red)(cancel(color(black)(35))) = color(black)(cancel(color(red)(35)))color(red)(5) xx color(blue)(cancel(color(black)(9)))/(color(blue)(cancel(color(black)(9))) xx color(red)(cancel(color(black)(7))))`

`w = 5`

Process 2:

In this process we can first factor the term on the right side of the equation:

`35/w = 63/9`

`35/w = (9 xx 7)/9`

`35/w = (color(red)(cancel(color(black)(9))) xx 7)/color(red)(cancel(color(black)(9)))`

`35/w = 7`

Now, we can multiply each side of the equation by `color(red)(w)/color(blue)(7)` to solve for `w` while keeping the equation balanced:

`color(red)(w)/color(blue)(7) xx 35/w = color(red)(w)/color(blue)(7) xx 7`

`cancel(color(red)(w))/cancel(color(blue)(7)) xx (color(blue)(cancel(color(black)(35)))5)/color(red)(cancel(color(black)(w))) = color(red)(w)/cancel(color(blue)(7)) xx color(blue)(cancel(color(black)(7)))`

`5 = w`

`w = 5`

Process 3:

We can cross product or cross multiply the equation:

`color(orange)(35)/color(orange)(w) = color(blue)(63)/color(blue)(9)`

`color(orange)(35) xx color(blue)(9) = color(blue)(63) xx color(orange)(w)`

`315 = 63w`

Now, divide each side of the equation by `color(red)(63)` to solve for `w` while keeping the equation balanced:

`(315)/color(red)(63) = (63w)/color(red)(63)`

`(63 xx 5)/color(red)(63) = (color(red)(cancel(color(black)(63)))w)/cancel(color(red)(63))`

`(color(red)(cancel(color(black)(63))) xx 5)/cancel(color(red)(63)) = w`

`5 = w`

`w = 5`