Question

How do you solve `(3t)/2+7=4t-3`?

Answer 1

See full solution process below:

Explanation:

First, multiply each side of the equation by `color(red)(2)` to eliminate the fraction and keep the equation balanced:

`color(red)(2)((3t)/2 + 7) = color(red)(2)(4t - 3)`

`(color(red)(2) xx 3t)/2 + (color(red)(2) xx 7) = (color(red)(2) xx 4t) - (color(red)(2) xx 3)`

`(cancel(color(red)(2)) xx 3t)/color(red)(cancel(color(black)(2))) + 14 = 8t - 6`

`3t + 14 = 8t - 6`

Next, add and subtract the necessary terms from each side of the equation to isolate the `t` terms on one side of the equation and the constants on the other side of the equation while keeping the equation balanced:

`3t + 14 - color(red)(3t) + color(blue)(6) = 8t - 6 - color(red)(3t) + color(blue)(6)`

`3t - color(red)(3t) + 14 + color(blue)(6) = 8t - color(red)(3t) - 6 + color(blue)(6)`

`0 + 14 + color(blue)(6) = 8t - color(red)(3t) - 0`

`20 = (8 - 3)t`

`20 = 5t`

Now, divide each side of the equation by `color(red)(5)` to solve for `t` while keeping the equation balanced:

`20/color(red)(5) = (5t)/color(red)(5)`

`4 = (color(red)(cancel(color(black)(5)))t)/cancel(color(red)(5))`

`4 = t`

`t = 4`

Answer 2

`t=4`

Explanation:

We can 'eliminate' the fraction in the equation by multiplying ALL terms on both sides by 2, the denominator of the fraction term.

`(cancel(2)xx(3t)/cancel(2))+(2xx7)=(2xx4t)-(2xx3)`

`rArr3t+14=8t-6`

subtract 8t from both sides.

`3t-8t+14=cancel(8t)cancel(-8t)-6`

`rArr-5t+14=-6`

subtract 14 from both sides.

`-5tcancel(+14)cancel(-14)=-6-14`

`rArr-5t=-20`

To solve for t, divide both sides by - 5

`(cancel(-5) t)/cancel(-5)=(-20)/(-5)`

`rArrt=4" is the solution"`