Question

How do you solve `\frac { 4m + 7} { m + 10} = \frac { m } { 2} + \frac { 11} { 2m + 20}`?

Answer 1

`m_1 = (-3), m_2 = 1`

Explanation:

Solve it with the quadratic formula

`(4m + 7)/(m+10) = m/2 + 11/(2m+20)`

Find LCM of `m+10,2m+20,2`

Factor the expression

`2m + 20`
`2(m+10)`

Prime factorization of 2 is 2

Therefore the LCM is `2(m+10)` since it is divisible by all and is one of the no. of which we are finding the LCM

Multiply with LCM

`(4m + 7)/cancel(m+10) xx cancel(2(m+10))^2 = m/cancel2 xx cancel(2)(m+10) + 11/cancel(2m+20) xx cancel(2(m+10))`

`(4m + 7) xx 2 = m xx (m+10) + 11`

Divide the equation into two halves which would make the equation easier. Just to add, In chemistry you divide the ionic reactions to half reactions to make them easier.

First simplify

`(4m + 7) xx 2`

Remove parentheses and multiply with two all the no. in the parentheses

`= 8m + 14`

Now simplify the second part that is

`m xx (m+10) + 11`
`= m^2 + 10m + 11`

Add the equations

`8m + 14 = m^2 + 10m + 11`

`8m = m^2 + 10m + 11-14`

`8m = m^2 + 10m + (-3)`

`8m = m^2 + 10m + (-3)`

`8m + 3 = m^2 + 10m`

`8m - 10m + 3 = m^2`

`-2m + 3 = m^2`

Subtract `m^2` from both sides

`-2m + 3-m^2 = m^2-m^2`

`-2m + 3-m^2 = 0`

For a quadratic equation of form `ax^2 + bx + c = 0`

`x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}`

`a=-1,b=-2, c=3`
`quad m_{1,\:2}=\frac{-\ (-2\ )\pm \sqrt{\ (-2\ )^2-4\ (-1\ )3}}{2\ (-1\ )}`

`m_1 = \frac{-\ (-2\ )+\sqrt{\ (-2\ )^2-4\ (-1\ )\ 3}}{2\ (-1\ )}`

`"" "" = \frac{-\ (-2\ )+\sqrt{\ (-2\ )^2-4\ (-1\ )\ 3}}{2\ (-1\ )}`

`"" "" = \frac{2+\sqrt{4-\ (-1\ )\ 12}}{-2}`

`"" "" = \sqrt{4+1\ *12}/(-2)`

`"" "" = (2 +sqrt(16))/(-2)`

`"" "" = (2 +4)/(-2)`

`"" "" = 6/(-2) = -3`

`m_2 = \frac{-\ (-2\ )-\sqrt{\ (-2\ )^2-4\ (-1\ )\\:3}}{2\ (-1\ )}`

`"" "" = \frac{2-\sqrt{\ (-2\ )^2-4\ (-1\ )\ \3}}{-2*1}`

`"" "" = \frac{2-\sqrt{4-\ (-1\ )\*12}}{-2}`

`"" "" = (2 -sqrt(4 + 1 * 12)) /(-2)`

`"" "" = (2 -sqrt(4 + 12))/(-2)`

`"" "" = (2 -sqrt(16))/(-2)`

`"" "" = (2-4)/(-2)`

`"" "" = (-2)/(-2) = 1`

Answer 2

`m =-3 or m=1`

Explanation:

When you are working with an equation that has algebraic fractions, you can get rid of the denominators.

Do this by multiplying each term by the LCM of the denominators so that they can all cancel.

`(4m+7)/(m+10) = m/2 + 11/(2m+20)" "larr` factorise first

`(4m+7)/((m+10)) = m/2 + 11/(2(m+10))" "rarr LCM =color(blue)(2(m+10))`

Multiply by `color(blue)(2(m+10)` on both sides and cancel:

`(color(blue)(2cancel((m+10)))*(4m+7))/cancel((m+10))=(color(blue)(cancel2(m+10))*m)/cancel2 + color(blue)(cancel(2(m+10))*11)/(cancel(2(m+10))`

This leaves us with:

`2(4m+7) =m(m+10) +11" "`simplify and solve

`8m +14 =m^2 +10m +11" "larr` make `=0`

`m^2 +2m-3=0" "larr` factorise

`(m+3)(m-1)=0`

Setting each factor equal to `0` gives two solutions:

`m =-3 or m=1`