Question

How do you solve for `n` in the equation `4k+mn=n-3`?

Solve for `c`, `c/d+2= f/g`.
Solve for `c`, `3ab-2bc=12`.
solve for `y`, `z= (x+y/3)w`

Answer 1

`n=-(3+4k)/(m-1)`

Explanation:

As with any equation we collect the terms containing the variable we wish to solve for on the left side of the equation and place all other values on the right.

Begin by subtracting n from both sides of the equation.

`4k+mn-n=cancel(n)cancel(-n)-3`

`rArr4k+mn-n=-3`

now subtract 4k from both sides.

`cancel(4k)cancel(-4k)+mn-n=-3-4k`

`rArrmn-n=-3-4k`

n can now be taken out as a `color(blue)"common factor"`

`n(m-1)=-3-4k=-(3+4k)`

To solve for n divide both sides by (m - 1).

`(cancel((m-1)) n)/cancel((m-1))=-(3+4k)/(m-1)`

`rArrn=-(3+4k)/(m-1)" is the solution"`

Equations of this form are referred to as ' literal equations' since they contain mainly letters.

Answer 2

`c=(df)/g-2d` (first on list)

Explanation:

This is for the first one in the above list.

There are different ways of approaching this one.

(1) multiply through by d, to eliminate the fraction.

`cancel(d)xxc/cancel(d)+2d=(fd)/g`

`rArrc+2d=(fd)/g`

subtract 2d from both sides.

`c cancel(+2d)cancel(-2d)=(fd)/g-2d`

`rArrc=(fd)/g-2d`
`"-----------------------------------------------"`

(2) subtract 2 from both sides.

`c/d cancel(+2)cancel(-2)=f/g-2`

`rArrc/d=f/g-2`

now multiply all terms on both sides by d.

`rArrc=(fd)/g-2d`
`"-------------------------------------------------------"`

(3) multiply through by dg, to eliminate both fractions.

`cancel(d)gxxc/cancel(d)+2dg=cancel(g)d xxf/cancel(g)`

`rArrcg+2dg=df`

subtract 2dg from both sides.

`cg cancel(+2dg)cancel(-2dg)=df-2dg`

`rArrcg=d(f-2g)larr" common factor of d"`

divide both sides by g, to solve for c.

`(cancel(g) c)/cancel(g)=(d(f-2g))/g`

`rArrc=(d(f-2g))/g`

The choice of approach is yours.

Answer 3

`c=(3ab-12)/(2b)` (second part)

Explanation:

As with equations isolate the term in c on the left and collect all other terms on the right.

subtract 3ab from both sides.

`cancel(3ab)cancel(-3ab)-2bc=12-3ab`

`rArr-2bc=12-3ab`

multiply through by - 1 , not totally necessary but eliminates the leading terms negative sign.

`rArr2bc=3ab-12`

to solve for c, divide both sides by 2b.

`(cancel(2b) c)/cancel(2b)=(3ab-12)/(2b)`

`rArrc=(3ab-12)/(2b)`

Answer 4

`y=(3(z-wx))/w` (third part)

Explanation:

Start by distributing the bracket.

`z=wx+(wy)/3`

`rArr(wy)/3+wx=zlarr" reversing the equation"`

subtract wx from both sides.

`(wy)/3cancel(+wx)cancel(-wx)=z-wx`

`rArr(wy)/3=z-wx`

multiply both sides by 3.

`cancel(3)xx(wy)/cancel(3)=3(z-wx)`

`rArrwy=3(z-wx)`

To solve for y, divide both sides by w.

`(cancel(w) y)/cancel(w)=(3(z-wx))/w`

`rArry=(3(z-wx))/w`