Question

How do you solve `x-5=-3/2x+5/2`?

Answer 1

`x=3`

Explanation:

`"to eliminate fractions multiply all terms on both sides of"`
`"the equation by 2"`

`rArr2x-10=cancel(2)xx-(3x)/cancel(2)+cancel(2)xx5/cancel(2)`

`rArr2x-10=-3x+5larrcolor(blue)"no fractions"`

`"add "3x" to both sides"`

`2x+3x-10=cancel(-3x)cancel(+3x)+5`

`rArr5x-10=5`

`"add 10 to both sides"`

`5xcancel(-10)cancel(+10)=5+10`

`rArr5x=15`

`"divide both sides by 5"`

`(cancel(5) x)/cancel(5)=15/5`

`rArrx=3" is the solution"`

Answer 2

`x=3`

Explanation:

`x-5=-3/2x+5/2`

`"multiply" L.H.S and R.H.S. by" 2`

`:.2x-10=-cancel 6^3/cancel2^1x+cancel 10^5/cancel 2^1`

`:.2x-10=-3x+5`

`:.2x+3x=5+10`

`:.5x=15`

`:.x=cancel 15^3/cancel 5^1`

`:.x=3`

~~~~~~~~~~~~

`"check:-"`

`"substitute " x=3`

`:.(3)-5=-3/2(3)+5/2`

`:.-2=-9/2+5/2`

`:.-2=-cancel 4^2/cancel 2^1`

`:.-2=-2`