Question 196538
should show up like x-3 divided by 5 and the 1/3 as a fraction (x)
Assume the problem is:
{{{(x-3)/5}}} = 2 + {{{1/3}}}x
multiply equation by 15
15*{{{(x-3)/5}}} = 15(2) + 15*{{{1/3}}}x
Cancel the denominators and you have:
3(x-3) = 30 + 5x
:
3x - 9 = 30 + 5x
:
-9 - 30 = 5x - 3x
:
-39 = 2x
:
x = {{{(-39)/2}}}
x = -19.5
:
Check solution in original equation
{{{(-19.5-3)/5}}} = 2 + {{{1/3}}}(-19.5)
{{{(-22.5)/5}}} = 2 + {{{(-19.5)/3}}}
-4.5 = 2 - 6.5
:
:
{{{-2x^2y^3 - 2xy}}} x = -3 y = 4
Substitute for x and y
-2(-3^2)(4^3) - (2*-3*4)
:
-2(9)*64 - (-24)
:
-18 * 64 + 24 = -1128