If the value of #(x(y-4))/5 = -2#, what are possible values of x and y?

1 Answer
Jul 23, 2017

#(-1,14), (-2,9), (1,-6)#

There are infinitely many solutions, but #x!=0 and y!=4#

Explanation:

There are two variables, but only one equation, so there is not only one solution for the equation.

Make one of the variables the subject:

Isolating #x# gives:

#(x xx(y-4))/5 = -2#

#x = (-2xx5)/((y-4)) = (-10)/(y-4)#

You can now choose any value of #y# and calculate a corresponding value for #x#.

Note that #color(red)(y!=4)#, because the denominator would be #0#

If #y=14, " "rarr x = (-10)/(14-4) = (-10)/(10) =-1#

which gives #(-1,14)#

If #y = 9 " "rarr x = (-10)/(9-4) = (-10)/(5) =-2#

which gives #(-2,9)#

Making #y# the subject:

#(x(y-4))/5 = -2#

#y-4 = (-2xx5)/x" "rarr y = (-10)/x +4#

Choose any value for #x# and calculate a value for #y#

Note that #color(red)(x!=0)# becasue the denominator would be #0#

If #x = 1" "rarry = (-10)/(1) +4 " "rarr -10+4 =-6#

which gives: #(1,-6)#

In this way there are infinitely many solutions which can be found.