To find solutions to this problem, substitute a value for #x# and calculate #y#. To make the calculations easier, you may want to solve the equation for #y# first:
#4x - 2y = 8#
#-color(red)(4x) + 4x - 2y = -color(red)(4x) + 8#
#0 - 2y = -4x + 8#
#-2y = -4x + 8#
#(-2y)/color(red)(-2) = (-4x + 8)/color(red)(-2)#
#(color(red)(cancel(color(black)(-2)))y)/cancel(color(red)(-2)) = (-4x)/color(red)(-2) + 8/color(red)(-2)#
#y = 2x - 4#
Solution 1 Substitute #color(red)(-5)# for #color(red)(x)# and calculate #y#:
#y = 2color(red)(x) - 4# becomes:
#y = (2 * color(red)(-5)) - 4#
#y = -10 - 4#
#y = -14#
One solution is: (-5, -14)
Solution 2 Substitute #color(red)(0)# for #color(red)(x)# and calculate #y#:
#y = 2color(red)(x) - 4# becomes:
#y = (2 * color(red)(0)) - 4#
#y = 0 - 4#
#y = -4#
A second solution is: (0, -4)
Solution 3 Substitute #color(red)(5)# for #color(red)(x)# and calculate #y#:
#y = 2color(red)(x) - 4# becomes:
#y = (2 * color(red)(5)) - 4#
#y = 10 - 4#
#y = 6#
A third solution is: (5, 6)
Try a fourth solution. Substitute #color(red)(10)# for #color(red)(x)# and calculate y:
