#4x +color(red)(y) =15" and " 7x -2y =15#
There is a single #color(red)(y)# in the first equation.
Re-arrange it to find #y# in terms of #x#.
#color(red)(y =(15-4x))#
Replace #color(red)(y)# in the second equation with #(color(red)(15-4x))#
#color(white)(................................................)7x - 2color(red)(y)=15#
#color(white)(........................................................)color(red)(darr)#
#color(white)(.........................................)7x - 2(color(red)(15-4x))=15#
#color(white)(...............................................)7x -30+8x=15#
#color(white)(...............................................................)15x=45#
#color(white)(..................................................................)color(blue)(x=3)#
Now that you know the value for #x#, find the value for #y#.
#y=15-4color(blue)(x)#
#y=15-4color(blue)((3))#
#y = 15-12#
#y =3#