How do solve the following linear system?: # 4y-16=x , 2x+7=-5y #?

The first equation already has #x# as the subject. Therefore you substitute that in place of #x# in the second equation.

#2(4y - 16) + 7 = -5y -> 8y - 32 + 7 = -5y#

Rearrange the equation to have all unknowns on one side.

#13y = 25#

#y = 25/13#

Now substitute the #y# value into the first equation to get #x#.

#4(25/13) - 16 = x#

#x = -8.31#

Given
[1]#color(white)("XXX")4y-16=x#
[2]#color(white)("XXX")2x+7=-5y#

[1]#rarr#
[3]#color(white)("XXX")x=4y-16#

Using [3] we can replace #x# in [2] with #4y-16#
[4]#color(white)("XXX")2(4y-16)+7=-5y#

Expanding and simplifying the left side of [4]
[5]#color(white)("XXX")8y-25=-5y#

Adding #(5y+25)# to both sides of [5]
[6]#color(white)("XXX")13y=25#

Dividing both sides of [6] by #13#
[7]#color(white)("XXX")y=25/13#

Using [7] we can replace #y# in [1] with #25/13#
[8]#color(white)("XXX")4 * 25/13-16=x#

Simplifying [8]
[9]#color(white)("XXX")x=-108/13#

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The ugliest part is probably verifying this result by putting the derived values for #x# and #y# back into the original equations to check that the results are correct.

...with values like these, I would suggest you go to the effort to verify the results (despite the ugliness). Personally I used a spread sheet for this but you might want to use some other method.

#4y-16=x#----#color(red)((1))#

#2x+7=-5y#----#color(red)((2)#

#4y-x=16#----#color(red)((3)#

#-5y-2x=7#----#(color(red)(4))#

#color(red)((3) xx 5#

#20y-5x=80#----#color(red)((5)#

#color(red)((4) xx 4#

#-20y-8x=28#----#color(red)((6)#

#color(red)((5)+(6)#

#-13x=108#

#color(red)(x=-108/13# or#color(red)(-8 4/13#

Substitute #color(red)(x=-108/13# in (2)

#2(color(red)(-108/13))+7=-5y#

#-216/13+7=-5y#

Multiply both sides by 13

#-65y=-216+91#

#-65y=-125#

#y=cancel125^25/cancel65^13#

#color(red)(y=25/13# or #color(red)(1 12/13#

Check:

Substitute #color(red)(y=25/13# and #color(red)(x=-108/13# in (1)

#4(color(red)(25/13))-16=color(red)(-108/13#

#100/13-16=-108/13#

#(100-208=-108)/13#

Multiply both sides by 65

#100-208=-108#

#color(red)(-108=-108#