Given #(1/4)x + y = 3/40#, subtract #(1/4)x# from both sides to get:

#y = 3/40 - (1/4)x#

Given #(3/2)x - y = -5/2#, add #y + 5/2# to both sides to get:

#(3/2)x+5/2 = y#

So we have

#(3/2)x + 5/2 = y = 3/40 - (1/4)x#

Just keeping the two ends and multiplying both sides by 40 we get:

#60x + 100 = 3 - 10x#

Add #10x# to both sides to get:

#70x + 100 = 3#

Subtract 100 from both sides to get:

#70x = -97#

Divide both sides by 70 to get:

#x = -97/70#

To calculate the value of #y#, substitute this value of #x# into one of our previous equations:

#y = (3/2)x + 5/2#

#= (3/2)(-97/70) + 5/2#

#= -(3*97)/(2*70) + (5*70)/(2*70)#

#= (-291+350)/140#

#= 59/140#

To check our calculation, try plugging these values of #x# and #y# back into the first equation:

#(1/4)x + y = (1/4)(-97/70) + 59/140#

#= -97/280 + 118/280#

#= 21/280#

#= 3/40#

I will leave checking the second one to you.