How do you solve #15x - 20y = 1# and #5y = 1 + 10x# using substitution?

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Answer 1 · 2016-03-20T11:22:05

#x,y=-1/5#

#color(blue)(15x-20y=1#

#color(blue)(5y=1+10x#

Use the second equation

#rarr5y=1+10x#

Divide both sides by #5#

#rarr(cancel5y)/cancel5=(1+10x)/5#

#rarry=1/5+2x#

Now substitute this value to the first equation

#rarr15x-20(1/5+2x)=1#

Use the distributive property #color(brown)(a(b+c)=ab+ac#

#rarr15x-20/5-40x=1#

#rarr15x-4-40x=1#

#rarr-25x-4=1#

Add #4# both sides

#rarr-25xcancel(-4+4)=1+4#

#rarr-25x=5#

#color(green)(rArrx=-5/25=-1/5#

Substitute the value of #x# to the second equation

#rarr5y=1+10(-1/5)#

#rarr5y=1-10/5#

#rarr5y=1-2#

#rarr5y=-1#

#color(green)(y=-1/5#

#:. color(indigo)(ul bar |x=y=-1/5| #