How do you solve #8y-1=x, 3x=2y#?

2 Answers
Gió
Nov 26, 2016

I found: #x=1/11 and y=3/22#

Explanation:

We can trysubstituting the first equation, for #x#, intothe second to get:
#3*(color(red)(8y-1))=2y#
Solve for #y#:
#24y-3=2y#
#22y=3#
So: #y=3/22#
Using this value back into the first equation:
#8*(3/22)-1=x#
#x=(24-22)/22=2/22=1/11#

MathFact-orials.blogspot.com
Nov 26, 2016

#(x,y)=(1/11,3/22)#

Explanation:

We have two equations:

#8y-1=x#
#3x=2y#

We can substitute in the value of x (in terms of y) from the first equation into the second, and then solve for y. Like this:

#3(8y-1)=2y#

#24y-3=2y#

#22y=3#

#y=3/22#

And then we substitute into either of the initial equations (I'll do both to show we'll get the same answer for x):

#8y-1=x#

#8(3/22)-1(1)=x#

#24/22-1(22/22)=x#

#24/22-22/22=2/22=1/11=x#

or:

#3x=2y#

#3x=2(3/22)#

#3x=3/11#

#x=1/11#

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