How do you find the solution of the system of equations #3x+4y=10# and #x-y=1#?

2 Answers

#x=2# and #y=1#

Explanation:

For such a simple system, you can use substitution: #x=y+1# from the second equation. Substitute this value of #x# in the first equation to the effect that
#3(y+1)+4y=10#. this simplifies to #7y=7#, that is #y=1#.

Substitute back in the second equation to get #x=y+1=1+1=2#.

Jun 7, 2018

#x=2,y=1#

Explanation:

The key insight here is that for the second equation, we can easily solve for a variable in term of the other variable.

Let's just add #y# to both sides to solve for #x# in terms of #y#. We get

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

We can plug this value of #x# into the first equation in the system. We get

#3(y+1)+4y=10#

Distributing the #3# to both terms in the parenthesis, we get

#3y+3+4y=10#

Combining like terms, we now have

#7y+3=10#

Subtracting #3# from both sides, we get

#7y=7#

Dividing both sides by #7#, we find that

#color(red)(y=1)#

We can plug this value into the blue expression to get

#x=1+1#

#color(red)(=>x=2)#

Hope this helps!