How do you solve #y+6=2x# & #4x-10y=4#?

2 Answers
May 25, 2017

Answer:

#x=7/2# and #y=1#

Explanation:

Strategy: Solve for #y# in terms of #x#. Plug in that #y# equation into the other equation to solve for #x#. With the solution for #x#, you can back track and find the solution for #y#.

Step 1. Solve for #y# in terms of #x#.
#y+6=2x# given
#y=2x-6# subtract 6 from both sides

Step 2. Plug this #y# equation into the other equation.
#4x-10color(red)(y)=4# ; given
#4x-10(color(red)(2x-6))=4# ; replace variable #y# with #2x-6#
#4x-20x+60=4# ; distribute #-10# through
#-16x=-56# ; combine #x# terms and subtract #60# both sides
#x=7/2# ; divide both sides by #-16 and reduce#

Step 3. Plug this solution back into the equation of step 1.
#y=2color(red)(x)-6# ; final part of step 1
#y=2(color(red)(7/2))-6# ; solution for #x# in step 2
#y=7-6=1#

So your solution is #x=7/2# and #y=1#

May 25, 2017

Answer:

#(7/2,1)#

Explanation:

#color(red)(y)+6=2xto(1)#

#4x-10color(red)(y)=4to(2)#

#"note that in " (1)" y can be expressed in terms of x"#

#rArrcolor(red)(y)=2x-6to(3)#

#"substitute into " (2)#

#rArr4x-10(2x-6)=4#

#rArr4x-20x+60=4larr" distributing"#

#rArr-16x+60=4larr" simplifying left side"#

#"subtract 60 from both sides"#

#-16xcancel(+60)cancel(-60)=4-60#

#rArr-16x=-56#

#"divide both sides by - 16"#

#(cancel(-16) x)/cancel(-16)=(-56)/(-16)#

#rArrx=56/16=7/2#

#"substitute this value in " (3)" and evaluate for y"#

#y=(2xx7/2)-6=7-6=1#

#(7/2,1)" is the point of intersection of the 2 equations"#
graph{(y-2x+6)(y-2/5x+2/5)=0 [-10, 10, -5, 5]}