How do you solve #y=4x# and #x+y=5#?

1 Answer
May 29, 2016

Answer:

The common point for these two equations ( where the graphs cross) is:

#x=1#
#y=4#

Explanation:

Given:
#color(blue)(y=4x)# ..................................(1)
#color(brown)(x+y=5)#................................(2)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
To solve for a variable your equation has to end up with just 1 of that variable and no others. That variable has to be on one side of the equals sign and everything else on the others side.

If we substitute for y in equation (2) using what y is worth from equation (1) we have just 1 variable. Which is #x#.

#color(brown)(x+y=5" "->" "x+color(blue)(4x)=5)color(white)(..)...................(2_a)#

But #4x+x=5x#

#x+4x=5" "->" "5x=5#

Divide both sides by 5

#5/5xx x = 5/5#

But #5/5=1#

#x=1#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute for #x# in equation (1) where #x=1#

#color(brown)(y=4x" "->" "y=4(color(blue)(1))#

In algebra #4x# is the same as #4 xx x#

But #x# has the value of 1 so we have

#y=4xx1#

#y=4#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Note that #x+y=5# is another way of writing
#y=-x+5#

Tony B