How do you solve the system of equations #y= 10x - 10# and #y = \frac { - 2} { 3} x + \frac { 1} { 7}#?

1 Answer
Mar 6, 2017

#x = 213/224 and y = -55/112#

Explanation:

This type of question is often asked when you are working with straight line equations.

This is a GREAT way to have the question asked because both equations have #y# as the subject.

Note that the value of #y# will be the same in each case.

#color(blue)(y = (-2)/3x +1/7)" and "color(red)( y = 10x-10)#

#color(white)(............................)color(blue)(y) = color(red)( y )#

#color(white)(.........)color(blue)((-2)/3x +1/7)" = "color(red)(10x-10)#

Multiply through by #21# to cancel the denominators

#color(green)(cancel21^7xx)(-2)/cancel3x +color(green)(cancel21^3xx)1/cancel7" = "color(green)(21xx)10x-color(green)(21xx)10#

#color(white)(.........................)-14x+3" = "210x-210#

#color(white)(.............................)3+210" = "210x+14x#

#color(white)(...................................)213" = "224x#

#color(white)(...................................)213/224" = "x#

Now that you have a value for #x#, substitute into either of the original equations to find #y#.

#color(red)( y = 10x-10)#

#y = 10(213/224)-10#

#y = -55/112#