How do you solve the following system: #5x + 8y = -2, 4x + 7y = 6x#?

1 Answer
Aug 5, 2017

#x = - frac(14)(39) and y = - frac(4)(39)#

Explanation:

We have: #5 x + 8 y = - 2 and 4 x + 7 y = 6 x#

Let's solve both equations for #y#:

#Rightarrow 8 y = - 5 x - 2 and 7 y = 6 x - 4 x#

#Rightarrow y = frac(- 5 x - 2)(8) and y = frac(2 x)(7)#

Then, let's set the two expressions for #y# equal to each other:

#Rightarrow frac(- 5 x - 2)(8) = frac(2 x)(7)#

#Rightarrow 7 times (- 5 x - 2) = 8 times 2 x#

#Rightarrow - 35 x - 14 = 4 x#

#Rightarrow - 35 x - 4 x - 14 = 0#

#Rightarrow - 39 x - 14 = 0#

#Rightarrow - 39 x = 14#

#therefore x = - frac(14)(39)#

Now that we have a value of #x#, let's find #y# using one of the expressions:

#Rightarrow y = frac(2 x)(7)#

#Rightarrow y = frac(2 times - frac(14)(39))(7)#

#Rightarrow y = frac(- frac(28)(39))(7)#

#Rightarrow y = frac(- frac(28)(39))(frac(7)(1))#

#Rightarrow y = - frac(28)(39) times frac(1)(7)#

#Rightarrow y = - frac(28 times 1)(39 times 7)#

#Rightarrow y = - frac(28)(273)#

#therefore y = - frac(4)(39)#

Therefore, the solutions to the equation are #x = - frac(14)(39)# and #y = - frac(4)(39)#.