How do you solve the following linear system: #y = 3x - 2 , 14x - 3y = 0#?

Question

How do you solve the following linear system: #y = 3x - 2  , 14x - 3y = 0#?

Isaiah Anthony · Accepted Answer

The solution is #(-6/5,-28/5)# or #(-1.2,-5.6)#.

Work out the linear system: #"Equation 1":# #y=3x-2# #"Equation 2":# #14x-3y=0# The solution is the point #(x,y)# that the two lines have in common, which is the point of intersection. I'm going to use substitution to solve the system. Equation 1 is already solved for #y#. Substitute #3x-2# for #y# in Equation 2 and solve for #x#. #14x-3(3x-2)=0# Expand. #14x-9x+6=0# Simplify. #5x+6=0# Subtract #6# from both sides. #5x=-6# Divide both sides by #5#. #x=-6/5# or #-1.2# Substitute #-6/5# for #x# in Equation 1. Solve for #y#. #y=3(-6/5)-2# Expand. #y=-18/5-2# Multiply #2# by #5/5# to get an equivalent fraction with #5# as the denominator. #y=-18/5-2xx5/5# #y=-18/5-10/5# Simplify. #y=-28/5# or #-5.6# The solution is #(-6/5,-28/5)# or #(-1.2,-5.6)#. y-3x+2)(14x-3y+0)=0 graph{[-6.366, 4.73, -8.243, -2.696]}

Kennedy Adams · Answer

undefined To solve the linear system:
1. Substitute the expression for y from the first equation into the second equation.
2. Solve the resulting equation for x.
3. Once you have found the value of x, substitute it back into either of the original equations to find the corresponding value of y.