How do you solve the system of linear equations: #x+3y=-4, 2x+6y=5#?

1 Answer
Nov 9, 2017

See a solution process below:

Explanation:

Both these equations are in Standard Linear Form. The standard form of a linear equation is: #color(red)(A)x + color(blue)(B)y = color(green)(C)#

Where, if at all possible, #color(red)(A)#, #color(blue)(B)#, and #color(green)(C)#are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

The slope of an equation in standard form is: #m = -color(red)(A)/color(blue)(B)#

The #y#-intercept of an equation in standard form is: #color(green)(C)/color(blue)(B)#

#m_1 = -1/3 = -1/3#

#m_2 -2/6 = -1/3#

Both these lines have the same slope so either they are parallel but distinct lines and have no points in common. Or, they are the same line and have every point in common.

For #y = 0#, #x# for each equation is:

Equation 1:

#x + (3 * 0) = -4#

#x + 0 = -4#

#x = -4#

Equation 2:

#2x + (6 * 0) = 5#

#2x + 0 = 5#

#2x = 5#

#(2x)/2 = 5/2#

#x = 2.5#

Because the #x# value for both equations at #y = 0# are different these are parallel and distinct lines and have no points in common.

Therefore the solution is the null or empty set: #(x, y) = {O/}#