How do you solve the following system: #4x+3y= -7, 2x-5y=-19 #?

1 Answer
Nov 2, 2017

Answer:

See a solution process below:

Explanation:

Step 1) Solve each equation for #4x#:

Equation 1:

#4x + 3y - color(red)(3y) = -7 - color(red)(3y)#

#4x + 0 = -7 - 3y#

#4x = -7 - 3y#

Equation 2:

#2x - 5y + color(red)(5y) = -19 + color(red)(5y)#

#2x - 0 = -19 + 5y#

#2x = -19 + 5y#

#color(red)(2) xx 2x = color(red)(2)(-19 + 5y)#

#4x = (color(red)(2) xx -19) + (color(red)(2) xx 5y)#

#4x = -38 + 10y#

Step 2) Because the left sides of each equation is equal, we can equate the right sides of each equation and solve for #y#:

#-7 - 3y = -38 + 10y#

#color(blue)(38) - 7 - 3y + color(red)(3y) = color(blue)(38) - 38 + 10y + color(red)(3y)#

#31 - 0 = 0 + (10 + color(red)3)y#

#31 = 13y#

#31/color(red)(13) = (13y)/color(red)(13)#

#31/13 = (color(red)(cancel(color(black)(13)))y)/cancel(color(red)(13))#

#31/13 = y#

#y = 31/13#

Step 3) Substitute #31/13# for #y# in the solution to either equation in Step 1 and calculate #x#:

#4x = -7 - 3y# becomes:

#4x = -7 - (3 xx 31/13)#

#4x = -7 - 93/13#

#4x = -(13/13 xx 7) - 93/13#

#4x = -91/13 - 93/13#

#4x = -184/13#

#color(red)(1/4) xx 4x = color(red)(1/4) xx -184/13#

#4/color(red)(4)x = color(red)(1/color(black)(cancel(color(red)(4)))) xx -(color(red)(cancel(color(black)(184)))46)/13#

#x = -46/13#

The Solution Is: #x = -46/13# and #y = 31/13# or #(-46/13, 31/13)#