How do you solve the system #2p - 5q = 14# and #p + 3/2 q = 5#?

1 Answer
Aug 21, 2017

Answer:

See a solution process below:

Explanation:

Step 1) Solve each equation for #2p#:

Equation 1

#2p - 5q = 14#

#2p - 5q + color(red)(5q) = 14 + color(red)(5q)#

#2p - 0 = 14 + 5q#

#2p = 14 + 5q#

Equation 2

#p + 3/2q = 5#

#p + 3/2q - color(red)(3/2q) = 5 - color(red)(3/2q)#

#p + 0 = 5 - 3/2q#

#p = 5 - 3/2q#

#color(red)(2) xx p = color(red)(2)(5 - 3/2q)#

#2p = (color(red)(2) xx 5) - (color(red)(2) xx 3/2q)#

#2p = 10 - 3q#

Step 2) Substitute #10 - 3q# from the second equation for #2p# in the first equation and solve for #p#:

#2p = 14 + 5q# becomes:

#10 - 3q = 14 + 5q#

#-color(blue)(14) + 10 - 3q + color(red)(3q) = -color(blue)(14) + 14 + 5q + color(red)(3q)#

#-4 - 0 = 0 + (5 + color(red)(3))q#

#-4 = 8q#

#-4/color(red)(8) = (8q)/color(red)(8)#

#-1/2 = (color(red)(cancel(color(black)(8)))q)/cancel(color(red)(8))#

#-1/2 = q# or #q = -1/2#

Step 3) Substitute #-1/2# for #q# in either of the equations from Step 1 and calculate #p#:

#2p = 14 + 5q# becomes:

#2p = 14 + (5 * -1/2)#

#2p = 14 - 5/2#

#2p = (2/2 * 14) - 5/2#

#2p = 28/2 - 5/2#

#2p = 23/2#

#color(red)(1/2) * 2p = color(red)(1/2) * 23/2#

#1p = 23/4#

#p = 23/4#

The Solution Is: #p = 23/4# and #q = -1/2#