How do you solve the system of equations #2p + 3q = 1# and #4p - 5q = 13#?

2 Answers
smendyka
Aug 13, 2017

See a solution process below:

Explanation:

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

Equation 1:

#2p + 3q = 1#

#2p + 3q - color(red)(3q) = 1 - color(red)(3q)#

#2p + 0 = 1 - 3q#

#2p = 1 - 3q#

#color(red)(2) xx 2p = color(red)(2)(1 - 3q)#

#4p = (color(red)(2) xx 1) - (color(red)(2) xx 3q)#

#4p = 2 - 6q#

Equation 2:

#4p - 5q = 13#

#4p - 5q + color(red)(5q) = 13 + color(red)(5q)#

#4p - 0 = 13 + 5q#

#4p = 13 + 5q#

Step 2) Because the left side of each equation is #4p# we can equate the right side of each equation and solve for #x#:

#2 - 6q = 13 + 5q#

#-color(blue)(13) + 2 - 6q + color(red)(6q) = -color(blue)(13) + 13 + 5q + color(red)(6q)#

#-11 - 0 = 0 + (5 + color(red)(6))q#

#-11 = 11q#

#(-11)/color(red)(11) = (11q)/color(red)(11)#

#-1 = (color(red)(cancel(color(black)(11)))q)/cancel(color(red)(11))#

#-1 = q#

#q = -1#

*Step 3) Substitute #-1# for #q# in the solution to either equation in Step 1 and calculate #p#:

#4p = 2 - 6q# becomes:

#4p = 2 - (6 * -1)#

#4p = 2 - (-6)#

#4p = 2 + 6#

#4p = 8#

#(4p)/color(red)(4) = 8/color(red)(4)#

#(color(red)(cancel(color(black)(4)))p)/cancel(color(red)(4)) = 2#

#p = 2#

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

Rina
Aug 13, 2017

#p=2#
#q=-1#

Explanation:

#2p=1-3q#
#p=1/2-3/2q#
simultaneous: substitute p into the 2nd equation
#4(1/2-3/2q)-5q=13#
#2-6q-5q=13#
#-11q=13-2#
#-11q=11#
#q=-1#
therefore substitute q into equation p
#p=1/2-3/2(-1)#
#p=1/2+3/2#
#p=2#

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