How do you solve this system of equations: #2t + 4z = 68 and 4t + 3z = 76#?

1 Answer
smendyka
Jun 30, 2018

See a solution process below:

Explanation:

Step 1) Solve both equations for #4t#:

  • Equation 1:

#2t + 4z = 68#

#color(red)(2)(2t + 4z) = color(red)(2) xx 68#

#(color(red)(2) xx 2t) + (color(red)(2) xx 4z) = 136#

#4t + 8z = 136#

#4t + 8z - color(red)(8z) = 136 - color(red)(8z)#

#4t + 0 = 136 - 8z#

#4t = 136 - 8z#

  • Equation 2:

#4t + 3z = 76#

#4t + 3z - color(red)(3z) = 76 - color(red)(3z)#

#4t + 0 = 76 - 3z#

#4t = 76 - 3z#

Step 2) Because the left side of both equations are now equal we can equate the right side of both equations and solve for #z#:

#136 - 8z = 76 - 3z#

#136 - color(blue)(76) - 8z + color(red)(8z) = 76 - color(blue)(76) - 3z + color(red)(8z)#

#60 - 0 = 0 + (-3 + color(red)(8))z#

#60 = 5z#

#60/color(red)(5) = (5z)/color(red)(5)#

#12 = (color(red)(cancel(color(black)(5)))z)/cancel(color(red)(5))#

#12 =z#

#z = 12#

Step 3) Substitute #12# for #z# in the solution to either equation in Step 1 and solve for #t#:

#4t = 136 - 8z# becomes:

#4t = 136 - (8 xx 12)#

#4t = 136 - 96#

#4t = 40#

#(4t)/color(red)(4) = 40/color(red)(4)#

#(color(red)(cancel(color(black)(4)))t)/cancel(color(red)(4)) = 10#

#t = 10#

The Solution Is:

#t = 10# and #z = 12#

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