How do you solve the system of equations #90a+34d=162# and #120a+66d=288#?

1 Answer
smendyka
Jan 2, 2018

See a solution process below:

Explanation:

Step 1) Multiply each side of both equations by the necessary number to make the coefficient of the #a# term equal #360#:

  • Equation 1:

#4(90a + 34d) = 4 xx 162#

#(4 xx 90a) + (4 xx 34d) = 648#

#360a + 136d = 648#

  • Equation 2:

#3(120a + 66d) = 3 xx 288#

#(3 xx 120a) + (3 xx 66d) = 864#

#360a + 198d = 864#

Step 2) Solve each equation for #360a#:

  • Equation 1:

#360a + 136d - color(red)(136d) = 648 - color(red)(136d)#

#360a + 0 = 648 - 136d#

#360a = 648 - 136d#

  • Equation 2:

#360a + 198d - color(red)(198d) = 864 - color(red)(198d)#

#360a + 0 = 864 - 198d#

#360a = 864 - 198d#

Step 3) Because the left side of each equation is the same we can equate the right side of the two equations and solve for #d#:

#648 - 136d = 864 - 198d#

#648 - color(red)(648) - 136d + color(blue)(198d) = 864 - color(red)(648) - 198d + color(blue)(198d)#

#0 + (-136 + color(blue)(198))d = 216 - 0#

#62d = 216#

#(62d)/color(red)(62) = 216/color(red)(62)#

#(color(red)(cancel(color(black)(62)))d)/cancel(color(red)(62)) = (108 xx 2)/color(red)(31 xx 2)#

#d = (108 xx 2)/color(red)(31 xx 2)#

#d = (108 xx color(red)(cancel(color(black)(2))))/color(red)(31 xx color(black)(cancel(color(red)(2))))#

#d = 108/31#

Step 4) Substitute #108/31# for #d# in the solution to either equation in Step 2 and solve for #a#:

#360a = 648 - 136d# becomes:

#360a = 648 - (136 xx 108/31)#

#360a = (31/31 xx 648) - (14688/31)#

#360a = 20088/31 - 14688/31#

#360a = 5400/31#

#(360a) xx 1/color(red)(360) = 5400/31 xx 1/color(red)(360)#

#(color(red)(cancel(color(black)(360)))a) xx 1/cancel(color(red)(360)) = (color(red)(cancel(color(black)(5400)))" "15)/31 xx 1/cancel(color(red)(360))#

#a = 15/31#

The Solution is: #a = 15/31# and #d = 108/31#

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