Question 697881
Let A = Cost of type A PC
Let B = Cost of type B PC
Equation 1: {{{8A + 12B = 21200}}}
Equation 2: {{{12A + 5B = 16200}}}
I am going to solve this by using the substitution method.
Note that a common mutiple of 8A and 12A is 24A.
Start by setting both equations to zero
Equation 1: {{{8A + 12B - 21200 = 0}}}
Equation 2: {{{12A + 5B - 16200 = 0}}}
Multiply equation 1 by 3 to get 24A
Equation 1: {{{24A + 36B - 63600 = 0}}}
Multiply equation 2 by 2 to get 24A
Equation 2: {{{24A + 10B - 32400 = 0}}}
Now set both equations equal to each other
{{{24A + 36B - 63600 = 24A + 10B - 32400}}}
Subtract 24A from both sides
{{{36B - 63600 = 10B - 32400}}}
Add 63600 to both sides
{{{36B = 10B + 31200}}}
Subtract 10B from both sides
{{{26B = 31200}}}
Divide both sides by 26
{{{highlight(B = 1200)}}}
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Now plug 1200 into one of the given equations for B
Equation 1: {{{8A + 12B = 21200}}}
{{{8A + 12*(1200) = 21200}}}
{{{8A + 14400 = 21200}}}
Subtract 14400 from btoh sides
{{{8A = 6800}}}
Divide both sides by 8
{{{highlight_green(A = 850)}}}