4801=24B+24A ........(1)
5304=34B+20A.........(2)
Divide both sides of (1) by 24
200 1/24=A+B
A=200 1/24 - B substitute in 2
20(200 1/24 -B)+34B=5304
4,000.83 -20B+34B =5304
14B=5304 - 4000.83
14B=1303.17 divide both sides by 14
B=1303.17/14
B=93.083.......(rounded)
A=106.9583...(rounded)
Solve the following system: {4801 = 24 A+24 B | (equation 1) 5304 = 20 A+34 B | (equation 2) Express the system in standard form: {-(24 A)-24 B = -4801 | (equation 1) -(20 A)-34 B = -5304 | (equation 2) Subtract 5/6 × (equation 1) from equation 2: {-(24 A)-24 B = -4801 | (equation 1) 0 A-14 B = (-7819)/6 | (equation 2) Multiply equation 1 by -1: {24 A+24 B = 4801 | (equation 1) 0 A-14 B = -7819/6 | (equation 2) Multiply equation 2 by -6/7: {24 A+24 B = 4801 | (equation 1) 0 A+12 B = 1117 | (equation 2) Divide equation 2 by 12: {24 A+24 B = 4801 | (equation 1) 0 A+B = 1117/12 | (equation 2) Subtract 24 × (equation 2) from equation 1: {24 A+0 B = 2567 | (equation 1) 0 A+B = 1117/12 | (equation 2) Divide equation 1 by 24: {A+0 B = 2567/24 | (equation 1) 0 A+B = 1117/12 | (equation 2) Collect results: Answer: | | {A = 2567/24 B = 1117/12
