Linear Algebra and Linear Equation

Question

Mark invested a total of P33 000.00 in three parts : the first at 5%, the second at 6%, and the third at 8%. His profit at the end of the year was P 2230. Had he interchanged the rates of the first and the third parts, he would have received P270.00 less. How much did he invest at each rate?

Answer 1

Let x,y and z be the three parts

x+y+z = 33000 eqn1

.05x + .06y + .08z = 2230 eqn2

.08x + .06y + .05z = 2230-270 = 1960 eqn3

Subtract eqn2 from eqn3

.03x -.03z = -270 --> z -x = 9000 --> z = 9000+x

substitute z into eqn1

x+y+9000+x = 33000

2x + y = 24000 --> y = 24000-2x

Substitute y and z into eqn1

.05x + .06(24000-2x) +0.08(9000-x) = 2230

5x+6(24000-2x)+8(9000+x) = 223000

5x+144000-12x+72000+8x = 223000

x = 7000

y = 24000-14000 = 10000

z = 9000+7000 = 16000

The three amounts invested are P7000.00, P10000.00 and P16000.00