how to solve 0.01x+0.07y=0.22 and 0.03x-0.05y=0.14 by substitution

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03-10-2014
Mathematics

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how to solve 0.01x+0.07y=0.22 and 0.03x-0.05y=0.14 by substitution

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flavour5 flavour5 · Answer 1 · 2014-10-03T05:23:00+02:00

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03-10-2014

0.01x + 0.07y = 0.22
0.03x -0.05y = 0.14

x + 7y = 22
3x - 5y = 14

x = 22 -7y

3*(22-7y) -5y =14
66 - 21y -5y =14
-26y = - 52
y =2

x + 14 = 22

x= 8

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