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Use substitution to solve s+a=4 and 35+5a=2550
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Oct 3, 2018
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Use substitution to solve s+a=4 and 35+5a=2550
Mathematics
high-school
Quinton Pike
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Quinton Pike
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s + a = 4
35 + 5a = 2550
From the second equation:
35 + 5a = 2550
5a = 2550 - 35
5a = 2515
a = 2515/5
a = 503
From s + a = 4
s + 503 =4
s = 4 - 503
s = -499
Therefore a = 503, s = -499
Chris Sedlmayr
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Oct 4, 2018
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Chris Sedlmayr
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35+5(4-s)=2550
35+20-5s=2550
55-5s=2550
-55 -55
-5s = 2495
s = - 499
-499 + a = 4
+499 +499
a = 503
RonyLoud
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Oct 7, 2018
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RonyLoud
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