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Question

Solve the following equations by substitution method.
3a2b=10;2a+3b=2

A
a=2,b=2
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B
a=1,b=2
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C
a=4,b=2
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D
a=7,b=2
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Solution

The correct option is A a=2,b=2
3a2b=10;2a+3b=2
a=2b103 Put it in second equation.
2×2b103+3b=2
4b20+9b=6
13b=26
b=2

a=2b103=2(2)103=2
a=2 and b=2

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