Question

Solve the following simultaneous equations.
(1) 3a + 5b = 26; a + 5b = 22
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Answer 1

3a +5b = 26

a +5b =22

This could be easily done by elimination method

Now, if we multiply the second equation by 3 then

3(a+5b)=3×22

3a+15b = 66

Now if we will subtract eq.1 from eq.2 then, one of the variable 'a' will get cancelled out,

(3a+5b)- (3a+15b) =26 - 66

So 5b - 15 b= -40

-10b =-40

so b=4

therefore 3a + 5× 4 =22

3a + 20 =22

so 3a = 2

so a =2/3

Answer 2

Step-by-step explanation:

I hope it will help you.....