how to solve the problem of simultaneous equation 3a+5b =26 ;a+5b=22
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multiplying the whole equation by 3
3a+15b=66 subtracting them
3a+5b=26
10b=40
b=4
3a+15b=66 subtracting them
3a+5b=26
10b=40
b=4
urmik:
plz follow
Answered by
1
Multiple 3 in equation a+5b=22;
that is 3a+15b=66
Now subtract 3a+5b=26 from 3a+15b=66;
10b=40;
b=4;
putting b=4 in a+5b=22;
a+20=22;
so a=22-20=2
that is 3a+15b=66
Now subtract 3a+5b=26 from 3a+15b=66;
10b=40;
b=4;
putting b=4 in a+5b=22;
a+20=22;
so a=22-20=2
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