solve the following pair of linear equations by substitution method s-7t+42=0,s-3t=6
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Answer:
s-7t+42=0......(1) & s-3t=6.
s-3t-6=0.
s=3t+6.......(2).
Substitute s=3t+6 in equation (1).
s-7t+42=0.
3t+6-7t+42=0.
-4t+48=0.
-4t=-48.
t=-48/-4.
t=12.
Substitute t=12 in equation (2).
s=3t+6.
s=3(12)+6.
s=36+6.
s=42.
s-7t+42=0.
42-7(12)+42=0.
84-84=0.
0=0.
s-3t=6.
42-3(12)=6.
42-36=6.
6=6.
(or).
s-3t-6=0.
42-3(12)-6=0.
36-36=0.
0=0.
I think this is your answer.
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