How do solve the following linear system?: 3s+4=−4t,7s+6t+11=0?

2 Answers
Cem Sentin
Jul 5, 2018

s=−2 and t=12

Explanation:

From first equation, s=−4t−43.

After plugging value of s into second one,

7⋅−4t−43+6t+11=0

−28t−28+18t+333=0

5−10t3=0

5−10t=0

10t=5, so t=510=12

Thus, s=(−4)⋅12−43=−2

Abhishek Malviya
Jul 5, 2018

Your answer is −2810=t

Explanation:

Here we have two equation
1) 3s+4=−4t
2) 7s+6t=0

using equation 2 to find value of s

7s+6t=0
7s=−6t
s=−6t7 ....(equation3)

Now put value of s in eq 1

3(−6t7)+4=−4t

−18t+287=−4t

−18t+28=−28t

28=−28t+18t

28=−10t

−2810=t

Now put value of t in equation 3

s=−6(−2810)7

s=6(2870)

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