\(\left \{ {{4t+3c=81.00\:(I)} \atop {10t+3c=135.00\:(II)}} \right.\)
Simplify the first equation by (-1)
\(\left \{ {{4t+3c=81.00\:*(-1)} \atop {10t+3c=135.00\:\:\:\:\:}} \right.\)
Once simplified, cancel the opposing terms.
\(\left \{ {{-4t-\diagup\!\!\!\!3c=-81.00} \atop {10t+\diagup\!\!\!\!3c=135.00}} \right.\)
Now find the value of "t".
\(\left \{ {{-4t=-81.00} \atop {10t=135.00}} \right.\)
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\(6t = 54.00\)
\(t = \frac{54.00}{6}\)
\(\boxed{\boxed{t = 9.00}}\end{array}}\qquad\quad\checkmark\)
Now find the value of "c", replace the found value of "t" in the first equation:
\(4t+3c=81.00\:(I)\)
\(4*9+3c=81.00\)
\(36 + 3c = 81.00\)
\(3c = 81.00 - 36\)
\(3c = 45.00\)
\(c = \frac{45.00}{3}\)
\(\boxed{\boxed{c = 15.00}}\end{array}}\qquad\quad\checkmark\)
Answer:
The values of "t" and "c" satisfying the linear system are successively: $ 9.00 and $ 15.00
