Question

4t+3c=$81.00 10t+3c=$135.00

Answer 1

$\left \{ {{4t+3c=81.00\:(I)} \atop {10t+3c=135.00\:(II)}} \right.$
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