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 = (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 = (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