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