Question

Solve each system by substitution x+12y=68

Answer 1

$\large\begin{array}{l} \textsf{Solve the system by substitution:}\\\\ \left\{\! \begin{array}{rrrrrc} \mathsf{x}&\!\!\!+\!\!\!&\mathsf{12y}&\!\!\!=\!\!\!&\mathsf{68}&\quad\mathsf{(i)}\\ \mathsf{x}&\!\!\!=\!\!\!&\mathsf{8y}&\!\!\!-\!\!\!&\mathsf{12}&\quad\mathsf{(ii)} \end{array} \right. \end{array}$


$\large\begin{array}{l} \textsf{Substitute x from (ii) into (i):}\\\\ \mathsf{(8y-12)+12y=68}\\\\ \mathsf{8y+12y=68+12}\\\\ \mathsf{20y=80}\\\\ \mathsf{y=\dfrac{80}{20}}\\\\ \boxed{\begin{array}{c}\mathsf{y=4} \end{array}} \end{array}$


$\large\begin{array}{l} \textsf{Substitute y back into (ii):}\\\\ \mathsf{x=8\cdot 4-12}\\\\ \mathsf{x=32-12}\\\\ \mathsf{x=20}\\\\ \boxed{\begin{array}{c} \mathsf{x=20} \end{array}}\\\\\\ \textsf{Solution: }\mathsf{S={\{(20;\,4)\}}} \end{array}$


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Answer 2

Substitute x=8y-12 into first equation
8y-12+12y=68
20y=68+12
20y=80
y=80/20
y=4

substitute y=4 in 2nd equ
x=8(4)-12
x=32-12
x=20