Question

x/Isl.exe/1o u-lgNslkr7 8P3)H- Solving a word problem using a system of linear equations of the.. Charlie the trainer has two solo workout plans that he offers his dients: Plan A and Plan B. Each client does either one or the other (not both). On Monday there were 3 dlients who did Plan A and 2 who did Plan B. On Tuesday there were 5 dients who did Plan A and 6 who did Plan B. Charlie trained his Monday dients for a total of 3 hours and his Tuesday dients for a total of 7 hours. How long does each of the workout plans last? Length of each Plan A workout: hour(s) 2 DOLL

Answer 1

Solution :

$\mathrm{and\:length\:of\:each\:plan\:B\:workout\:be\:}t_{2}\:\mathrm{hours}.$

$\mathrm{Then,\:on\:Monday,\:\:3\:clients\:who\:did\:plan\:A\:\:and\:\:2\:clients\:who\:did\:\:plan\:B\:for\:a\:total\:of\:3\:hours.}$

$3t_1+2t_2=3$

$\mathrm{Then,\:on\:Tuesday,\:\:5\:clients\:who\:did\:plan\:A\:\:and\:\:6\:clients\:who\:did\:\:plan\:B\:for\:a\:total\:of\:7\:hours.}$

$5t_1+6t_2=7$

$\mathrm{Following\:are\:the\:two\:systems\:of\:equations\::}$

$\begin{bmatrix}3t_1+2t_2=3\\ 5t_1+6t_2=7\end{bmatrix}$

$\mathrm{Using\:the\:substitution\:method\:for\:systems\:of\:equations,}$

$\mathrm{Isolate}\:t_1\:\mathrm{for}\:3t_1+2t_2=3:$

$3t_1+2t_2=3$

$\mathrm{Subtract\:}2t_2\mathrm{\:from\:both\:sides}$

$3t_1+2t_2-2t_2=3-2t_2$

$\mathrm{Simplify}$

$3t_1=3-2t_2$

$\mathrm{Divide\:both\:sides\:by\:}3$

$\frac{3t_1}{3}=\frac{3}{3}-\frac{2t_2}{3}$

$\mathrm{Simplify}$

$t_1=\frac{3-2t_2}{3}$

$\mathrm{Substitute\:}t_1=\frac{3-2t_2}{3}$

$\begin{bmatrix}5* \frac{3-2t_2}{3}+6t_2=7\end{bmatrix}$

$\mathrm{Isolate}\:t_2\:\mathrm{for}\:5*\frac{3-2t_2}{3}+6t_2=7:$

$5* \frac{3-2t_2}{3}+6t_2=7$

$\frac{5\left(3-2t_2\right)}{3}+6t_2=7$

$\mathrm{Multiply\:both\:sides\:by\:}3$

$\frac{5\left(3-2t_2\right)}{3}*3+6t_2*3=7*3$

$\mathrm{Refine}$

$5\left(3-2t_2\right)+18t_2=21$

$\mathrm{Add\:similar\:elements:}\:-10t_2+18t_2=8t_2$

$15+8t_2=21$

$\mathrm{Subtract\:}15\mathrm{\:from\:both\:sides}$

$15+8t_2-15=21-15$

$\mathrm{Simplify}$

$8t_2=6$

$\mathrm{Divide\:both\:sides\:by\:}8$

$\frac{8t_2}{8}=\frac{6}{8}$

$\mathrm{Simplify}$

$t_2=\frac{3}{4}$

$\mathrm{For\:}t_1=\frac{3-2t_2}{3}$

$\mathrm{Substitute\:}t_2=\frac{3}{4}$

$t_1=\frac{3-2* \frac{3}{4}}{3}=\frac{3-\frac{3}{2}}{3}=\frac{\frac{3*2-3}{2}}{3}=\frac{\frac{6-3}{2}}{3}=\frac{\frac{3}{2}}{3}=\frac{3}{6}=\frac{1}{2}$

$\mathrm{Substitute\:}t_{1}=\frac{1}{2}\:\mathrm{\:in\:the\:equation\:\:}3t_1+2t_2=3\::$

$3*\left ( \frac{1}{2} \right )+2t_2=3$

$\frac{3}{2}+2t_2=3$

$\mathrm{Subtract\:}\frac{3}{2}\mathrm{\:from\:both\:sides}$

$\frac{3}{2}+2t_2-\frac{3}{2}=3-\frac{3}{2}$

$\mathrm{Simplify}$

$2t_2=\frac{3}{2}$

$\mathrm{Divide\:both\:sides\:by\:}2$

$\frac{2t_2}{2}=\frac{\frac{3}{2}}{2}$

$\mathrm{Simplify}$

$t_2=\frac{3}{4}$

Hence

$\mathrm{Length\:of\:each\:plan\:A\:workout\:=\:\frac{1}{2}\:hour(s).}$

$\mathrm{and\:length\:of\:each\:plan\:B\:workout\:=\:\frac{3}{4}\:hour(s).}$