How would you determine the value of $c$?

Question

$$3a+2b = 4$$

$$\dfrac{1}{5a-2b} = \dfrac{2}{a+6b+5} = \dfrac{3}{c-4}$$

How would you determine the value of $c$?

Regards!

Answer 1

$$\frac { 1 }{ 5a-2b } =\frac { 2 }{ a+6b+5 } =\frac { 3 }{ c-4 } \Rightarrow \frac { 1+2 }{ 5a-2b+a+6b+5 } =\frac { 3 }{ c-4 } \Rightarrow \\ \Rightarrow \frac { 3 }{ 2\underset { 4 }{ \underbrace { \left( 3a+2b \right) } } +5 } =\frac { 3 }{ c-4 } \Rightarrow \frac { 3 }{ 13 } =\frac { 3 }{ c-4 } \Rightarrow c=17\\ \\ $$ Note that we have used here the property of the proportions:

$$\frac { a }{ b } =\frac { c }{ d } =\frac { a\pm c }{ b\pm d } $$

Answer 2

$3a+2b=4 (1)$

$\frac{1}{5a-2b}=\frac{2}{a+6b+5} (2)$

$\frac{2}{a+6b+5}=\frac{3}{c-4} (3)$

From (2) $2(5a-2b)=a+6b+5$ => $10a-4b=a+6b+5$ => $9a=10b+5$ => $a=\frac{10b+5}{9} (4)$

Put (4) into (1) $3*\frac{10b+5}{9} +2b =4$ => $\frac{10b+5}{3} +2b =4$ => $\frac{10b+5 +6b}{3}=4$ => $16b+5=12$ => $b=\frac{7}{16} (5)$

Get $a$ by using (5) and (4) $a=\frac{10b+5}{9}=\frac{10*\frac{7}{16}+5}{9}=\frac{\frac{35}{8}+5}{9} = \frac{75}{72}=\frac{25}{24} (6)$

Calculate $a+6b+5$ by using (5) and (6)

$a+6b+5=\frac{25}{24}+6*\frac{7}{16}+5=\frac{25}{24}+\frac{21}{8}+5=\frac{25+63+120}{24}=\frac{208}{24}=\frac{26}{3} (7)$

Put (7) to (3)

$\frac{2}{a+6b+5}=\frac{3}{c-4}$ => $2(c-4)=3*\frac{26}{3}$ => $2(c-4)=26$ => $c=13+4=17$

Thanks to NickD for finding my mistake.

Answer 3

\begin{align} 3a+2b &= 4 ,\\ \frac{1}{5a-2b} &= \frac{2}{a+6b+5} = \frac{3}{c-4} . \end{align}

This system is equivalent to the linear $4\times4$ system (assuming $5a-2b\ne0$, $a+6b+5\ne0$, $c\ne4$):

\begin{align} 3a+2b &= 4 ,\\ 5a-2b-d&=0 ,\\ \tfrac12a+3b-d&=-\tfrac52 ,\\ \tfrac13c-d&=\tfrac43 . \end{align}

which has a solution \begin{align} a &= \tfrac{25}{24} ,\\ b &= \tfrac{7}{16} ,\\ c &= 17 ,\\ d &= \tfrac{13}{3} . \end{align}

Edit

Aforenamed system of equations can be presented in matrix form as follows.

\begin{align} \begin{bmatrix} 3 & \phantom{-}2 & 0 & \phantom{-}0 \\ 5 & -2 & 0 & -1 \\ 1 & \phantom{-}6 & 0 & -2 \\ 0 & \phantom{-}0 & 1 & -3 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} &= \begin{bmatrix} \phantom{-}4 \\ \phantom{-}0 \\ -5 \\ \phantom{-}4 \end{bmatrix} , \end{align}

And one way of solving it is by using Gauss elimination

$[2]\ \to\ -2\times[2]+[3]$ to eliminate $A_{24}$

\begin{align} \begin{bmatrix} \phantom{-}3 & 2 & 0 & \phantom{-}0 \\ -9 & 10 & 0 & \phantom{-}0 \\ \phantom{-}1 & 6 & 0 & -2 \\ \phantom{-}0 & 0 & 1 & -3 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} &= \begin{bmatrix} \phantom{-}4 \\ -5 \\ -5 \\ \phantom{-}4 \end{bmatrix} , \end{align}

$[2]\ \to\ [2]+3\times[1]$ to eliminate $A_{21}$

\begin{align} \begin{bmatrix} 3 & 2 & 0 & \phantom{-}0 \\ 0 & 16 & 0 & \phantom{-}0 \\ 1 & 6 & 0 & -2 \\ 0 & 0 & 1 & -3 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} &= \begin{bmatrix} \phantom{-}4 \\ \phantom{-}7 \\ -5 \\ \phantom{-}4 \end{bmatrix} . \end{align}

It follows that $b=\tfrac7{16}$ and $a=\tfrac13(4-2\times\tfrac7{16})=\tfrac{25}{24}$, $d=-\tfrac12(-5-\tfrac{25}{24}-6\times\tfrac7{16})=\tfrac{13}3$, and finally, $c=4-(-3)\times\tfrac{13}3=17$.