Nonlinear System (Part 2)

Note first that $a,b,c$ must all be positive, since as $abc$ is positive, either all three variables are positive or precisely two are negative, in which cases

• if $a,b$ were negative and $c$ positive then $a^{2}b^{3}c^{4}$ would be negative,

• if $a,c$ were negative and $b$ positive then $a^{-1}b^{7}c^{-2}$ would be negative, and

• if $b,c$ were negative and $a$ positive then $a^{2}b^{3}c^{4}$ would be negative.

So, as $a,b,c \gt 0$ we can let $x = \ln(a), y = \ln(b)$ and $z = \ln(c)$ , transforming the system to

(i) $x + y + z = \ln(2)$ , (ii) $2x + 3y + 4z = \ln(5)$ and (iii) $-x + 7y - 2z = \ln(9)$ .

Then (ii) + 2*(iii) yields $17y = \ln(5) + 2\ln(9) = \ln(405) \Longrightarrow y = \dfrac{\ln(405)}{17}$ .

Next, (i) + (iii) gives us that $8y - z = \ln(2) + \ln(9) = \ln(18) \Longrightarrow z = 8y - \ln(18) = \dfrac{8}{17}\ln(405) - \ln(18)$ .

Finally, using equation (i) we find that $x = \ln(2) - \dfrac{\ln(405)}{17} - \dfrac{8}{17}\ln(405) + \ln(18) = \ln(36) - \dfrac{9}{17}\ln(405)$ .

Thus $a + b + c = e^{x} + e^{y} + e^{z} \approx 1.4236 + 0.9371 + 1.4993 = \boxed{3.86}$ to 2 decimal places.

(Note that $abc \approx 2.000$ as expected.)

I originally solved it by linearizing using natural logarithms (see Brian C's solution). But we can also directly apply Gaussian elimination in a multiplicative way. Equations are repeated and numbered below.

$\large \begin {cases} abc = 2 (Eq 1) \\ a^{2}b^{3}c^{4} = 5 (Eq 2) \\ a^{-1}b^{7}c^{-2} = 9 (Eq 3) \end {cases}$

$(Eq 3)^{2}$ multiplied by $(Eq 2)$ yields $b = (81*5)^{\frac{1}{17}} \approx 1.42357$

$(Eq 3)$ multiplied by $(Eq 1)$ yields $b^{8} c^{-1} = 18$ . Therefore, $c = \frac{(81*5)^\frac{8}{17}}{18} \approx 0.93706$

Plugging $b$ and $c$ into $(Eq1)$ , $a$ is easily found to be $\frac{36}{(81*5)^\frac{9}{17}} \approx 1.49929$

$\begin {cases} abc = 2 \ ...(1) \\ a^{2}b^{3}c^{4} = 5 \ ...(2) \end {cases}$

Rearranging $(2)$ .

$\implies (abc)^2 × bc^2=5$

This gives:

$\implies \color{#D61F06}{b}=\dfrac{5}{4c^2}$ and $c=\sqrt{\dfrac{5}{4b}} \ ...(3)$

Putting $\color{#D61F06}{b}$ in $(1)$ .

$\implies a×\dfrac{5}{4c^2}×c=2$

This gives:

$\implies \color{#20A900}{c}$ $= \dfrac{5a}{8}$

Putting $\color{#20A900}{c}$ in $\color{#D61F06}{b}$ .

$\implies \color{#D61F06}{b}$ $= \dfrac{5}{4×\frac{25a^2}{4}}=\dfrac{16}{5a^2}$

This gives:

$\implies a=\sqrt{\dfrac{16}{5b}} \ ...(4)$

Now, putting $a$ and $c$ in $(1)$ .

$\implies \sqrt{\dfrac{16}{5b}}×b×\sqrt{\dfrac{5}{4b}}=2$

This gives:

$\implies b=1$

Putting $b$ in $(3)$ and $(4)$ .

$\implies c=\sqrt{\dfrac{5}{4}}$

$\implies a=\sqrt{\dfrac{16}{5}}$

Now,

$\implies a+b+c=\sqrt{\dfrac{16}{5}}+1+\sqrt{\dfrac{5}{4}}\approx \boxed{3.86}$