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 Oct 7, 2019
edited by SoulSlayer615  Oct 23, 2019
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\(3x+4y=7\\ 6x+4y=c-4y\\~\\ 6x+8y=c\\ \text{letting $c=14$ makes equation 2 twice equation 1 and thus the system has infinite solutions}\)

 

 

 

\(ab^4 = 12\\ a^5 b^5 = 7776\\ a^5b^{20} = 12^5\\ b^{15}=\dfrac{12^5}{7776} = 32\\ b^3 = 2\\ b= 2^{1/3}\\ 2^{4/3} a= 12\\ a = 3 \cdot 2^{2/3}\)

 

 

 

\(\text{Add the two equations}\\ 2a = 8b\\ b=\dfrac a 4\\ a + a\left(\dfrac a 4\right)^2 = 40\left(\dfrac a 4\right)\\ \dfrac{a^3}{16} =9a\\ a^2 = 144\\ a= \pm 12\\ b=\pm 3\\ \text{$a=0,b=0$ is also a solution} \)

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 Oct 7, 2019

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