Question

Find all values of c such that c/(c - 5) = 4/(c - 4). If you find more than one solution, then list the solutions you find separated by commas.

Answer 1

`(c)/(c - 5) = (4)/(c - 4)`

`(c - 5)(c - 4) * (c)/(c - 5) = (4)/(c - 4) * (c - 5)(c - 4)`

`(c - 4) * c = 4(c - 5)`

With c≠4, 5

`{c}^(2) - 4c = 4c - 20`

`{c}^(2) - 8c + 20 = 0`

`c = (2 - + √(2 - 20) )/(2)`

But 2-20 is negative, therefore there isn't any real solution to this equation.

Answer 2

`\large\boxed{\bold{NO\ REAL\ SOLUTIONS}}\\\boxed{x=4-2i,\ x=4+2i}`

Explanation:

`Domain:\\c-4\\eq0\ \wedge\ c-5\\eq0\Rightarrow c\\eq4\ \wedge\ c\\eq5\\\\(c)/(c-5)=(4)/(c-4)\qquad\text{cross multiply}\\\\c(c-4)=4(c-5)\qquad\text{use the distributive property}\\\\c^2-4c=4c-20\qquad\text{subtract}\ 4c\ \text{from both sides}\\\\c^2-8c=-20\qquad\text{add 20 to both sides}\\\\c^2-8c+20=0\qquad\text{use the quadratic formula}`

`\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac<0,\ \text{then the equation has no solution}\\\\\text{if}\ b^2-4ac=0,\ \text{then the equation has one solution}\ x=(-b)/(2a)\\\\\text{if}\ b^2-4ac>0,\ \text{then the equation has two solutions}\ x=(-b\pm√(b^2-4ac))/(2a)`

`x^2-8x+20=0\\\\a=1,\ b=-8,\ c=20\\\\b^2-4ac=(-8)^2-4(1)(20)=64-80=-16<0\\\\\bold{NO\ REAL\ SOLUTIONS}`

`\text{In the set of complex numbers:}\\\\i=√(-1)\\\\\text{therefore}\ √(b^2-4ac)=√(-16)=√((16)(-1))=√(16)\cdot√(-1)=4i\\\\x=(-(-8)\pm4i)/(2(1))=(8\pm4i)/(2)=(8)/(2)\pm(4i)/(2)=4\pm2i`