Question

Algebra: Find all values of x satisfying the given conditions.

Answer 1

\[\Large {\color{red}{y_{1}}}={\color{red}{\frac{2}{x+4}}}\]

\[\Large {\color{blue}{y_{2}}}={\color{blue}{\frac{5}{x+3}}}\]

\[\Large {\color{green}{y_{3}}}={\color{green}{\frac{12x-4}{x^2+7x+12}}}\]


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\[\Large {\color{red}{y_{1}}}+{\color{blue}{y_{2}}} = {\color{green}{y_{3}}}\]


\[\Large {\color{red}{\frac{2}{x+4}}}+{\color{blue}{\frac{5}{x+3}}} = {\color{green}{\frac{12x-4}{x^2+7x+12}}}\]


\[\Large \frac{2}{x+4}+\frac{5}{x+3} = \frac{12x-4}{x^2+7x+12}\]


\[\Large \frac{2}{x+4}{\color{purple}{*\frac{x+3}{x+3}}}+\frac{5}{x+3}{\color{purple}{*\frac{x+4}{x+4}}} = \frac{12x-4}{x^2+7x+12}\]


\[\Large \frac{2(x+3)}{(x+4)(x+3)}+\frac{5(x+4)}{(x+3)(x+4)} = \frac{12x-4}{x^2+7x+12}\]


\[\Large \frac{2x+6}{x^2+7x+12}+\frac{5x+20}{x^2+7x+12} = \frac{12x-4}{x^2+7x+12}\]


\[\Large \frac{2x+6+5x+20}{x^2+7x+12} = \frac{12x-4}{x^2+7x+12}\]

I'll let you finish up

Answer 2

i ended up with -3, -4 which when put back in the problem comes back as 0. it has no solution set.

Answer 3

\[\large \frac{2x+6+5x+20}{x^2+7x+12} = \frac{12x-4}{x^2+7x+12}\]

\[\large \frac{7x+26}{x^2+7x+12} = \frac{12x-4}{x^2+7x+12}\]

\[\large \frac{7x+26}{x^2+7x+12}{\color{red}{*(x^2+7x+12)}} = \frac{12x-4}{x^2+7x+12}{\color{red}{*(x^2+7x+12)}}\]

\[\large \frac{7x+26}{\cancel{x^2+7x+12}}{\cancel{(x^2+7x+12)}} = \frac{12x-4}{\cancel{x^2+7x+12}}{\cancel{(x^2+7x+12)}}\]

\[\large 7x+26 = 12x-4\]

Do you see what to do next?

Answer 4

thank you so much for helping! i finished it out and got x=6, checked it and it was right.

Answer 5

yep x = 6 is the final answer