Question 270998
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A gallant effort, Elizabeth, but not quite right. Here's how I know:


You say *[tex \Large m\ =\ 3].  If that is true, then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5\ -\ (3)}{7\ -\ (3)}\ =\ \frac{2}{3}]


must be a true statement.  But


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{4}\ =\ \frac{1}{2}\ \neq\ \frac{2}{3}]


You were on the right track on the second step, but where you went from there I cannot fathom.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5\ -\ m}{7\ -\ m}\ =\ \frac{2}{3}]


You multiplied both sides of the equation by the denominator on the left to get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\ -\ m\ =\ \frac{2\left(7\ -\ m\right)}{3}]


Your next step should have been to distribute the 2 in the numerator on the right, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\ -\ m\ =\ \frac{14\ -\ 2m}{3}]


Then you can multiply both sides by 3, the denominator on the right:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\left(5\ -\ m\right)\ =\ 14\ -\ 2m]


Distribute the 3:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 15\ -\ 3m\ =\ 14\ -\ 2m]


Add *[tex \Large 2m] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 15\ -\ 3m\ +\ 2m\ =\ 14]


Add -15 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -m\ =\ 14\ -\ 15]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -m\ =\ -1]


Multiply both sides by -1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ 1]


Check the answer:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5\ -\ (1)}{7\ -\ (1)}\ =^?\ \frac{2}{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4}{6}\ =\ \frac{2}{3}]


Yep, it checks.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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