algebra 2 literal equations

algray09

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so my class is doing literal equations and there is one problem on the homework that I keep getting wrong even though I got the same answer as others.
the problem is g=c/x + d/r, solve for x.

the answer I keep getting is cr/gr-d but that's not the right answer. how do I solve this?
 
so my class is doing literal equations and there is one problem on the homework that I keep getting wrong even though I got the same answer as others.
the problem is g=c/x + d/r, solve for x.

the answer I keep getting is cr/gr-d but that's not the right answer. how do I solve this?
Your response is ALMOST correct - but it has a 'fatal flaw'.

Please share your work - so that we can catch where you are slipping off.
 
The equation is g= c/x+ d/r. If you were given x how would you calculate g? What the equation says is "first divide c by x then add d/r". To solve for x do the opposite of each step in the reverse order! The last done was "add d/r" so subtract d/r. Of course you have to do the same thing so g- d/r= c/x+ d/r- d/r= c/x. The opposite of "divide by x" is multiply by x: (g- d/r)x= c. Now x is multiplied by g- d/r so do the opposite of that- divide both side by g- d/r= (gr- d)/r. That is the same as multiplying by r/(gr- d). Notice the parentheses!
 
so my class is doing literal equations and there is one problem on the homework that I keep getting wrong even though I got the same answer as others.
the problem is g=c/x + d/r, solve for x.

the answer I keep getting is cr/gr-d but that's not the right answer. how do I solve this?
Even if what you have above is correct, why didn't you continue and cancel out the r's?? Then x=c/g-d.
 
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Even if what you have above is correct, why didn't you continue and cancel out the r's?? Then x=c/g-d.
If
[MATH] \frac{c}{g-d}=\frac{cr}{gr-d}[/MATH]then [MATH]c(gr-d)=cr(g-d) \longrightarrow cgr-dc=cgr-dcr \longrightarrow -dc=-dcr \therefore r=1? [/MATH]
 
so my class is doing literal equations and there is one problem on the homework that I keep getting wrong even though I got the same answer as others.
the problem is g=c/x + d/r, solve for x.

the answer I keep getting is cr/gr-d but that's not the right answer. how do I solve this?
A week went by - the OP is AWOL and Jomo is getting ......... well .... So the solution:

\(\displaystyle \ g \ = \ \frac{c}{x} \ + \ \frac{d}{r}\) ..... there are multiple ways to solve - but I like to isolate the "find"

\(\displaystyle \ g \ - \ \frac{d}{r} \ = \ \frac{c}{x} \ \) ....... eliminate fraction by multiplying all the terms LCM of the denominators (= r*x*1)

\(\displaystyle \ g * r * x\ - \ \frac{d}{r} *r * x \ = \ \frac{c}{x} * r * x \ \)

\(\displaystyle \ g * r * x \ - \ d * x \ = \ c * r \ \)........................."factor out" x

\(\displaystyle \ x * (g * r \ - \ d ) \ = \ c * r \ \)......................and we get

\(\displaystyle \ x = \frac {c * r}{g * r \ - \ d } \ \).

can also be written as:

x = c * r / (g*r -d) ..........................Those () are super-important to display correct answer.
 
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