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# Help

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Solve

$$\frac{x^2+2x+3}{x+4}=x+5$$

for x.

Jan 5, 2021

#1
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If we multiply both sides by x+4,we get $$x^2+2x+3=x^2+9x+20$$ which can be simplified to $$0=7x+17$$ubtract 17 from both sides and divide by 7. we get -17/7.

Jan 5, 2021

#1
+412
+1

If we multiply both sides by x+4,we get $$x^2+2x+3=x^2+9x+20$$ which can be simplified to $$0=7x+17$$ubtract 17 from both sides and divide by 7. we get -17/7.

MooMooooMooM Jan 5, 2021
#2
+312
0

Cross-multiplication gives

$$x^2+2x+3=(x+4)(x+5)=x^2+9x+20.$$

Therefore

$$0=7x+17$$

and $$x=\boxed{-\frac{17}7}$$.

Jan 6, 2021