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Question Number 155631 by aunzo last updated on 03/Oct/21

(x+3(x−2)=x+10

$$\left({x}+\mathrm{3}\left({x}−\mathrm{2}\right)={x}+\mathrm{10}\right. \\ $$

Commented by aunzo last updated on 03/Oct/21

plss

$${plss} \\ $$

Commented by Ar Brandon last updated on 03/Oct/21

syntax error !

$$\mathrm{syntax}\:\mathrm{error}\:! \\ $$

Answered by Rasheed.Sindhi last updated on 03/Oct/21

If it′s as (x+3)(x−2)=x+10            x^2 +x−6=x+10            x^2 −6=10            x^2 =10+6=16             x=±4  If it′s as (x+3(x−2))=x+10                       x+3x−6=x+10                        4x−x=10+6=16                         3x=16                          x=((16)/3)=5(1/3)

$${If}\:{it}'{s}\:{as}\:\left({x}+\mathrm{3}\right)\left({x}−\mathrm{2}\right)={x}+\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +{x}−\mathrm{6}={x}+\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{6}=\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} =\mathrm{10}+\mathrm{6}=\mathrm{16} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}=\pm\mathrm{4} \\ $$$${If}\:{it}'{s}\:{as}\:\left({x}+\mathrm{3}\left({x}−\mathrm{2}\right)\right)={x}+\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}+\mathrm{3}{x}−\mathrm{6}={x}+\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}{x}−{x}=\mathrm{10}+\mathrm{6}=\mathrm{16} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{x}=\mathrm{16} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{16}}{\mathrm{3}}=\mathrm{5}\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Answered by peter frank last updated on 03/Oct/21

x+3(x−2)=x+10  x+3x−6=x+10  4x−6=x+10  3x=16  x=((16)/3)

$$\mathrm{x}+\mathrm{3}\left(\mathrm{x}−\mathrm{2}\right)=\mathrm{x}+\mathrm{10} \\ $$$$\mathrm{x}+\mathrm{3x}−\mathrm{6}=\mathrm{x}+\mathrm{10} \\ $$$$\mathrm{4x}−\mathrm{6}=\mathrm{x}+\mathrm{10} \\ $$$$\mathrm{3x}=\mathrm{16} \\ $$$$\mathrm{x}=\frac{\mathrm{16}}{\mathrm{3}} \\ $$

Commented by ajfour last updated on 03/Oct/21

good notice!

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