Answer to Question #236101 in Linear Algebra for quack

Question

Michael is a bicycle commuter. he has observed that the reduces the time for his 9 km commute by 15 min when he increases his average speed by 3 km/h. what is Michael's faster speed?

Answer 1

Let $x=$ Michael's slower speed. Then his faster speed will be $x+3.$

When Michael travels with slower speed his time is $t_1=\dfrac{9}{x}.$

When Michael travels with faster speed his time is $t_2=\dfrac{9}{x+3}.$

Michael reduces the time for his 9 km commute by 15 min

$$t_1-t_2=\dfrac{1}{4}$$

$$\dfrac{9}{x}-\dfrac{9}{x+3}=\dfrac{1}{4}$$

$$36(x+3)-36x=x(x+3)$$

$$x^2+3x-108=0, x>0$$

$$D=(3)^2-4(1)(108)=441$$

$$x=\dfrac{-3\pm\sqrt{441}}{2(1)}=\dfrac{-3\pm21}{2}$$

Since $x>0,$ we take

$$x=\dfrac{-3+21}{2}$$

$$x=9$$

Therefore Michael's faster speed is $12$ km/h.