Respuesta :
Let:
x = biking speed
y = running speed
To solve this question, follow the steps below.
Step 01: Write an equation that relates the running and biking speed.
Given: His biking speed is 6 mph faster than his running speed
Then,
x = y + 6
Step 02: Write an equation to the total hours trained.
Given: speed = distance/time
Then, time = distance/speed
4 hours = (distance/speed)biking + ((distance/speed)running
Then,
4 = 15/x + 10/y
Step 03: Substitute x by (y + 6) in the equation from step 02.
[tex]\begin{gathered} 4=\frac{15}{x}+\frac{10}{y} \\ 4=\frac{15}{y+6}+\frac{10}{y} \end{gathered}[/tex]Step 04: Solve the equation above.
[tex]\begin{gathered} \frac{4\left(y+6\right)\left(y\right)=15y+10(y+6)}{(y+6)(y)} \\ \begin{equation*} 4\left(y+6\right)\left(y\right)=15y+10(y+6) \end{equation*} \\ (4y+24)y=15y+10y+60 \\ 4y^2+24y=25y+60 \\ 4y^2+24y-25y-60=0 \\ 4y-y-60=0 \end{gathered}[/tex]Use the quadratic formula to solve the equation. For a equation ax² + bx+ c = 0, the roots are:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]Then, substituting the values from this question:
[tex]\begin{gathered} y=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ y=\frac{-(-1)\pm\sqrt{(-1)^2-4*4*(-60)}}{2*4} \\ y=\frac{1\pm\sqrt{1+960}}{8} \\ y=\frac{1\operatorname{\pm}\sqrt{961}}{8} \\ y=\frac{1\operatorname{\pm}31}{8} \\ y_1=\frac{1-31}{8}=-\frac{30}{8}=3.75 \\ y_2=\frac{32}{8}=4 \end{gathered}[/tex]Since y must be positive, y = 4 mph.
Step 06: Find x.
x = y + 6
[tex]\begin{gathered} x=4+6 \\ x=10mph \end{gathered}[/tex]Since y = running speed:
Answer: the running speed was 4 mph.
