Question 31099
This question seems unsensable because it doesn't have a consant in its equation:
Let the speed be x
Going speed = x and returning speed = x
Equation:
*[tex \frac{29}{x}-\frac{24}{x}=1]
<pre><b>29(x)-24(x)=(x)(x)
{{{5x=x^2}}}
{{{x^2-5x=0}}}
Solve by comlpleting the square:
{{{x^2-5x+(-5/x)^2=0+(-5/2)^2}}}
Simpplify:
{{{x^2-5x+25/4=25/4}}}
You get x=5
Or you can factor the left side:
You get x(x-5)=0
x=0 or x=5
<font color =purple>Hence, the speed going and reutrning is 5mph.</font>
Paul.