Answer :
Let v be the average speed from Ayville to Beeville.
Let t_1 be the time that Russell took to travel from Ayville to Beevile, and t_2 the time that he took to travel from Beeville to Charlwoods.
Remember the formula that relates the average speed of a moving object with the distance that it traveled and the time that it takes to travel that distance:
[tex]v=\frac{d}{t}[/tex]Isolate t from the equation:
[tex]t=\frac{d}{v}[/tex]For the first part of the journey, the distance from Ayville to Beeville is 119 miles. Then:
[tex]t_1=\frac{119mi}{v}[/tex]For the second part of the journey, the distance from Beeville to Charlwoods is 90 miles, and Russell drives 10 miles per hour faster. Then:
[tex]t_2=\frac{90mi}{v+10\frac{mi}{h}}[/tex]Since the combined travel time is 5.4 hours, then:
[tex]t_1+t_2=5.4h[/tex]Substitute the expression for t_1 and t_2:
[tex]\Rightarrow\frac{119mi}{v}+\frac{90mi}{v+10\frac{mi}{h}}=5.4h[/tex]Add both fractions on the left member of the equation:
[tex]\Rightarrow\frac{(119mi)(v+10\cdot\frac{mi}{h})+(90mi)(v)}{(v)(v+10\frac{mi}{h})}=5.4h[/tex]Simplify the expression on the numerator:
[tex]\begin{gathered} \Rightarrow\frac{119mi\cdot v+1190\cdot\frac{mi^2}{h}+90mi\cdot v}{(v)(v+10\frac{mi}{h})}=5.4h \\ \Rightarrow\frac{209mi\cdot v+1190\cdot\frac{mi^2}{h}}{(v)(v+10\frac{mi}{h})}=5.4h \end{gathered}[/tex]Multiply the factors on the denominator:
[tex]\Rightarrow\frac{209mi\cdot v+1190\cdot\frac{mi^2}{h}}{v^2+10\frac{mi}{h}v}=5.4h[/tex]Multiply both sides of the equation by (v^2+10(mi/h)v):
[tex]\begin{gathered} \Rightarrow209mi\cdot v+1190\cdot\frac{mi^2}{h}=(5.4h)(v^2+10\frac{mi}{h}v) \\ \Rightarrow209mi\cdot v+1190\cdot\frac{mi^2}{h}=5.4h\cdot v^2+54mi\cdot v \end{gathered}[/tex]Notice that we obtained a quadratic equation for v. Write it in standard form and use the quadratic formula to find the values of v. Recall the following:
[tex]\begin{gathered} ax^2+bx+c=0 \\ \Rightarrow x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]Then, from the equation for v:
[tex]\begin{gathered} \Rightarrow209mi\cdot v+1190\cdot\frac{mi^2}{h}=5.4h\cdot v^2+54mi\cdot v \\ \Rightarrow0=5.4h\cdot v^2+54mi\cdot v-209mi\cdot v-1190\cdot\frac{mi^2}{h} \\ \Rightarrow5.4h\cdot v^2-155mi\cdot v-1190\cdot\frac{mi^2}{h}=0 \end{gathered}[/tex]To use the quadratic formula, notice that a=5.4h, b=-155mi and c=-1190(mi^2/h):
[tex]\begin{gathered} \Rightarrow v=\frac{-(-155mi)\pm\sqrt[]{(-155mi)^2-4(5.4h)(-1190\cdot\frac{mi^2}{h})}}{2(5.4h)} \\ =\frac{155mi\pm\sqrt[]{24,025mi^2+25,704mi^2}}{10.8h} \\ =\frac{155mi\pm223mi}{10.8h} \end{gathered}[/tex]Using only the positive value for the speed, we get that:
[tex]\begin{gathered} v=\frac{155mi+223mi}{10.8h} \\ =\frac{378mi}{10.8h} \\ =35\frac{mi}{h} \end{gathered}[/tex]Since v is the speed from Ayville to Beevile and the speed from Beevile to Charlwoods is 10 miles per hour faster, then, the average speeds for each part, are:
From Ayville to Beevile: 35 miles per hour.
From Beevile to Charlwoods: 45 miles per hour.
