The distance travelled is given by the product between the average speed and the time used to travel.
[tex]d=vt[/tex]If we call the average speed to travel to her friend's house as r1 and the time she took travelling to go as t1, the distance between between Carmen's house and her friend's house is given by
[tex]r_1_.t_1=12[/tex]Analogously, if we call the average speed to travel to back to her house as r2 and the time she took travelling to go as t2, we have
[tex]r_2t_2=12[/tex]On her way there, her average speed was 7 miles per hour faster than on her way home, therefore, we have
[tex]r_1=r_2+7[/tex]And also, Carmen spent a total of 1 hour bicycling
[tex]t_1+t_2=1[/tex]Then, we have a system with 4 equations and 4 variables.
[tex]\begin{gathered} r_1t_1=12 \\ r_{2}t_{2}=12 \\ r_{1}=r_{2}+7 \\ t_{1}+t_{2}=1 \end{gathered}[/tex]If we solve the second equation for r2 and substitute in the third equation, we're going to have r1 in terms of t2.
[tex]\begin{gathered} r_2t_2=12\Rightarrow r_2=\frac{12}{t_2} \\ r_1=r_2+7\Rightarrow r_1=\frac{12}{t_2}+7 \end{gathered}[/tex]Then, if we solve the last equation for t1, we're going to have the
[tex]t_1=1-t_2[/tex]Using those two expressions for r1 and t1 on the first equation of our system, we're going to have a new equation for t2.
[tex]\begin{gathered} r_{1}t_{1}=12 \\ \lparen\frac{12}{t_2}+7)\left(1-t_2\right)=12 \\ -7t_2^2-17t_2+12=0 \end{gathered}[/tex]Using the quadratic formula, we can find our value for t2.
[tex]t_2=\frac{4}{7}[/tex]Using this value on the second equation, we can find r2.
[tex]\begin{gathered} r_2.\left(\frac{4}{7}\right?=12 \\ r_2=\frac{12\cdot7}{4} \\ r_2=21 \end{gathered}[/tex]Then, using this value on the third equation
[tex]r_1=\left(21\right)+7=28[/tex]Carmen's average speed to her friend's house was 28 mph, and
Carmen's average speed from her friend's house was 21 mph.
