Question 768700
Mackenzie can bicycle 69miles in the same time as it takes her to walk 27miles. 

{{{69mil/s=t}}}...............{{{s}}}= speed when she bicycle

{{{27mil/s[1]=t}}}..............{{{s[1]}}}= speed when she

since time is same,  =>

{{{69miles/s=27mil/s[1]}}}


since she can ride {{{7mph}}} faster than she can walk, then speed 

{{{s=s[1]+7mph}}}.......plug it in equation above


{{{69mil/(s[1]+7mil/h)=27mil/s[1]}}}....solve for {{{s[1]}}}


{{{69mil*s[1]=27mil(s[1]+7mil/h)}}}


{{{69mil*s[1]=27mil*s[1]+189(mil^2/h)}}}..cancel mil


{{{69*s[1]-27*s[1]=189(mil/h)}}}


{{{42s[1]=189(mil/h)}}}


{{{s[1]=(189/42)(mil/h)}}}


{{{s[1]=4.5(mil/h)}}}.......she walks with a speed of {{{4.5(mil/h)}}}


{{{s=s[1]+7mph}}}=>{{{s=4.5(mil/h)+7mil/h}}}=> {{{s=11.5(mil/h)}}}


let’s check is time same for given distances above:


{{{69mil/s=t}}}=>{{{69mil/11.5(mil/h)=t}}}=>{{{6h=t}}}

{{{27mil/s[1]=t}}}=> {{{27mil/4.5(mil/h)=t}}}=>{{{6h=t}}}