Question 947686
let man’s present age be {{{x}}} and son’s age {{{y}}}}


if one year ago, a man was {{{8}}} times as old as his son, then we have

{{{x-1=8(y-1)}}} ...............eq.1 solve for x
{{{x=8y-8+1}}}
{{{x=8y-7}}}

Now his age is equal to the square of his son’s age:

{{{x=y^2}}} ...............eq.2

go to
{{{x=8y-7}}} ..........substitute in eq.2

{{{8y-7=y^2}}} ...............eq.2..1 solve for y

{{{0=y^2 -8y+7}}} ...factor

{{{0=y^2 -y-7y+7}}}

{{{(y^2 -y)-(7y-7)=0}}}

{{{y(y -1)-7(y-1)=0}}}

{{{(y -7)(y-1)=0}}}

solutions:

if {{{(y -7) =0}}} => {{{highlight(y=7)}}}
if {{{(y-1)=0}}} => {{{y=1}}} .........we can disregard {{{y=1}}} because it will give us same present age for both, father and son 

now find {{{x}}}
{{{x=y^2}}} ...............eq.2
plug in {{{y=7}}}

{{{x=7^2}}}

{{{highlight(x=49)}}}

so, now father is {{{highlight(49)}}} and son {{{highlight(7)}}}

check:
year ago father is {{{48}}} and son {{{6}}}, and  a man was {{{8}}} times as old as his son which is true because {{{6*8=48}}}