Question 1197668


let  two nubbers be {{{x}}} and {{{y}}}

if the product of two numbers is {{{10}}} we have 
 {{{xy=10}}}.....eq1

if one number is {{{3}}} more than the other number we have
 {{{x=y+3}}}........solve for {{{y}}}

 {{{y=x-3}}}.......eq.2

a. What is the quadratic equation that models these numbers?
{{{xy=10}}}....eq1, substitute {{{y}}} from eq.2
{{{x(x-3)=10}}}
{{{x^2-3x=10}}}
{{{x^2-3x-10=0}}}


b. Solve the quadratic equation from "a" to find the pairs of numbers

{{{x^2-3x-10=0}}}....factor
{{{x^2+2x-5x-10=0}}}
{{{(x^2+2x)-(5x+10)=0}}}
{{{x(x+2)-5(x+2)=0}}}
{{{(x - 5) (x + 2) = 0}}}

solutions:

{{{(x - 5)= 0}}}=>{{{x=5}}}
{{{ (x + 2) = 0}}}=>{{{x=-2}}}

go to

 {{{y=x-3}}}.......eq.2, substitute {{{x}}}
 {{{y=5-3}}} =>{{{y=2}}}
{{{y=-2-3}}} =>{{{y=-5}}}

you have two pairs of numbers:
{{{x=5}}},{{{y=2}}}

or
{{{x=-2}}},{{{y=-5}}}