Question 807305

let a number be {{{x}}} and another number {{{y}}}
number times another number equals {{{49}}}:

{{{x*y=49}}}........eq.1

 and the sum of {{{28}}}:

{{{x+y=28}}}........eq.2

solve the system:

{{{x*y=49}}}........eq.1
{{{x+y=28}}}........eq.2
__________________________

{{{x+y=28}}}........eq.2.....solve for {{{x}}}

{{{x =28-y}}}........eq.2a.....substitute in eq.1


{{{(28-y)*y=49}}}........eq.1...solve for {{{y}}}

{{{28y-y^2=49}}}

{{{0=y^2-28y+49}}}......use quadratic formula


{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{y = (-(-28) +- sqrt( (-28)^2-4*1*49 ))/(2*1) }}}

{{{y = ( 28  +- sqrt( 784-196 ))/2 }}}

{{{y = ( 28  +- sqrt( 588 ))/2 }}}

{{{y = ( 28  +- 24.25)/2 }}}

solutions:

{{{y = ( 28  + 24.25)/2 }}}

{{{y = 26.125}}}

{{{y = ( 28  -24.25)/2 }}}

{{{y = 1.875}}}

now find {{{x =28-y}}}........eq.2a

{{{x =28-26.12}}}

{{{x =1.875}}}

or

{{{x =28-1.875}}}

{{{x =26.125}}}

so, one solution is {{{x =1.875}}} and {{{y = 26.125}}}

and another is vs, {{{x =26.125}}} and {{{y = 1.875}}}

check:
{{{x*y=49}}}........eq.1=>{{{1.875*26.125=48.984375}}}...rounded {{{49}}}
{{{x+y=28}}}........eq.2=>{{{1.875+26.125=28}}}