Question 953068

Leslie has {{{10}}} gift cards . 

Some are {{{SQUARE}}} in shape => these have {{{4}}} sides
and some are {{{TRIANGULAR}}} => these have {{{3}}} sides


let  {{{SQUARE}}} be {{{x}}} and {{{TRIANGULAR}}}be {{{y}}}:

if Leslie has {{{10}}} gift cards, means {{{x+y=10}}}....eq.1

if the cards have {{{34}}} sides altogether, means {{{4x+3y=34}}}...eq.2

now we have a system to solve:

{{{x+y=10}}}....eq.1
{{{4x+3y=34}}}...eq.2
----------------------------

{{{x+y=10}}}....eq.1...solve for {{{x}}}
{{{x=10-y}}}.........substitute in eq.2

{{{4(10-y)+3y=34}}}...eq.2..solve for {{{y}}}

{{{40-4y+3y=34}}}

{{{40-y=34}}}

{{{40-y+y=34+y}}}

{{{40=34+y}}}

{{{40-34=y}}}

{{{highlight(y=6)}}}

now find {{{x}}}:

{{{x=10-y}}}
{{{x=10-6}}}
{{{highlight(x=4)}}}

so, Leslie has {{{4}}} square cards