text: Question 178816
  Elimination is the easiest way to solve this system.
  
  {{{system(3x-5y=7,2x+5y=13)}}}
  
  Add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:
  
  {{{(3x-5y)+(2x+5y)=(7)+(13)}}}
  
  {{{(3x+2x)+(-5y+5y)=7+13}}} Group like terms.
  
  {{{5x+0y=20}}} Combine like terms.
  
  {{{5x=20}}} Simplify.
  
  {{{x=(20)/(5)}}} Divide both sides by {{{5}}} to isolate {{{x}}}.
  
  {{{x=4}}} Reduce.
  
  ------------------------------------------------------------------
  
  {{{3x-5y=7}}} Now go back to the first equation.
  
  {{{3(4)-5y=7}}} Plug in {{{x=4}}}.
  
  {{{12-5y=7}}} Multiply.
  
  {{{-5y=7-12}}} Subtract {{{12}}} from both sides.
  
  {{{-5y=-5}}} Combine like terms on the right side.
  
  {{{y=(-5)/(-5)}}} Divide both sides by {{{-5}}} to isolate {{{y}}}.
  
  {{{y=1}}} Reduce.
  
  So our answer is {{{x=4}}} and {{{y=1}}}.
  
  Which form the ordered pair *[Tex \LARGE \left(4,1\right)].
  
  This means that the system is consistent and independent.
  
  Notice when we graph the equations, we see that they intersect at *[Tex \LARGE \left(4,1\right)]. So this visually verifies our answer.
  
  {{{drawing(500,500,-6,14,-9,11,
  grid(1),
  graph(500,500,-6,14,-9,11,(7-3x)/(-5),(13-2x)/(5)),
  circle(4,1,0.05),
  circle(4,1,0.08),
  circle(4,1,0.10)
  )}}} Graph of {{{3x-5y=7}}} (red) and {{{2x+5y=13}}} (green)