Systems of equations

Subtract equation 2 from equation 1 to get:

5y = 20   or  y =4

Sustitute y = 4 into EITHER equation to find 'x'   5x + 3 (4) = 17   5x = 5    x = 1

Solve the following system:
{5 x-2 y = -3 |     (equation 1)
5 x+3 y = 17 |     (equation 2)
Subtract equation 1 from equation 2:
{5 x-2 y = -3 |     (equation 1)
0 x+5 y = 20 |     (equation 2)
Divide equation 2 by 5:
{5 x-2 y = -3 |     (equation 1)
0 x+y = 4 |     (equation 2)
Add 2 × (equation 2) to equation 1:
{5 x+0 y = 5 |     (equation 1)
0 x+y = 4 |     (equation 2)
Divide equation 1 by 5:
{x+0 y = 1 |     (equation 1)
0 x+y = 4 |     (equation 2)
Collect results:
Answer: | {x =1  and y=4}
 

Systems of equations 

5x+3y=17

5x-2y=-3 

5x+3y=17

minus

5x-2y=-3

5x+3y-(5x-2y)=17-(-3)

5x+3y-5x+2y=20

5y=20

y=4

5x+3y=17

+

-5x+2y=3

=

5y=20

y=4

x=1