+0

+1
248
2
+1405

Solve the Systems of equations

{x+y=5}

{x2+y2=13}

Mar 6, 2018

#1
+7604
+3

x + y  =  5     Subtract  y  from both sides of this equation.

x  =  5 - y      Use this value of  x  in the second equation.

x2 + y2   =  13

(5 - y)2 + y2   =  13

(5 - y)(5 - y)  +  y2   =  13

25 - 10y + y2  +  y2   =  13

25 - 10y + 2y2   =  13              Subtract  13  from both sides of the equation.

12 - 10y + 2y2   =  0                Rearrange.

2y2 - 10y + 12   =  0                Divide through by  2 .

y2 - 5y + 6   =   0                     Factor left side.

(y - 3)(y - 2)  =  0                      Set each factor equal to zero.

y - 3  =  0     or     y - 2  =  0

y  =  3          or     y  =  2

Use both of these values of  y  to find  x .

When   y  =  3 ,   x  =  5 - y  =  5 - 3  =  2

When   y  =  2 ,   x  =  5 - y  =  5 - 2  =  3

So...

One solution is   x = 2   and   y = 3

Another solution is   x = 3   and   y = 2

The solution set is  { (2, 3), (3, 2) }

Mar 6, 2018

#1
+7604
+3

x + y  =  5     Subtract  y  from both sides of this equation.

x  =  5 - y      Use this value of  x  in the second equation.

x2 + y2   =  13

(5 - y)2 + y2   =  13

(5 - y)(5 - y)  +  y2   =  13

25 - 10y + y2  +  y2   =  13

25 - 10y + 2y2   =  13              Subtract  13  from both sides of the equation.

12 - 10y + 2y2   =  0                Rearrange.

2y2 - 10y + 12   =  0                Divide through by  2 .

y2 - 5y + 6   =   0                     Factor left side.

(y - 3)(y - 2)  =  0                      Set each factor equal to zero.

y - 3  =  0     or     y - 2  =  0

y  =  3          or     y  =  2

Use both of these values of  y  to find  x .

When   y  =  3 ,   x  =  5 - y  =  5 - 3  =  2

When   y  =  2 ,   x  =  5 - y  =  5 - 2  =  3

So...

One solution is   x = 2   and   y = 3

Another solution is   x = 3   and   y = 2

The solution set is  { (2, 3), (3, 2) }

hectictar Mar 6, 2018
#2
+1405
+2

Thanks Hect

Mar 6, 2018