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9
avatar+1450 

Help Me please.

 

Solve for all values of x and y

 

\(\left \{x^2-3y^2=13 \right \} \left \{x-2y=1 \right \}\)

 Apr 9, 2018
 #1
avatar+428 
0

sorry man idk

 Apr 9, 2018
 #2
avatar+8439 
+2

x - 2y  =  1

                         Add  2y  to both sides of the equation.

x  =  1 + 2y

 

x2 - 3y2  =  13

                                               Since  x  =  1 + 2y  we can replace  x  with  1 + 2y .

(1 + 2y)2 - 3y2  =  13

                                               Multiply out  (1 + 2y)2  .

(1 + 2y)(1 + 2y) - 3y2  =  13

 

(1)(1) + (1)(2y) + (2y)(1) + (2y)(2y) - 3y2  =  13

 

1 + 2y + 2y + 4y2 - 3y2  =  13

 

1 + 4y + y2  =  13

 

y2 + 4y + 1  =  13

                                    Subtract  13  from both sides.

y2 + 4y - 12  =  0

                                    Factor the left side.

(y + 6)(y - 2)  =  0

                                    Set each factor equal to zero and solve for  y .

 

y + 6  =  0       or       y - 2  =  0

  y  =  -6          or         y  =  2

 

Now let's use the equation   x  =  1 + 2y   to find  x .

 

If   y  =  -6   then   x  =  1 + 2(-6)  =  1 + -12  =  -11


So one solution is   x  =  -11   and   y =  -6  .

 

If   y  =  2   then   x  =  1 + 2(2)  =  1 + 4  =  5

 

So another solution is   x  =  5   and   y  =  2  .

 

The solution set is   { (-11, -6), (5, 2) }

 Apr 9, 2018
 #3
avatar+101872 
+2

x^2 - 3y^2  = 13

 x - 2y  =  1  

Rearrange the second equation as

x  = 2y + 1

 

Sub this into  the first equation for x

 

(2y + 1)^2 - 3y^2  =  13      simplify

4y^2 + 4y + 1  - 3y^2  = 13

y^2 + 4y + 1  = 13     subtract 1 from both sides

y^2 + 4y - 12  =  0      factor

(y + 6) (y - 2)  = 0

Setting each factor to 0 and solving for y we have

 

y + 6   =  0              y  - 2  = 0

y  = -6                     y = 2

 

And when y  = -6 ....  x  = 2(-6) + 1  =  -11

And when y  = 2.....  x  = 2(2) + 1   =  5

 

So....the solutions are

(x,y)  =  ( 5, 2)  and  ( -11, -6)

 

 

cool cool cool

 Apr 9, 2018
 #4
avatar+428 
0

i would agree with hectictar

 Apr 9, 2018
 #5
avatar+428 
0

and cphill maybe cphill more

 Apr 9, 2018
 #6
avatar+101872 
0

LOL, Jakesplace....!!!

 

I'd say that hectictar is as "correct"  as I am......!!!!!

 

 

cool cool cool

 Apr 9, 2018
 #7
avatar+428 
+1

i agree cphill

 Apr 9, 2018
 #8
avatar+1450 
+2

Thank You!! all!!!

 Apr 9, 2018
 #9
avatar+428 
0

cphill and hectictar did all the work 

 Apr 9, 2018

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