+0  
 
+1
227
9
avatar+1317 

Help Me please.

 

Solve for all values of x and y

 

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

ManuelBautista2019  Apr 9, 2018
 #1
avatar+428 
0

sorry man idk

jakesplace  Apr 9, 2018
 #2
avatar+7324 
+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) }

hectictar  Apr 9, 2018
 #3
avatar+90023 
+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

CPhill  Apr 9, 2018
 #4
avatar+428 
0

i would agree with hectictar

jakesplace  Apr 9, 2018
 #5
avatar+428 
0

and cphill maybe cphill more

jakesplace  Apr 9, 2018
 #6
avatar+90023 
0

LOL, Jakesplace....!!!

 

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

 

 

cool cool cool

CPhill  Apr 9, 2018
 #7
avatar+428 
+1

i agree cphill

jakesplace  Apr 9, 2018
 #8
avatar+1317 
+2

Thank You!! all!!!

ManuelBautista2019  Apr 9, 2018
 #9
avatar+428 
0

cphill and hectictar did all the work 

jakesplace  Apr 9, 2018

10 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.