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Solve for all values of x and y

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

Apr 9, 2018

#1
0

sorry man idk

Apr 9, 2018
#2
+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
+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)   Apr 9, 2018
#4
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i would agree with hectictar

Apr 9, 2018
#5
0

and cphill maybe cphill more

Apr 9, 2018
#6
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LOL, Jakesplace....!!!

I'd say that hectictar is as "correct"  as I am......!!!!!   Apr 9, 2018
#7
+1

i agree cphill

Apr 9, 2018
#8
+2

Thank You!! all!!!

Apr 9, 2018
#9
0

cphill and hectictar did all the work

Apr 9, 2018