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Solve alebratically the simultaneous equation 

x^2 + y^2 = 25 

y - 2x = 5

 

please explain every step 

Guest Aug 21, 2017

Best Answer 

 #1
avatar+4478 
+5

Here's how to do it using substitution. There might be a quicker way though....

 

The problem tells us...

y - 2x  =  5               Add  2x  to both sides of this equation.

y  =  5 + 2x

 

The problem tells us...

x2 + y2  =  25                  And since  y = 5 + 2x  , we can replace  y  with  5 + 2x  .

x2 + (5 + 2x)2  =  25

x2 + (5 + 2x)(5 + 2x)  =  25                                 Multiply out the parenthesees.

x2 + (5)(5) + (5)(2x) + (2x)(5) + (2x)(2x)  =  25

x2 + 25 + 10x + 10x + 4x2  =  25                        Combine like terms.

5x2 + 25 + 20x  =  25                                         Subtract  25  from both sides of the equation.

5x2 + 20x  =  0               Factor out an  x  from both terms.

x(5x + 20)  =  0              Set each factor equal to  0  and solve for  x  .

 

x  =  0     or     5x + 20  =  0     Subtract  20  from both sides.

                       5x  =  -20          Divide both sides by  5  .

                       x  =  -4

 

Now plug these values for  x  into the second equation given. (The first one will give you two answers for  y  , but only one answer for  y  works in the second equation.)

 

y - 2x  =  5       Plug in  0  for  x  .

y - 2(0)  =  5

y - 0  =  5

y  =  5

                        

y - 2x  =  5       Plug in  -4  for  x  .

y - 2(-4)  =  5

y - -8  =  5

y + 8  =  5        Subtract  8  from both sides.

    y  =  -3

So the two solutions are:

x = 0,  y = 5          and          x = -4, y = -3

hectictar  Aug 21, 2017
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1+0 Answers

 #1
avatar+4478 
+5
Best Answer

Here's how to do it using substitution. There might be a quicker way though....

 

The problem tells us...

y - 2x  =  5               Add  2x  to both sides of this equation.

y  =  5 + 2x

 

The problem tells us...

x2 + y2  =  25                  And since  y = 5 + 2x  , we can replace  y  with  5 + 2x  .

x2 + (5 + 2x)2  =  25

x2 + (5 + 2x)(5 + 2x)  =  25                                 Multiply out the parenthesees.

x2 + (5)(5) + (5)(2x) + (2x)(5) + (2x)(2x)  =  25

x2 + 25 + 10x + 10x + 4x2  =  25                        Combine like terms.

5x2 + 25 + 20x  =  25                                         Subtract  25  from both sides of the equation.

5x2 + 20x  =  0               Factor out an  x  from both terms.

x(5x + 20)  =  0              Set each factor equal to  0  and solve for  x  .

 

x  =  0     or     5x + 20  =  0     Subtract  20  from both sides.

                       5x  =  -20          Divide both sides by  5  .

                       x  =  -4

 

Now plug these values for  x  into the second equation given. (The first one will give you two answers for  y  , but only one answer for  y  works in the second equation.)

 

y - 2x  =  5       Plug in  0  for  x  .

y - 2(0)  =  5

y - 0  =  5

y  =  5

                        

y - 2x  =  5       Plug in  -4  for  x  .

y - 2(-4)  =  5

y - -8  =  5

y + 8  =  5        Subtract  8  from both sides.

    y  =  -3

So the two solutions are:

x = 0,  y = 5          and          x = -4, y = -3

hectictar  Aug 21, 2017

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