Solve by elimation y=-5x-1 the other equation 25x^2+y^2=100

Question

Need help solving this elimation problem. should have 2 x's 2y 's

Answer 1

Here is how we solve this by elimination.  We need both equations in standard form.  That means the x and y terms are on one side of the equation.  The constant terms on the other side.  By rearranging the equations, we have

 

5x + y = -1                  eq1      ---->  linear equation part

25x² + y² = 100            eq2     ---->  ellipse equation part

 

 

Next, we need both equations to have either x² and y² terms, or x and y terms.  It will be easier to have all the equations to have x² and y².  So we are going to square both sides of eq1 and keep the eq2 as it is.

 

(5x + y)(5x + y) = (-1)²               eq1

 

25x² + 10xy + y² = 1                  eq1

25x²             + y² = 100               eq2

 

 

Notice that the coefficient of x² and y² of both equation are the same.  Subtract eq2 from eq1 to eliminate the x² and y² terms.

 

10xy = -99

 

xy = -9.9

 

 

From this equation, solve for  either x or y.

 

x = -9.9 / y            

 

 

Substitute this value of x into eq2 (bolded) to solve for y.

 

25(-9.9 / y)² + y² = 100

 

(2450.25 / y²) + y² = 100

 

(2450.25 + y⁴) / y² = 100y² / y²

 

 

Equating numerators, we have

 

2450.25 + y⁴ = 100y²

 

 

Rearranging the equation, we get

 

y⁴ - 100y² + 2450.25 = 0

 

 

Letting  q=y² ,

 

q² - 100q + 2450.25 = 0      ---->  quadratic equation

 

 

Solve for q using the quadratic formula:

 

q = (-b ± √(b² - 4ac)) / 2a

 

where:

a = 1

b = -100

c = 2450.25

 

Plug in these values to solve for q.  Accept the positive value of q.  Once you have q, then substitute that value of q into

 

y = ±√q

 

 

Finally, use the equation      x = -9.9 / y           to solve for x.

 

You should have 2 x values because you will have 2 y values.

 

 

I know this process is a bit lengthy, but this is how you would solve by elimination.

Answer 2

Ok, I can see how you could use elimination, now.  Although I'm thinking that there might also be a typo because the solutions are not "pretty" numbers (have to use the quadratic formula).

 

Square both sides of the first equation:

 

y^{2 =} (-5x-1)^2

 

Subtract this from the first equation like so:

 

     25x²+y² = 100

 

-   (          y² = (-5x-1)²   )

 

===================

 

      25x² =   100  -   (-5x-1)²

 

(This is where the elimination came from, since the y is gone now, leaving only x's).

 

     25x²  = 100 - (25x²+10x+1)

 

     25x² = 100 - 25x² - 10x - 1

 

     50x² + 10x - 99 = 0.

 

And this brings you to the same place as the other solution on this page, which is done much faster with substitution, to be sure.

Answer 3

y = -5x-1

25x²+y²=100

 

Solving this by elimination leads to xy = -9.9

We need to solve by substitution given that we

have y in terms of x.  We can use xy=-9.9 to

verify the final answer.

 

Substitute -5x-1 in place of y in the 2nd equation and solve for x

 

25x²+(-5x-1)²=100

25x² + 25x² + 10x + 1 = 100

50x² + 10x - 99 = 0

Solve for x using the quadratic equation

x = [-b ±√(b²-4ac)]/2a,   a=50, b=10, c=-99

 

x = [-10 ±√(10²-4(50)(-99))]/2(50)

x = [-10 ±√19900]/100

x = [-10 ±√((100)(199))]/100

x = [-10±10√199]/100

x = -1 ±√(199)/10

x = (-1 ± 14.107)/10

 

x = 13.107/10 = 1.3107

OR

x = -15.107/10 = -1.5107

 

We have 2 values for x.  Use those to find the corresponding

values of y=-5x-1

 

For x = 1.3107

y = -5(1.3107)-1 = -7.5535

 

For x = -1.5107

y = -5(-1.5107)-1 = 6.5535

 

Our two solutions to the system are

1.3107, -7.5535   and   -1.5107, 6.5535

 

Check that our answers multiply out to xy = -9.9

1.3107(-7.5535) = -9.9

(-1.5107)(6.5535) = -9.9