Solve the system, 4x^2+y^2=20 and y=4x?

Question

how would you solve this

Answer 1

y = 4x

4x² + y² = 20

4x² + (4x²) = 20

4x² + 16x² = 20

20x² = 20

20x²/20 = 20/20

Dividing both sides by 20

x² = ± √1

x = ± 1

The answer is x = ± 1

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Solving for y

4x² + y² = 20

4(1)² + y² = 20

4 + y² = 20
-4 -4

Subtracting - 4 from both sides

y² = √16

y = ± 4

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x = ± 1

y = ± 4

Insert the x value and y values in to the formuls

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Solving for ( + )

4x² + y² = 20

4(1)² + 4² = 20

4 + 16 = 20

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Solving for ( - )

4x² + y² = 20

4(-1)² + (- 4)² = 20

4 + 10 = 20

Answer 2

If y = 4x then y^2=16x^2. Substitute y^2 in the first equation by 16x^2 and solve for x. 4x^2 + y^2 = 4x^2 +16x^2 = 20x^2 = 20. Therefore, x^2 = 1 or x = +/-1. From 2nd equation, y = 4x = 4 * +/-1 = +/-4

Answer 3

Use substitution. You know that y = 4x, so replace y in the other equation with 4x so it will look like:
4x^2 + (4x)^2 = 20 -----> 4x^2 + 16x^2 = 20
Then you can solve that equation with respect to x. Once you find the value of x, you know what y is (since y is 4 times x.)

Answer 4

Substituting, 4x^2 + 14x^2 = 20, implies 20 x^2 = 20, and so x = -a million, or a million Now, 455/ninety = 5. fifty 5, so, Q1 80 + ok(360) = 440 even as ok = a million, so, 440 is coterminal with 80 i'll't comprehend the subsequent question in case you rearrange the equation, you'll see that's the equation of an ellipse (sq. words with diverse sensible coefficients, and linear ones)

Answer 5

4x^2 + 16x^2 = 20 <=========sub in 4^x
x^2 = 1
x = +1, -1

sub in to y = 4x
get y = 4, y = -4

easy 2 points

Answer 6

I think it is like that substitute the 2nd eq to 1 i.e y=4x in to the first to give
4x^2+16x^2=20
or 20x^2=20
or x^2=1
or x=+-1
hence y=4 for x=1 and y=-4 for x=-1

Answer 7

4x^2 + y^2 = 20
y = 4x

4x^2 + (4x)^2 = 20
4x^2 + 16x^2 = 20
20x^2 = 20
x^2 = 1
x = -1 or 1

y = 4(-1) = -4
y = 4(1) = 4

ANS : (-1,-4) and (1,4)

Answer 8

Dibakar is right : the system is quadratic so it has 2 sets of solutions

Answer 9

4x^2+4x^2=20
4x^2=10
4x=10^.5
x=(10^.5)/4
y=10^.5