Question

(12) Solve.
2x² + 4y² = 4
2x²-4y² = 25x-10

Answer 1

Explanation:

We can solve this system of equations by using the method of substitution. First, we can isolate one of the variables in terms of the other from one of the equations. Let's isolate y from the second equation:

2x² - 4y² = 25x - 10

-4y² = -2x² + 25x - 10

y² = (1/2)x² - (25/4)x + 5/2

y = ±√[(1/2)x² - (25/4)x + 5/2]

Now we can substitute this expression for y into the first equation:

2x² + 4y² = 4

2x² + 4[(1/2)x² - (25/4)x + 5/2] = 4

3x² - 25x + 7 = 0

We can solve this quadratic equation by factoring or using the quadratic formula:

3x² - 25x + 7 = (3x - 1)(x - 7) = 0

Therefore, we have two solutions:

3x - 1 = 0 → x = 1/3

x - 7 = 0 → x = 7

We can plug each of these values of x back into either of the two equations to find the corresponding values of y:

For x = 1/3:

y = ±√[(1/2)(1/3)² - (25/4)(1/3) + 5/2] ≈ ±1.807

For x = 7:

y = ±√[(1/2)(7)² - (25/4)(7) + 5/2] ≈ ±2.426

Therefore, the solutions to the system of equations are:

(1/3, ±1.807)

(7, ±2.426)

Answer 2

• (1/4, -√14/4)
• (1/4, √14/4)

Explanation:

You want the solution to the system of quadratic equations ...

• 2x² +4y² = 4
• 2x² -4y² = 25x -10

Elimination

Adding the equations together gives ...

(2x² +4y²) +(2x² -4y²) = (4) +(25x -10)

4x² -25x +6 = 0

(4x -24)(4x -1)/4 = 0

(x -6)(4x -1) = 0

Values of x that satisfy this equation are ...

x = 6, x = 1/4 . . . . . . . x = 6 is an extraneous solution

Substitution

For x = 1/4, we have ...

y² = 1 -x²/2 . . . . . from the first equation

y = ±√(7/8) = ±(√14)/4

The solutions are (1/4, -√14/4) and (1/4, √14/4).

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Additional comment

Only one of the branches of the hyperbola of the second equation intersects the ellipse of the first equation. That ellipse in standard form is ...

x²/2 +y² = 1

Which has a domain of |x|≤√2 and a range of |y|≤1. This is why x=6 is extraneous (not in the domain).