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39 votes
39 votes
4x² +9y² = 72

x-y²=-1
Select all of the following that are solutions to the system shown.
(-3,-2)
(3.-2)
(-3,2)
(3,2)

User Ytg
by
3.2k points

2 Answers

17 votes
17 votes

Answer:Hope this helps ;)

Explanation:

To find the solutions to a system of equations, we need to find the values of the variables that make both equations in the system true at the same time. One way to do this is to solve each equation for one of the variables, and then substitute that expression into the other equation.

For the first equation, 4x² + 9y² = 72, we can rearrange the terms to get x² = (72 - 9y²)/4.

For the second equation, x - y² = -1, we can rearrange the terms to get y² = x + 1.

Substituting the expression for y² from the second equation into the first equation, we get:

x² = (72 - 9(x + 1))/4

This simplifies to:

x² = (72 - 9x - 9)/4

Which simplifies to:

x² - 9x - 27 = 0

We can use the quadratic formula to solve this equation for x:

x = (-(-9) ± √((-9)² - 4(1)(-27)))/(2(1))

This simplifies to:

x = (9 ± √(81 + 108))/2

Which simplifies to:

x = (9 ± 15)/2

This gives us the two solutions for x: x = 6 and x = -3.

We can substitute these values back into either of the original equations to find the corresponding values of y.

If x = 6, then substituting into the second equation gives us y² = 6 + 1, or y² = 7. Taking the square root of both sides, we get y = √7 or y = -√7.

If x = -3, then substituting into the second equation gives us y² = -3 + 1, or y² = -2. Taking the square root of both sides, we get y = √-2 or y = -√-2. Since the square root of a negative number is not a real number, this means there is no solution for y in this case.

Therefore, the solutions to the system of equations are (-3,-2) and (3,2).

User Duoc Tran
by
3.4k points
12 votes
12 votes

Answer:

b and d are the solutions

User Wynn
by
3.1k points
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