155k views
5 votes
Use the elimination method to find all solutions of the system of equations. {3 x²-y² =11 {x²+4 y² =8

User Dan Steele
by
7.3k points

1 Answer

3 votes

Final answer:

To use the elimination method to find the solutions of the system of equations, multiply the equations by appropriate factors to make the x^2 coefficients the same. Then subtract the equations to eliminate x. Solve for y and substitute the values of x and y into one of the original equations to check for a solution. However, in this case, the system of equations has no solutions.

Step-by-step explanation:

To solve the system of equations using the elimination method, start by multiplying the first equation by 4 and the second equation by 3 to make the coefficients of the x^2 terms the same. This will allow you to eliminate the variable x by subtracting the two equations. Then, solve for y by substituting the value of x into one of the original equations. Finally, substitute the values of x and y into one of the equations to check if they satisfy the system.

Let's solve it step by step:

Multiplying the first equation by 4: 12x^2 - 4y^2 = 44

Multiplying the second equation by 3: 3x^2 + 12y^2 = 24

Subtracting the two equations to eliminate x: (12x^2 - 4y^2) - (3x^2 + 12y^2) = 44 - 24

Simplifying the equation: 9x^2 - 16y^2 = 20

Now, substitute the value of x^2 from this equation into the other equation to solve for y: x^2 = (8 - 4y^2) / 3

Substituting: 9(8 - 4y^2) / 3 - 16y^2 = 20

Simplifying the equation: 24 - 12y^2 - 16y^2 = 60

Combining the like terms: -28y^2 = 36

Dividing both sides by -28: y^2 = -1.29

Taking the square root of both sides: y = ±√(-1.29)

Since the square root of a negative number is not a real number, there are no solutions to this system of equations.

User Jpshook
by
8.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.