Solving the following system of equations

[Spensy]

2x+3y = 7, -> equation1
2x^2-3y^2 = -25; -> equation2

first of all tell me the type of equations

first i solve for y equation 1

y = (-2x +7)/3

now what should i do ?

 

[phyzguy]

Now plug that expression for y into equation 2 and solve for x. It will be a quadratic equation for x, so you will need the quadratic formula.

 

[Spensy]

please i need more help can any 1 solve this equations my textbook has a solution set
{(-1,3),(-13,11)}

please help me I'd be thankful to you.

 

[sjb-2812]

2x+3y = 7, -> equation1
2x^2-3y^2 = -25; -> equation2

first of all tell me the type of equations

first i solve for y equation 1

y = (-2x +7)/3

now what should i do ?

So if y = (7-2x)/3, what is 3y²?

 

[CellCoree]

The equations provided are a system of two polynomial equations, one linear and one quadratic. To solve this system, we can use substitution or elimination methods.

Substitution method:

Substitute the expression for y from equation 1 into equation 2:

2x^2 - 3((-2x + 7)/3)^2 = -25

Simplify and solve for x:

2x^2 - (4x^2 - 28x + 49) = -25

2x^2 - 4x^2 + 28x - 49 = -25

-2x^2 + 28x - 24 = 0

x^2 - 14x + 12 = 0

(x - 2)(x - 12) = 0

x = 2 or x = 12

Substitute these values back into equation 1 to solve for y:

When x = 2, y = (-2(2) + 7)/3 = 1

When x = 12, y = (-2(12) + 7)/3 = -5

Therefore, the solutions to this system of equations are (2, 1) and (12, -5).

Elimination method:

Multiply equation 1 by 3 and equation 2 by 2 to eliminate the y terms:

6x + 9y = 21

4x^2 - 6y^2 = -50

Rearrange the terms:

9y = 21 - 6x

6y^2 = 4x^2 + 50

Substitute the expression for y from the first equation into the second equation:

6(21 - 6x)^2 = 4x^2 + 50

Simplify and solve for x:

252 - 252x + 36x^2 = 4x^2 + 50

32x^2 - 252x + 202 = 0

8x^2 - 63x + 51 = 0

(8x - 17)(x - 3) = 0

x = 17/8 or x = 3

Substitute these values back into equation 1 to solve for y:

When x = 17/8, y = (-2(17/8) + 7