solve each system of equations
6x^2+3y^2-111x-y+128=0
3x-y=1
from the 2nd:
y = 3x - 1
then
6x^2+3y^2-111x-y+128=0
6x^2 + 3(3x - 1)^2 - 111x - (3x - 1) + 128 = 0
simplify, and solve the equation.
(comes apart nicely)
6x^2+3y^2-111x-y+128=0 , 3x-y=1
3 x - y = 1
Subtract 3x to both sides:
- y = 1 - 3 x
- y = - 3 x + 1
Multiply both sides by - 1
y = 3 x - 1
plug y = 3 x - 1 into 6 x² + 3 y² -111 x - y +128 = 0
6 x² + 3 ∙ ( 3 x - 1 )² -111 x - ( 3 x - 1 ) + 128 = 0
6 x² + 3 ∙ ( 9 x² - 6 x + 1 ) - 111 x - 3 x + 1 + 128 = 0
6 x² + 27 x² - 18 x + 3 - 111 x - 3 x + 1 + 128 = 0
33 x² -132 x + 132 = 0
Divide both sides by 33
x² - 4 x + 4 = 0
Solution of this equation:
x = 2
y = 3 x - 1 = 3 ∙ 2 - 1 = 6 - 1 = 5
Solution x = 2 , y = 5
( 2 , 5 )
To solve the system of equations, we can use the method of substitution.
1. Start by solving the second equation, "3x - y = 1", for one variable in terms of the other.
- Add "y" to both sides: "3x = 1 + y".
- Divide both sides by 3: "x = (1 + y)/3".
2. Substitute the expression for "x" into the first equation, "6x^2 + 3y^2 - 111x - y + 128 = 0".
- Replace "x" with "(1 + y)/3": "6( (1 + y)/3 )^2 + 3y^2 - 111( (1 + y)/3 ) - y + 128 = 0".
3. Simplify the equation.
- Start by expanding the square: "6(1 + y)/3 * 6(1 + y)/3 + 3y^2 - 111(1 + y)/3 - y + 128 = 0".
--> Note: The fraction "(1 + y)/3" is squared because it's being multiplied by itself.
- Simplify further: "(1 + y)^2/2 + 3y^2 - 111(1 + y)/3 - y + 128 = 0".
--> Note: The fraction "(1 + y)/3" is divided by 2 because it's being multiplied by 6 and 3.
- Multiply through by the common denominator, which is 6.
- Multiply each term by 6: "3(1 + y)^2 + 18y^2 - 222(1 + y) - 6y + 768 = 0".
4. Expand and simplify the equation further.
- Expand: "3(1 + 2y + y^2) + 18y^2 - 222 - 222y - 6y + 768 = 0".
- Simplify: "3 + 6y + 3y^2 + 18y^2 - 222 - 222y - 6y + 768 = 0".
- Combine like terms: "21y^2 - 228y + 549 = 0".
5. Solve the quadratic equation, "21y^2 - 228y + 549 = 0", for "y".
- This equation can be factored or solved using the quadratic formula.
6. After finding the values of "y", substitute each value back into the equation "3x - y = 1" to solve for "x".
- Plug the value of "y" into "3x - y = 1": "3x - (y value) = 1".
- Solve for "x".
Once we have the "x" and "y" values, we will have obtained the solutions to the system of equations.