how do I solve quadratic equations by elimination?
eg: 6x^2-3x=2y-5
2x^2+x=y-4
6x^2-3x-2y = -5
4x^2+2x-2y = -8
now subtract to get
2x^2-5x = 3
2x^2-5x-3 = 0
(2x+1)(x-3) = 0
...
To solve quadratic equations by elimination, you need to eliminate one variable by manipulating the equations and then solve for the other variable. Let's solve the given equations step-by-step:
1. Given equations:
6x^2 - 3x = 2y - 5 ........(Equation 1)
2x^2 + x = y - 4 ........(Equation 2)
2. Rearrange Equation 1 to isolate the term containing 'y':
2y = 6x^2 - 3x + 5
3. Rearrange Equation 2 to isolate the term containing 'y':
y = 2x^2 + x + 4
4. Now, equate the two expressions for 'y':
6x^2 - 3x + 5 = 2x^2 + x + 4
5. Simplify the equation by subtracting 'x' from both sides:
6x^2 - 2x^2 - 3x - x + 5 - 4 = 0
6. Combine like terms:
4x^2 - 4x + 1 = 0
7. Factor the equation, if possible. In this case, the equation is a perfect square trinomial and can be factored as:
(2x - 1)(2x - 1) = 0
8. Apply the zero product property:
2x - 1 = 0
9. Solve for 'x':
2x = 1
x = 1/2
10. Substitute the value of 'x' back into either Equation 1 or Equation 2 to solve for 'y'. Using Equation 2:
y = 2(1/2)^2 + (1/2) + 4
y = 1/2 + 1/2 + 4
y = 5
11. The solution to the system of equations is:
x = 1/2
y = 5
So, the solution to the quadratic equations by elimination is x = 1/2 and y = 5.
To solve a system of quadratic equations by elimination, you need to eliminate one of the variables by adding or subtracting the equations. Then, rearrange the resulting equation into a quadratic equation and solve for the remaining variable.
Let's solve the given system of equations:
1. 6x^2 - 3x = 2y - 5
2. 2x^2 + x = y - 4
Step 1: Eliminate the y variable.
Since both equations have y terms, you can eliminate y by subtracting equation 2 from equation 1.
(6x^2 - 3x) - (2x^2 + x) = (2y - 5) - (y - 4)
Simplifying the equation, we get:
4x^2 - 4x - 1 = 0
Step 2: Solve the resulting quadratic equation.
Now, we need to solve the quadratic equation 4x^2 - 4x - 1 = 0. To do this, we can use factoring, completing the square, or quadratic formula.
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 4, b = -4, and c = -1. Substituting these values into the quadratic formula:
x = (-(-4) ± √((-4)^2 - 4(4)(-1))) / (2(4))
Simplifying further:
x = (4 ± √(16 + 16)) / 8
x = (4 ± √32) / 8
x = (4 ± 4√2) / 8
x = (1 ± √2) / 2
So, the solutions for x are:
x₁ = (1 + √2) / 2
x₂ = (1 - √2) / 2
Step 3: Solve for y.
To find the corresponding values of y, substitute the values of x back into one of the initial equations. Let's use equation 1.
For x₁ = (1 + √2) / 2:
6((1 + √2) / 2)^2 - 3((1 + √2) / 2) = 2y - 5
Simplify the equation and solve for y.
For x₂ = (1 - √2) / 2:
6((1 - √2) / 2)^2 - 3((1 - √2) / 2) = 2y - 5
Simplify the equation and solve for y.
After substituting the values, you will find the corresponding values of y.
In summary, to solve quadratic equations by elimination, you eliminate one variable by adding or subtracting the equations, rearrange the resulting equation into a quadratic form, solve for the remaining variable using factoring, completing the square, or quadratic formula, and substitute the solutions into one of the original equations to find the values of the other variable.