Can somebody show I how to solve simultaneouosly x^2 -3y=0 and -3x+3y^2=0 to get the point (0,0) and (1,1). I can't cancel anything out with those two equations so I'm stuck.
If your system of equations is:
3 x² - 3 y = 0 and - 3 x + 3 y² = 0
then
3 x² - 3 y = 0
Divide both sides by 3
x² - y = 0
Add y to both sides
x² = y
y = x²
- 3 x + 3 y² = 0
Divide both sides by 3
- x + y² = 0
Replace y with x²
- x + ( x²)² = 0
- x + x⁴ = 0
x ( - 1 + x³ ) = 0
x ( x³ - 1 ) = 0
x ( x - 1 ) ( x² + x + 1 ) = 0
This equation is equal zero when:
x = 0 , x - 1 = 0 and x² + x + 1 = 0
x - 1 = 0
Add 1 to both sides
x = 1
The solutions of x² + x + 1 = 0 are:
x1 = - 1 / 2 + i √3 / 2 and x 2 = - 1 / 2 - i √3 / 2
So the real solutions are:
x = 0 and x =1
Replace this values in equation 3 x² - 3 y = 0
x = 0
3 ∙ 0² - 3 y = 0
- 3 y = 0
Divide both sides by - 3
y = 0
x = 1
3 ∙ 1² - 3 y = 0
3 - 3 y = 0
Add 3 y to both sides
3 = 3 y
3 y = 3
Divide both sides by 3
y = 1
This is correct only if your system of equations is 3 x² - 3 y = 0 and - 3 x + 3 y² = 0
since y = x^2/3, use that in the 2nd equation to get
-3x + 3x^4/9 = 0
-3x + x^4/3 = 0
x^4 - 9x = 0
x = 0,∛9
so y = 0,∛3
That does not give the point (0,0) and (1,1). Something isn't right.
well, clearly (1,1) does not work!
x^3 - 3y = 0
??? 1^3 is not 3
To solve the system of equations x^2 - 3y = 0 and -3x + 3y^2 = 0, we can use the method of substitution. Here's how you can do it step by step:
Step 1: Solve one equation for one variable in terms of the other variable. Let's start with the first equation, x^2 - 3y = 0. Rearrange it to get x^2 = 3y.
Step 2: Substitute this expression for x in the second equation. Replace x^2 with 3y in the equation -3x + 3y^2 = 0. It becomes -3(√(3y)) + 3y^2 = 0.
Step 3: Simplify the equation and rearrange it to solve for y. Distribute -3 to get -3√(3y) + 3y^2 = 0. You can rewrite this equation as 3y^2 = 3√(3y) to eliminate the negatives.
Step 4: Square both sides of the equation to get rid of the square root. After squaring, you will have 9y^4 = (3√(3y))^2. Simplifying this equation, you get 9y^4 = 9y^3.
Step 5: Divide both sides by 9y^3 to isolate y. This leads to y = 1.
Step 6: Substitute the value of y into either equation to solve for x. Using the equation x^2 = 3y, plug in y = 1. You will find that x^2 = 3(1), which simplifies to x^2 = 3. Taking the square root of both sides, you get x = ± √3.
So the first solution pair is (x, y) = (√3, 1). The second solution pair can be obtained with the negative square root, (x, y) = (-√3, 1).
To verify these solutions, substitute the values of x and y back into the original equations and check if they satisfy both equations. For example, for the first solution pair, substitute x = √3 and y = 1 into both equations:
1st equation: (√3)^2 - 3(1) = 3 - 3 = 0 (satisfied)
2nd equation: -3(√3) + 3(1)^2 = -3√3 + 3 = 0 (satisfied)
Both equations are satisfied, confirming that the solution pair (x, y) = (√3, 1) is correct. You can do the same for the second solution pair.
Keep in mind that the method of substitution may not always yield both solutions, but in this case, it does.