x-3y=1
x2-2xy+9y2=17 solve simalteneously
You probably figured it out yourself by at the end it should have said ...
if y = -4/3, x = 3(-4/3) + 1 = -3
if y = 1 , x = 3(1) + 1 = 4
I'm confused
To solve the given system of equations simultaneously, we'll use the method of substitution. Here's how to proceed:
Step 1: Solve the first equation for x in terms of y.
x - 3y = 1
x = 1 + 3y
Step 2: Substitute the expression for x from the first equation into the second equation.
(1 + 3y)^2 - 2(1 + 3y)y + 9y^2 = 17
Step 3: Expand and simplify the equation.
1 + 6y + 9y^2 - 2y - 6y^2 + 9y^2 = 17
15y^2 + 4y - 16 = 0
Step 4: Solve the quadratic equation for y.
To solve the quadratic equation, we can either factor it, complete the square, or use the quadratic formula. In this case, the quadratic equation doesn't factor easily, so we'll use the quadratic formula.
The quadratic formula is:
y = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 15, b = 4, and c = -16. Let's substitute these values into the quadratic formula:
y = (-4 ± √(4^2 - 4 * 15 * -16)) / (2 * 15)
Simplifying further:
y = (-4 ± √(16 + 960)) / 30
y = (-4 ± √976) / 30
y = (-4 ± 2√61) / 30
Step 5: Simplify the solutions for y.
y = (-2 ± √61) / 15
Step 6: Substitute the values of y back into the first equation to find the corresponding values of x.
For y = (-2 + √61) / 15:
x = 1 + 3((-2 + √61) / 15)
For y = (-2 - √61) / 15:
x = 1 + 3((-2 - √61) / 15)
So, the solutions to the system of equations are:
x = 1 + 3((-2 + √61) / 15), y = (-2 + √61) / 15
x = 1 + 3((-2 - √61) / 15), y = (-2 - √61) / 15
from the 1st ----> x = 3y+1
(3y+1)^2 - 2y(3y+1) + 9y^2 = 17
9y^2 + 6y + 1 - 6y^2 - 2y + 9y^2 = 17
12y^2 + 4y -16 = 0
3y^2 + y - 4 = 0
(3y + 4)(y - 1) = 0
y = -4/3 or y = 1
if x = -4/3, y = 3(-4/3) + 1 = -3
if x = 1 , y = 3(1) + 1 = 4
btw, it is "simultaneously"