Solving Quadratic Systems
x+2y=4
3x-y=5
Please show work!!!
x+2y=4
3x-y=5
x = 4 - 2y
3 (4 - 2y) - y = 5
12 - 6y - y = 5
-7y = -7
y = 1
x + 1 = 4
x = 3
where did you get x+1=4 from
I goofed. :-( Sorry.
3x - y = 5
3x - 1 = 5
3x = 6
x = 2
I corrected your answer :D
Thank you
To solve the quadratic system of equations, we can use the elimination or substitution method. Let's use the elimination method in this case.
1. Start by multiplying the second equation by 2 to eliminate the y-term:
2 * (3x - y) = 2 * 5
6x - 2y = 10
Now we have:
x + 2y = 4 (Equation 1)
6x - 2y = 10 (Equation 2)
2. Add the two equations together to eliminate the y-term:
(x + 2y) + (6x - 2y) = 4 + 10
x + 2y + 6x - 2y = 14x
Combining like terms, we have:
7x = 14
3. Divide both sides of the equation by 7 to solve for x:
7x/7 = 14/7
x = 2
Now we have the value of x.
4. Substitute the value of x back into either of the original equations to find y. Let's substitute it into Equation 1:
x + 2y = 4
2 + 2y = 4
Subtracting 2 from both sides:
2y = 2
Dividing both sides by 2:
y = 1
So, the solution to the quadratic system of equations is x = 2 and y = 1.