math problem. Don't know how to solve the equation? Thank You?

Question

y=x^2-x-6
3x-4y+12=0
I need to solve these equations algebraically. Thanks

Answer 1

Your first step would be to substitute the "Y" from the first equation and put it into the second one.
It would look like this 3x - 4(x²-x-6) +12 = 0
then when you work it out you get
-4x²+7x+36=0 but you should multiply the entire thing by -1 because it is easier to deal with. So that would leave you with 4x²-7x-36=0. Then you factor it out and get (4x+9)(x-4)=0
From that you know that x=4 and x=-9/4.
Now you have to plug those values back into one of the original equations and find the y values.
That would look like y=(4)²-(4)-6 and you get 6. (That was the first equation, and all you did was plug the X values that you already found everywhere there was an "X" in the equation.
You do that same thing with the (-9/4).
So your anwers are (4,6) and (-9/4,21/16)

Answer 2

2 general methods to solve simultaneous equations:
A) Substitution method
In substitution method, you manipulate one of the variables to be the subject of any one equation:
e.g x=...
or y=...

Then substitute it into the other equation.

B) Subtraction method
You subtract off one of the variables such that only 1 variable is left.
e.g y= 3x + 5
y = 2x - 3
Subtracting equation 1 and 2 will eliminate the y.
LHS: y-y
RHS: 3x + 5 - (2x - 3)
= 3x + 5 - 2x + 3
= x + 8
Thus 0 = x + 8
x = -8

For your question it's quite impossible to use (B), because there's a squared variable, a bit difficult.

So we can use method (A).

y = x^2 - x - 6 ----------------- (1)
3x - 4y +12 = 0 ----------- (2)

Substitute (1) into (2)
3x - 4(x^2 - x - 6) + 12 = 0
3x - 4x^2 + 4x + 24 + 12 = 0
- 4x^2 + 7x + 36 = 0
4x^2 - 7x - 36 = 0
(4x + 9)(x - 4) = 0
x = -9/4 , 4

Answer 3

Take the value of y from the first equation (x^2 - x - 6) and substitute that for y in the second:

3x - 4*(x^2 - x - 6) + 12 = 0

Multiply it out and solve for x, using quadratic formula. Then put that value back into equation 2 and solve for y.

There will be two solutions because of the x^2 term.

Answer 4

1st step: - Eliminate 'y' from secoND equation and put its value in first equation
y=(12 + 3x) / 4
(12 + 3x) / 4 = x^2 - x - 6
2nd step : - convert equation into a quadratic equation
4x^2 - 7x - 36 = 0
4x^2 -16x + 9x - 36= 0
4x (x - 4) + 9(x - 4) = 0
(4x + 9) (x - 4) = 0
x = 4, -9 / 4
3 rd step : - put x = 4 and -9 / 4 in first equation
we get,
y=6, 21 / 16
4 Th step: - Cross check your answer in second equation by putting x=4,y=6 and x= - 9/4, y = 21/16 one by one.
you will see x= - 9/4, y = 21 / 16 doest not satisfy 2 ND equation. therefore your answer is x = 4 and y = 6.
Good Luck

Answer 5

if you want x,y that can use two equation
you must let's y=y

y=x^2-x-6 ----#1
3x-4y+12=0
4y =3x+12
y =(1/4)(3x+12) ----#2

let's y=y ------#1= #2
x^2-x-6 =(1/4)(3x+12)
4x^2-4x-24 = 3x+12
4x^2-7x-36 =0
(4x+9)(x-4)=0
then x = -9/4 or x = 4 and find y value from #2
when x= -9/4
y=(1/4)(3(-9/4)+12) =21/16
when x= 4
y=(1/4)(3(4)+12) =6