Question

solve the following system of equations:
9x^2-2x-y=-3
6x^2+3x-y=-1

i dont understand this at all.

Answer 1

what method u wanna use?

Answer 2

?

Answer 3

anything that works.

Answer 4

and makes sense. i've seen that the elimination method is what i've been TRYING to learn.

Answer 5

ok lets try that

Answer 6

subtract second equation from first
\[3x^2-5x=-2 \]
find x and then y from either equation.

Answer 7

\[3x^2-5x+2=0\]
\[3x^2-3x-2x+2=0\]
\[3x \left( x-1 \right)-2\left( x-1 \right)=0\]
\[\left( x-1 \right)\left( 3x+2 \right)=0\]
x=1,
\[\frac{ -2 }{ 3 }\]
find y for each

Answer 8

yup ^^

Answer 9

From the Mathematica program:\[\text{Solve}\left[\left\{9 x^2-2 x-y=-3,6 x^2+3 x-y=-1\right\},\{x,y\}\right] \]\[\left\{x\to \frac{2}{3},y\to \frac{17}{3},x\to 1,y\to 10\right\} \]

Answer 10

For each equation, solve it for y first.
Then set them equal and solve the quadratic for x.

9x^2-2x-y=-3 ----> 9x^2 - 2x + 3 = y

6x^2+3x-y=-1 ----> 6x^2 + 3x + 1 = y

9x^2 - 2x + 3 = 6x^2 + 3x + 1

Now subtract everything to the left side, and solve the quadratic equation for x.
You can try factoring, or just use the quadratic formula.
Then for each solution you get for x, plug it into one of the original equations to find a corresponding y value.

Answer 11

correction
type error
factors are (x-1) (3x-2)=0
x=1,2/3