Question

Solve the system:

7x=y+9
5x^2-y= 8 + 3x

Answer 1

You can use the substitution method.
Solve the first equation for y, and substitute what y equals into the second equation.

Answer 2

this is what i have so far:

7x = y+ 9 -> y = 7x- 9
5x^2 - y = 8 + 3x -> y = 5x^2 - 3x -8

7x- 9 = 5x^2 - 3x -8
=5x^2 - 3x - 8

Answer 3

form eq 1 find y as

y=2x-9

now put this value in 2nd eq. just plug in the value y

Answer 4

yes its correct step...now move ahead..

Answer 5

factorize it if possible otherwise use quadratic formula..

Answer 6

so will it be 5x^2 - 10x +1?

Answer 7

This is correct.
7x- 9 = 5x^2 - 3x -8
Now collect like terms and set it equal to zero.
Then solve the quadratic for x.

Answer 8

Yes, it's what you have just above, but set it equal to zero.
Solve that quadratic for x.
I suggest the quadratic formula.

Answer 9

i dont really know what to do at this point :$ lol

Answer 10

Have you heard of the quadratic formula?

Answer 11

yes. so i just plug that in and i have my final answer?

Answer 12

7x- 9 = 5x^2 - 3x -8
=5x^2 - 3x - 8
am i missing something right here? where the left side go...
subtract 7x get 5x^2-10x+1
now you use the quadratic formula for factor it
here a=5, b=-10 and c=1
so x=[-b+/-sqrt(b^2-4ac)]/2a
x=[10+/-sqrt((-10)^2-4(5)(1)]/2(5)
x=[10+/-sqrt(100-20)]/10
x=[10+/-sqrt(80)]/10
x=(10+/-8.944)/10
x=(10+8.944)/10=1.8944
x=(10-8.944)/10=0.1056
x=1.8944 or 0.1056

Answer 13

\(x=\dfrac{10 \pm \sqrt{80}}{10}\)

\(x=\dfrac{10 \pm 4\sqrt{5}}{10}\)

\(x=\dfrac{5 \pm 2\sqrt{5}}{5}\)

\(x=1 + \dfrac{2\sqrt{5}}{5}\) or \(x=1 - \dfrac{2\sqrt{5}}{5}\)

So far so good.
Now substitute each of those x values into one of the original equations to find the y-values.

Answer 14

ah thanks guys! i really appreciate all your help:)

Answer 15

You're welcome.