Question

What is the solution of the linear-quadratic system of equations?

Answer 1

use substitution

Answer 2

the first equation says
\[ \large y=x^2+5x-3 \]
and the second
\[ \large y-x=2 \]
so pluging the first into the second you get
\[ \large \color{red}{x^2+5x-3}-x=2 \]
and this is just a quadratic equation!!

Answer 3

So , now what after that ?

Answer 4

solve this last equation.

Answer 5

would it be x^ +4x -3 ?

Answer 6

no. u forgot the two that's on the RHS

Answer 7

What am i supposed to do with that ? would it be x^ +4x-3 =2 ?

Answer 8

add -2 to both sides

Answer 9

so would it be x^ +4x -5 = 0 ?

Answer 10

yes
\[ \large x^2+4x-5=0 \]

Answer 11

solve it

Answer 12

solve this last equation

Answer 13

I dont understand how to solve this last part ?

Answer 14

do u know the quadratic formula?
\[ \large x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \]

Answer 15

No , I dont know it . . .

Answer 16

really??

Answer 17

yeah , sadly :/

Answer 18

ok. in the equation
\[ \large x^2+4x-5=0 \]
when compared with the GENERAL equation
\[ \large \color{blue}{ax^2+bx+c=0} \]
we have than a=1, b=4 and c=-5. do u get this?

Answer 19

yes .

Answer 20

so pluging this into the formula we have
\[ \large x=\frac{-4\pm\sqrt{4^2-4(1)(-5)}}{2(1)} \]

Answer 21

do all the computations.

Answer 22

x= -4 ( plus or minus ) 4 -4 +20 / 2 ?

Answer 23

no

Answer 24

first powers, then products or division,, then addition or substraction.

Answer 25

-4 -4 + 20 /2 ? im confused . .

Answer 26

i'll do it (but u should be able to this kind of stuff)

Answer 27

\[ \large x=\frac{-4\pm\sqrt{4^2-4(1)(-5)}}{2(1)}=
\frac{-4\pm\sqrt{16+20}}{2}=
\frac{-4\pm\sqrt{36}}{2} \]
ok?

Answer 28

ohhh i see now !!! so -4 ( plus or minus) square root 36 / 2 is the answer ?

Answer 29

no wait

Answer 30

\[ \large x=\frac{-4\pm6}{2} \]

Answer 31

so we have to values for x
\[ \large x_1=\frac{-4+6}{2}=\frac{2}{2}=1 \]
and
\[ \large x_2=\frac{-4-6}{2}=\frac{-10}{2}=-5 \]

Answer 32

now use these to find the corresponding values of \(y\).