Question

How to solve systems of linear and quadratic equations...

-y^2 - 3x - 10y + 6 = 0
x + 3y - 2 = 0

Answer 1

substitution is you way out

Answer 2

have you tried?

Answer 3

from second x=-3y+2
put in first
\[-y^2-10y-3(-3y+2)+6=0\]
\[-y^2-10y+9y-6+6=0\
find values of y and then corresponding x.

Answer 4

ok give me a moment please

Answer 5

so i got
\(-y^2 + 6 + 9y - 10y + 6\)
\(-y^2 - 1y + 12\)

Answer 6

then i would find the values of y by factoring it?

Answer 7

yeah quadratic equation you can use any method you are most familiar with

Answer 8

\[-y^2+y=0,y(-y+1)=0,y=?\]

Answer 9

y = 4
y = -3

Answer 10

sorry i meant y = 3

Answer 11

y=0,1

Answer 12

but how?

Answer 13

you distributed -3 * +2 and got +6

Answer 14

okay so would i get now
\(-y^2 - y\)

Answer 15

equal to 0

Answer 16

ok

Answer 17

now i use the quadratic formula right?

Answer 18

yes
iwrote wrong it should be -y^2-y=0

Answer 19

you could multiply both sides by -1 to get
\[ y^2+y=0\]
now factor out a y

Answer 20

\(y(y + 1)\)

Answer 21

=0

you know either y is 0 or the factor (y+1) is 0

Answer 22

take- y common ,then y=0,-1
find corresponding x from second eq.

Answer 23

okay so i would need to find what value for both of the y that will make it equal to 0

Answer 24

are you asking how to solve
y(y+1)=0
?

Answer 25

or, are you asking is there a different x value for each of the solutions for y ?
(answer is yes, each y gives a different x)