Question

HELP:
How do you find the solutions to a system of equations involving quadratic functions and other functions?

Answer 1

what grade is this

Answer 2

pemdas

Answer 3

12

Answer 4

I am in 7th

Answer 5

8th

Answer 6

Example: Solve these two equations:
•y = x2 - 5x + 7
•y = 2x + 1



Make both equations into "y=" format:

They are both in "y=" format, so go straight to next step



Set them equal to each other

x2 - 5x + 7 = 2x + 1



Simplify into "= 0" format (like a standard Quadratic Equation)

Subtract 2x from both sides: x2 - 7x + 7 = 1

Subtract 1 from both sides: x2 - 7x + 6 = 0

Answer 7

Solve the Quadratic Equation!

(The hardest part for me)

You can read how to solve Quadratic Equations, but here we will factor the Quadratic Equation:

Start with: x2 - 7x + 6 = 0

Rewrite -7x as -x-6x: x2 - x - 6x + 6 = 0

Then: x(x-1) - 6(x-1) = 0

Then: (x-1)(x-6) = 0



Which gives us the solutions x=1 and x=6

Answer 8

do yall need help

Answer 9

yeaa

Answer 10

ok it is long

Answer 11

it is ?

Answer 12

is this ur qusetion Algebra a system of equations
Make both equations into "y =" format
Set them equal to each other
Simplify into "= 0" format (like a standard Quadratic Equation)
Solve the Quadratic Equation!
Use the linear equation to calculate matching "y" values, so we get (x,y) points as answers
An example will help:

Example: Solve these two equations:
y = x2 - 5x + 7
y = 2x + 1

Make both equations into "y=" format:
They are both in "y=" format, so go straight to next step

Set them equal to each other
x2 - 5x + 7 = 2x + 1

Simplify into "= 0" format (like a standard Quadratic Equation)
Subtract 2x from both sides: x2 - 7x + 7 = 1
Subtract 1 from both sides: x2 - 7x + 6 = 0

Solve the Quadratic Equation!
(The hardest part for me)
You can read how to solve Quadratic Equations, but here we will factor the Quadratic Equation:
Start with: x2 - 7x + 6 = 0
Rewrite -7x as -x-6x: x2 - x - 6x + 6 = 0
Then: x(x-1) - 6(x-1) = 0
Then: (x-1)(x-6) = 0

Which gives us the solutions x=1 and x=6

Answer 13

was that it

Answer 14

yeah thank you so much guys ;)

Answer 15

no problem

Answer 16

ur welcome

Answer 17

i have a next question tho

Answer 18

tag mehhhh

Answer 19

ok whats up