What are the solutions to the following system? startlayout enlarged left-brace 1st row negative 2 x squared + y = negative 5 2nd row y = negative 3 x squared + 5 endlayout (0, 2) (1, –2) (startroot 2 endroot, negative 1) and (negative startroot 2 endroot, negative 1) (startroot 5 endroot, negative 10) and (negative startroot 5 endroot, negative 10)

(StartRoot 2 EndRoot, negative 1) and (negative StartRoot 2 EndRoot, negative 1)

Step-by-step explanation:

we have

----> equation A

-----> equation B

solve by substitution

substitute equation B in equation A

solve for x

Find the value of y

For ---->

For ---->

therefore

The solutions are

(StartRoot 2 EndRoot, negative 1) and (negative StartRoot 2 EndRoot, negative 1)

5x + 3y = - 4

step-by-step explanation:

the equation of a line in standard form is

ax + by = c ( a is a positive integer and b, c are integers )

express the line first of all in slope- intercept form

y = mx + c ( m is the slope and c the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (7, - 3) and (x₂, y₂ ) = (4, - 8)

m = = -

y = - x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (4, - 8 ), then

- 8 = - + c ⇒ c = -

y = - x - ← in slope-intercept form

multiply all terms by 3 to eliminate the fractions

3y = - 5x - 4 ( add 5x to both sides )

5x + 3y = - 4 ← in standard form