How can x2 = 2x + 4 be set up as a system of equations? (1 point) a y = x2 − 2x + 4 y = x2 + 2x + 4 b y = x2 y = x2 + 2x + 4 c y = x2 + 2x y = x2 + 4 d y = x2 y = 2x + 4

x=3 and y=4

Step-by-step explanation:

x+y=7;x+2y=11

Step: Solvex+y=7for x:

x+y=7

x+y+−y=7+−y(Add -y to both sides)

x=−y+7

Step: Substitute−y+7forxinx+2y=11:

x+2y=11

−y+7+2y=11

y+7=11(Simplify both sides of the equation)

y+7+−7=11+−7(Add -7 to both sides)

y=4

Step: Substitute4foryinx=−y+7:

x=−y+7

x=−4+7

x=3(Simplify both sides of the equation)

x=3 and y=4

ddd

step-by-step explanation:

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