Mathematics, 18.10.2020 01:01 kendramiller3965

Which order pairs are solutions to the equation 4x+2y=4

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Answer from: TMeansStupidity

Slope=2.000/4.000

−2.000x−intercept= 22

=1y−intercept= 12

=2.00000

Step-by-step explanation:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

4*x+2*y-(4)=0

Pull out like factors :

4x + 2y - 4 = 2 • (2x + y - 2)

Solve : 2 = 0

This equation has no solution.

A non-zero constant never equals zero.

Solve 2x+y-2 = 0

Tiger recognizes that we have here an equation of a straight line. Such an equation is usually written y=mx+b ("y=mx+c" in the UK).

"y=mx+b" is the formula of a straight line drawn on Cartesian coordinate system in which "y" is the vertical axis and "x" the horizontal axis.

In this formula :

y tells us how far up the line goes

x tells us how far along

m is the Slope or Gradient i.e. how steep the line is

b is the Y-intercept i.e. where the line crosses the Y axis

The X and Y intercepts and the Slope are called the line properties. We shall now graph the line 2x+y-2 = 0 and calculate its properties

Notice that when x = 0 the value of y is 2/1 so this line "cuts" the y axis at y= 2.00000

y-intercept = 2/1 = 2.00000

When y = 0 the value of x is 1/1 Our line therefore "cuts" the x axis at x= 1.00000

x-intercept = 2/2 = 1

Slope is defined as the change in y divided by the change in x. We note that for x=0, the value of y is 2.000 and for x=2.000, the value of y is -2.000. So, for a change of 2.000 in x (The change in x is sometimes referred to as "RUN") we get a change of -2.000 - 2.000 = -4.000 in y. (The change in y is sometimes referred to as "RISE" and the Slope is m = RISE / RUN)

Slope = -4.000/2.000 = -2.000

Which order pairs are solutions to the equation 4x+2y=4

Answer from: Quest

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like, however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

$\bf e(\stackrel{x_1}{-2}~,~\stackrel{y_1}{2})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-2}{)}\implies \implies \cfrac{-2}{0+2}\implies \cfrac{-2}{2}\implies -1 \\\\[-0.35em] ~\dotfill$

$\bf d(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{)}{)}\implies \cfrac{0+1}{0+2}\implies \cfrac{1}{2} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope~of~ef}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{1}\implies 1}}$