Graph the following equations using the slope-intercept method.
4x + 3y = -15
y = x + 2
4x + 3y = -15
get y alone on the left, in form y = m x + b
3 y = -4 x - 15
so
y = (-4/3) x - 5
slope m = -4/3
y axis intercept at y = -5, thus at (0 , -5)
mark (0 , -5)
then go right 3 and down 4 for the second point ( 3 , -9 )
=======================
y = x + 2
is
y = (1) x + 2
slope m is +1
y axis intercept at y = 2 when x = 0
To graph the equations using the slope-intercept method, we need to rewrite them in the form y = mx + b, where m is the slope and b is the y-intercept.
Let's start with the first equation: 4x + 3y = -15
Step 1: Rearrange the equation to isolate y.
3y = -4x - 15
Step 2: Divide both sides of the equation by 3 to solve for y.
y = (-4/3)x - 5
Now we have the equation in slope-intercept form, where the slope is -4/3 and the y-intercept is -5.
To plot the graph, we can start with the y-intercept at (0, -5) and use the slope to find additional points on the line.
For the second equation: y = x + 2
The equation is already in slope-intercept form, with a slope of 1 and a y-intercept of 2.
We can start with the y-intercept at (0, 2) and use the slope to find the next point on the line.
Now that we have the necessary information, we can plot the points and draw the lines on a coordinate plane to graph the equations.
I am deppresed anyone know what I can do?
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